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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Find the roots of the following equations : (i) `x-1/x=3,x!=0` (ii) `1/(x+4)-1/(x-7)=(11)/(30),x!=-4,7` |
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Answer» (i) Given equation is `x-(1)/(x)=3` `implies(x^(2)-1)/(x)=3` `impliesx^(2)-1=3ximpliesx^(2)-3x-1=0` On comparing with `ax^(2)+bx+c=0` a=1,b=-3 and c=-1 `becausex=(-b+-sqrt(b^(2)-4ac))/(2a)` `impliesx=(-(3)+-sqrt((-3)^(2)-4xx1xx-1))/(2xx1)` `impliesx=(3+-sqrt13)/(2)` `becausex=(3+sqrt13)/(2) and (3-sqrt13)/(2)` Hence, roots of the equation are `(3+sqrt13)/(2) and (3-sqrt13)/(2)` (ii) Given equation is `(1)/(x+4)-(1)/(x-7)=(11)/(30)` `implies((x-7)-(x+4))/((x+4)(x-7))=(11)/(30)implies(-11)/(x^(2)-7x+4x-28)=(11)/(30)` `implies11(x^(2)-3x-28)=30xx(-11)` `impliesx^(2)-3x-28=-30` `impliesx^(2)-3x+2=0` `impliesx^(2)-(2+1)x+2=0` `impliesx^(2)-2x-x+2=0` `impliesx(x-2)-1(x-2)=0` `implies(x-2)(x-1)=0` `impliesx=2 and x=1` Hence, roots of the equation are 2 and 1. |
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| 152. |
The sum of the squares of three consecutive natural numbers is 149.Find the numbers. |
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Answer» Let three consecutive natural numbers be `x,(x+1)and(x+2)` respectively. According to given condition `x^(2)+(x+1)^(2)+(x+2)^(2)=149` `impliesx^(2)+(x^(2)+1+2x)+(x^(2)+4+4x)=149` `implies3x^(2)+6x+5=149` `implies3x^(2)+6x-144=0` `impliesx^(2)+2x-48=0` `impliesx^(2)+8x-6x-48=0` `impliesx(x+8)-6(x+8)=0` `implies(x+8)(x-6)=0` `impliesx=-8andx=6` `impliesx=6" "(x=-8"is not a natural number")` Hence, the required natural number are `6,(6+1),(6+2)=6,7,8` respectively. |
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| 153. |
A sailor can row a boat 8 km downstream and return back to the starting point in 1 hour 40 minutes. If the speed of the stream is 2 km/hr, find the speed of the boat in still water. |
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Answer» Let the speed of the boat in still water be x km/hr. Speed of the stream = 2 km/hr. `:." ""speed downstream"=(x+2)km/hr,` speed upstream `=(x-2)km/hr.` Time taken to cover 8 km downstream and return back to the starting point `=(8)/((x+2))+(8)/((x-2))`. But, this time is given as `1(40)/(60)` hours `=1(2)/(3)" hours "=(5)/(3)` hours. `:." "(8)/(x+2)+(8)/(x-2)=(5)/(3)` `implies" "(1)/(x+2)+(1)/(x-2)=(5)/(24)implies((x-2)+(x+2))/((x+2)(x-2))=(5)/(24)` `implies" "(2x)/((x^(2)-4))=(5)/(24)implies5x^(2)-20=48x` `implies" "5x^(2)-48x-20=0implies5x^(2)-50x+2x-20=0` `implies" "5x(x-10)+2(x-10)=0implies(x-10)(5x+2)=0` `implies" "x-10=0" or "5x+2=0` `implies" "x=10" or "x=(-2)/(5)` `implies" "x=10" "[because" speed of the boat cannot be negative"]` Hence, the speed of the boat in still water is 10 km/hr. |
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| 154. |
If `-5`is a root of the quadratic equation `2x^2+p x-15=0`and the quadratic equation `p(x^2+x)+k=0`has equal roots, find the value of `kdot` |
| Answer» Correct Answer - `k=(49)/(28)` | |
| 155. |
In a class test, the sum of Shefalis marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects. |
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Answer» Let maths marks=x Then english marks=30-x If she had got maths marks=x+2 Then she had got english marks=27-x According To Question (x+2)*(27-x)=210 27x-`x^2`+54-2x=210 `x^2`-25x+156=0 x(x-12)-13(x-12)=0 (x-12)(x-13)=0 x=12,13 If x=12 then english marks will be 18 if x=13 then english marks will be 17. |
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| 156. |
Some students planned a picnic.The budget for food was Rs.480. But , 8 of these failed to go and thus the cost of food for each member increased by Rs 10. How many students attended the picnic? |
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Answer» Let, no. of students who planned the picnic = x Given, the budget for food RS.480 `because` Shaer of each student `=RS.(480)/(x)` According to given condition `(480)/(x-8)-(480)/(x)=10` `implies(480x-480(x-8))/((x-8)x)=10implies(480x-480x+3840)/(x^(2)-8x)=10` `implies10x^(2)-80x=3840impliesx^(2)-8x-384=0` `impliesx^(2)-24x+16x-384=0impliesx(x-24)+16(x-24)=0` `implies(x-24)(x+16)=0impliesx=24orx=-16` Since, no. of students cannot be negative Hence, x=24 No. of students who went for picni =x-8=24-8=16 students. |
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| 157. |
The total cost of a certain length of cloth is `Rs 200`. If the piece was 5 m longer and each metre of cloth costs `Rs2` less, the cost of the piece would have remained unchanged. How longer is the piece and what is its original rate per metre? |
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Answer» Let, the length of the piece be x meters Since the cost of x meters of cloth =RS.200 `implies` Cost of each meter of cloth `=RS.(200)/(x)` New length of cloth =(x+5)m New cost of each more of cloth `=RS.(200)/(x+5)` Now, given `(200)/(x)-(200)/(x+5)=2` `implies(200)/(x)-(200)/(x+5)=2implies(200x+1000-200x)/(x^(2)+5x)=2` `1000=2x^(2)+10x` `implies2x^(2)+10x-1000=0` `x^(2)+5x+500=0` `x^(2)+25x-20x-500=0` `impliesx(x+25)-20(x+25)=0` `implies(x-20)(x+25)=0` `impliesx=20orx=-25` Since, length cannot be negative Hence, x=20m and the original rate per metre `=RS.(200)/(20)=RS.10` |
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| 158. |
Solve each of the following quadratic equations: `3x^(2)-243=0` |
| Answer» Correct Answer - `x=9" or "x=-9` | |
| 159. |
Solve: `6x^(2)+40=31x.` |
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Answer» The given equation may be written as `6x^(2)-31x+40=0.` We write, `-31x=-16x-15x" as "6x^(2)xx40=240x^(2)=(-16x)xx(-15x).` `:." "6x^(2)-31x+40=0` `implies" "6x^(2)-16x-15x+40=0implies2x(3x-8)-5(3x-8)=0` `implies" "(3x-8)(2x-5)=0implies3x-8=0" or "2x-5=0` `implies" "x=(8)/(3)" or "x=(5)/(2).` Hence, `(8)/(3)" and "(5)/(2)` are the roots of the given equation. |
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| 160. |
What are the roots of the quadratic equation `sqrt( 2x^(2)+9) = 9` ?A. `6,-6`B. 3,-3C. 2,-2D. `6,0` |
| Answer» Correct Answer - A | |
| 161. |
If the quadratic equation ` p x^2 - 2 sqrt(5) + 15 = 0` has two equal roots, then find value of p. |
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Answer» Correct Answer - p=3 The given equation is of the form `ax^(2)+bx+c=0,` where `a=pmb=-2sqrt(5)p" and "c=15.` `:." "D=(b^(2)-4ac)=(20p^(2)-60p).` So, `D=0implies20p^(2)-60p=0implies20p(p-3)=0impliesp=3." "[becausepne0]` |
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| 162. |
If the roots of the quadratic equation `2x^(2)+8x+k=0` are equal, find the value of k. |
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Answer» Correct Answer - k=8 `D=(8)^(2)-4xx2xxk=(64-8k).` `:." "D=0implies64-8k=0implies8k=64impliesk=8.` |
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| 163. |
Which of the following is not a quadratic equation ?A. `( y^(2))/(2) = 2y + 7`B. `(6)/( y) -5-y=0`C. `y-18=7y`D. `( y - 3) ( y+3) = 0` |
| Answer» Correct Answer - C | |
| 164. |
Solve the equation `x^(2)-(sqrt(3)+1)x+sqrt(3)=0` by the method of completing the square. |
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Answer» We have `x^(2)-(sqrt(3)+1)x+sqrt(3)=0` `implies" "x^(2)-(sqrt(3)+1)x=-sqrt(3)` `implies" "x^(2)-2xx x xx((sqrt(3)+1)/(2))+((sqrt(3)+1)/(2))^(2)=-sqrt(3)+((sqrt(3)+1)/(2))^(2)` `" ""[""adding "((sqrt(3)+1)/(2))^(2)" on both sides""]"` `implies" "{x-((sqrt(3)+1))/(2)}^(2)={((sqrt(3)+1)^(2))/(4)-sqrt(3)}` `implies" "("("sqrt(3)+1")"^(2)-4sqrt(3))/(4)=((sqrt(3)-1)/(2))^(2)` `implies" "{x-("("sqrt(3)+1")")/(2)}=+-("("sqrt(3)-1")")/(2)" "["taking square root on both sides"]` `implies" "x-("("sqrt(3)+1")")/(2)=("("sqrt(3)-1")")/(2)" or "x-("("sqrt(3)+1")")/(2)=(-"("sqrt(3)-1")")/(2)` `implies" "x=("("sqrt(3)-1")")/(2)+("("sqrt(3)+1")")/(2)=(2sqrt(3))/(2)=sqrt(3)` or `x=("("-sqrt(3)+1)/(2)+("("sqrt(3)+1")")/(2)=(2)/(2)=1` `implies" "x=sqrt(3)" or "x=1.` Hence, `sqrt(3)` and 1 are the roots of the given equation. |
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| 165. |
What is the nature of the rootsof the quadratic equation `4x^(2) - 8x + 9 =0` ?A. RealB. Not realC. Real and equalD. Real and unequal |
| Answer» Correct Answer - B | |
| 166. |
Show that the equation `9x^(2)+7x-2=0` has roots and solve it. |
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Answer» The given equation is `9x^(2)+7x-2=0.` Comparing it with `ax^(2)+bx+c=0`, we get `a=9,b=7" and "c=-2` `:." "D=(b^(2)-4ac)=(7^(2)-4xx9xx(-2)]=121gt0.` So, the given equation has real roots. Now, `sqrt(D)=sqrt(121)=11.` `:." "alpha=(-b+sqrt(D))/(2a)=((-7+11))/(2xx9)=(4)/(18)=(2)/(9),` `beta=(-b-sqrt(D))/(2a)=((-7-11))/(2xx9)=(-18)/(18)=-1.` Hence, the required roots are `(2)/(9)` and -1. |
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| 167. |
The positive value of `k`for which the equation `x^2+k x+64=0`and `x^2-8x+k=0`will both have real roots, is4 (b) 8(c) 12 (d) 16 |
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Answer» Let `D_(1)" and "D_(2)` be the discriminants of the first and second given equations respectively. For real roots, we must have `D_(1)ge0" and "D_(2)ge0.` Now, `D_(1)ge0" and "D_(2)ge0` `implies" "(4p^(2)-4xx64)ge0" and "(64-8p)ge0` `implies" "p^(2)-64ge0" and "64-8pge0` `implies" "p^(2)ge64" and "8pge64` `implies" "pge8" and "ple8" "[because" p is positive"]` `implies" "p=8.` Hence, p=8. |
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| 168. |
The sum of a number and its reciprocal is `2(1)/(20).` The number isA. `(5)/(4)" or "(4)/(5)`B. `(4)/(3)" or "(3)/(4)`C. `(5)/(6)" or (6)/(5)`D. `(1)/(6)" or "6` |
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Answer» Correct Answer - A Let the required number be x. Then, `x+(1)/(x)=(41)/(20)implies20x^(2)-41x+20=0` `implies20x^(2)-25x-16x+20=0` `implies5x(4x-5)-4(4x-5)=0` `implies(4x-5)(5x-4)=0` `impliesx=(5)/(4)" or "x=(4)/(5).` |
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| 169. |
Find the values of k for which the given equation has real roots: (i)`kx^(2)-6x-2=0" "(ii)" "3x^(2)+2x+k=0" "(iii)" "2x^(2)+kx+2=0` |
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Answer» (i) The given equation is `kx^(2)-6x-2=0.` `:." "D={(-6)^(2)-4xxkxx(-2)}=(36+8k).` The given equation will have real roots if `Dge0.` Now, `Dge0implies36+8kge0impliesimplieskge(-36)/(8)implieskge(-9)/(2).` (ii) The given equation is `3x^(2)+2x+k=0.` `:." "D=(2^(2)-4xx3xxk)=(4-12k).` The given equation will have real roots if `Dge0.` Now, `Dge0implies4-12kge0implies12kle4implieskle(1)/(3).` (iii) The given equation is `2x^(2)+kx+2=0.` `:." "D=(k^(2)-4xx2xx2)=(k^(2)-16).` The given equation will have real roots if `Dge0.` Now, `Dge0implies(k^(2)-16)ge0impliesk^(2)ge16implieskge4" or "kle-4.` |
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| 170. |
Show that the equation `x^(2)+6x+6=0` has real roots ad solve it. |
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Answer» The given equation is `x^(2)+6x+6=0.` Comparing it with `ax^(2)+bx+c=0,` we get `a-1,b=6" and "c=6.` `:." "D=(b^(2)-4ac)=(36-4xx1xx6)=12gt0.` So, the given equation has real roots. Now, `sqrt(D)=sqrt(12)=2sqrt(3).` `:." "alpha=(-b+sqrt(D))/(2a)=((-6+2sqrt(3)))/(2xx1)=((-6+2sqrt(3)))/(2)=(-3+sqrt(3)),` `beta=(-b-sqrt(D))/(2a)=((-6-2sqrt(3)))/(2xx1)=((-6-2sqrt(3)))/(2)=(-3-sqrt(3)).` Hence, `(-3+sqrt(3))` and `(-3-sqrt(3))` are the roots of the given equation. |
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| 171. |
For what values of k, the equation `kx^(2)-6x-2=0` has real roots?A. `kle(-9)/(2)`B. kge(-9)/(2)`C. `kle-2`D. None of these |
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Answer» Correct Answer - B For real roots, we must have `b^(2)-4acge0.` `:." "(-6)^(2)-4xxkxx(-2)ge0implies36+8kge0` `implies8kge-36implieskge(-9)/(2).` |
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| 172. |
both roots of the equation `(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0` are |
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Answer» The given equation may be written as `3x^(2)-2x(a+b+c)+(ab+bc+ca)=0.` `:." "D=4(a+b+c)^(2)-12(ab+bc+ca)` `=4[(a+b+c)^(2)-3(ab+bc+ca)]` `=4(a^(2)+b^(2)+c^(2)-ab-bc-ca)` `=2(2a^(2)+2b^(2)+2c^(2)-2ab-2bc-2ca)` `=2[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)]ge0` `[because(a-b)^(2)ge0,(b-c)^(2)ge0" and "(c-a)^(2)ge0].` This shows that both the roots of the given equation are real. For equal roots, we must have D=0. Now, `D=0implies(a-b)^(2)+(b-c)^(2)+(c-a)^(2)=0` `implies" "(a-b)=0,(b-c)=0" and "(c-a)=0` `implies" "a=b=c.` Hence, the roots are equal only when a=b=c. |
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| 173. |
If the roots of the equation `(b-c)x^2+(c-a)x+(a-b)=0`are equal, then prove that `2b=a+cdot` |
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Answer» Clearly, x=1 satisfies the given equation. Since its roots are equal, so 1 and 1 are its roots. `:." ""product of roots of the given equation "=(1xx1)=1.` But, product of roots `=(a-b)/(b-c)." "[because" product of roots "=(C)/(A)]` `:." "(a-b)/(b-c)=1impliesa-b=b-cimplies2b=a+c.` Hence, `2b=a+c.` |
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| 174. |
Show that the equation `3x^(2)+7x+8=0` is not true for any real value of x. |
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Answer» The given equation is `3x^(2)+7x+8=0.` `:." "D=(7^(2)-4xx3xx8)=(49-96)=-47lt0.` So, the given equation has no real roots. Hence, the given equation is not true for any real value of x. |
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| 175. |
Show that the equation `2(a^2+b^2)x^2+2(a+b)x+1=0` has no real roots when `a!=b` |
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Answer» The given equation is `2(a^(2)+b^(2))x^(2)+2(a+b)x+1=0.` `:." "D=4(a+b)^(2)-8(a^(2)+b^(2))` `=-4(a^(2)+b^(2)-2ab)=-4(a-b)^(2)lt0," when "a-bne0.` So, the given equation has no real roots, when `aneb.` |
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| 176. |
Find the value of `k`for which the quadratic equation`(k+4)x^2+(k+1)x+1=0` |
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Answer» The given equation is `(k+4)x^(2)+(k+1)x+1=0.` This is of the form `ax^(2)+bx+c=0,` where `a=(k+4), b=(k+1)" and "c=1.` `:." "D=(b^(2)-4ac)=(k+1)^(2)-4xx(k+4)xx1=(k+1)^(2)-4(k+4)` `=(k^(2)+1+2k-4k-16)=(k^(2)-2k-15).` For equal roots, we must have `D=0impliesk^(2)-2k-15=0` `implies" "k^(2)-5k+3k-15=0impliesk(k-5)+3(k-5)=0` `implies" "(k-5)(k+3)=0impliesk-5=0" or "k+3=0` `implies" "k=5" or "k=-3.` Hence, the required value of k is or -3. |
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| 177. |
If the equation `x^(2)+5kx+16=0` has no real roots then |
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Answer» The given equation is `x^(2)+5kx+16=0.` This is of the form `ax^(2)+bx+c=0,` where a=1, b=5k and c=16. `:." "D=(b^(2)-4ac)=(25k^(2)-4xx1xx16)=(25k^(2)-64).` Since, the given equation has no real root, we have `:." "D=(b^(2)-4ac)=(25k^(2)-4xx1xx16)=(25k^(2)-64).` `implies" "k^(2)lt(64)/(25)impliesk^(2)lt((8)/(5))^(2)` `implies" "(-8)/(5)ltklt(8)/(5).` Hence, the required real values of k are such that `(-8)/(5)ltklt(8)/(5).` |
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| 178. |
If the equation `x^(2)-kx+1=0` has nor real roots thenA. `klt-2`B. `kgt2`C. `-2ltklt2`D. None of these |
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Answer» Correct Answer - C For no real roots, we must have:` b^(2)-4aclt0.` `k^(2)-4lt0impliesk^(2)lt4implies-2ltklt2.` |
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| 179. |
For what value of k, are the roots of the quadratic equation kx (x-2) +6 = 0equal ? |
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Answer» The given equation is `kx^(2)-2kx+6=0.` This is of the form `ax^(2)+bx+c=0,` where a=k, b=-2k and c=6. `:." "D=(b^(2)-4ac)=(4k^(2)-4xxkxx6)=(4k^(2)-24k).` For equal roots, we must have `D=0implies4k^(2)-24k=0implies4k(k-6)=0impliesk=0" or "k=6.` Now, k=0, we get 6=0, which is absurd. `:." "kne0` andm hance k=6. |
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| 180. |
Solve: `(x+2)(3x-5)=0.` |
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Answer» We have `(x+2)(3x-5)=0impliesx+2=0" or "3x-5=0` `implies" "x=-2" or "x=(5)/(3).` Hence, the roots of the given equation are -2 and `(5)/(3).` |
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| 181. |
If the equation `x^(2)+5kx+16=0` has no real roots thenA. `kgt(8)/(5)`B. `klt(-8)/(5)`C. `(-8)/(5)ltklt2`D. None of these |
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Answer» Correct Answer - C For no real roots, we must have `b^(2)-4aclt0.` `:." "(25k^(2)-4xx16)lt0implies25k^(2)lt64impliesk^(2)lt(64)/(25)implies(-8)/(5)ltklt(8)/(5).` |
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| 182. |
Find the values `k` for which the quadratic equation `2x^2 + kx +3=0` has two real equal roots |
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Answer» The given equation is `2x^(2)+kx+3=0.` This is of the form `ax^(2)+bx+c=0,` where a=2, b=k and c=3. `:." "D=(b^(2)-4ac)=(k^(2)-4xx2xx3)=(k^(2)-24).` For real and equal roots, we much have `D=0impliesk^(2)-24=0impliesk=+-sqrt(24)impliesk=+-sqrt(24)=+-2sqrt(6).` Hence, `2sqrt(6)` and `-2sqrt(6)` are the required values of k. |
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| 183. |
Find the roots of the quadratic equation `6x^2-x-2=0`. |
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Answer» equation is `6x^2 - x-2=0 ` let `alpha & beta` be two terms such that `alpha+ beta= -x` `alpha* beta= (6x^2)(-2) = -12x^2` `alpha= -4x & beta= 3x` `6x^2 - 4x+3x-2=0` `2x[3x-2]+1[3x-2]=0` `(2x+1)(3x-2)=0` `2x+1=0` `x=-1/2` answer and `3x-2=0` `x=2/3` answer |
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| 184. |
Show that `sqrt(2)` and `-2sqrt(2)` are the roots of the equation `x^(2)+sqrt(2)x-4=0.` |
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Answer» The given equation is `x^(2)+sqrt(2)x-4=0.` Putting `x=sqrt(2)` in the given equation, we get `LHS=(sqrt(2))^(2)+(sqrt(2)xxsqrt(2))-4=(2+2-4)=0=RHS.` `:." "sqrt(2)` is a root of the given equation. Putting `x=-2sqrt(2)` in the given equation, we get `LHS=(-2sqrt(2))^(2)+sqrt(2)xx(-2sqrt(2))-4=(8-4-4)=0=RHS.` `:." "-2sqrt(2)` is a root of the given equation. Hence, `sqrt(2)` and `-2sqrt(2)` are the roots of the given equation. |
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| 185. |
If the roots of `5x^(2)-kx+1=0` are real and distinct thenA. `-2sqrt(5)ltklt2sqrt(5)`B. `kgt2sqrt(5)` onlyC. `klt-2sqrt(5)` onlyD. either `kgt2sqrt(5)" or "klt-2sqrt(5)` |
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Answer» Correct Answer - D The roots of `5x^(2)-kx+1=0` are real and distinct. `:." "(k^(2)-4xx5xx1)ge0impliesk^(2)ge20implieskgesqrt(20)" or "klt-sqrt(20)` `implieskgt2sqrt(5)" or "klt-2sqrt(5).` |
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| 186. |
Check whether the following are quadratic equations : (i) `(x-2)^2+1=2x-3` (ii) `x(x+1)+8=(x+2)(x 2)`(iii) `x(2x+3)=x^2+1` (iv) `(x+2)^3=x^3-4` |
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Answer» (i) `(x-2)^2+1 = 2x-3` `=>x^2+4-4x+1 = 2x-3` `=>x^2-6x+8 = 0` As coefficient of `x^2` is not `0`, so it is a quadratic equation. (ii)`x(x+1)+8 = (x+2)(x-2)` `=>x^2+x+8 = x^2-4` `=>x+12 = 0` As coefficient of `x^2` is `0`, so it is not a quadratic equation. (iii) `x(2x+3) = x^2+1` `=>2x^2+3x = x^2+1` `=>x^2+3x-1 = 0` As coefficient of `x^2` is not `0`, so it is a quadratic equation. (iv) `(x+2)^3 = x^3-4` `=>x^3+2^3+3(2)(x)(x+2) = x^3-4` `=>x^3+8+6x^2+12x = x^3-4` `=>6x^2+12x+12 = 0` `=>x^2+2x+2 = 0` As coefficient of `x^2` is not `0`, so it is a quadratic equation. |
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| 187. |
For the quadratic equation `2x^(2)-5x-3=0.` show that (i)` x=3" is its solution."`(ii) `x=(-1)/(2)" is its solution."` (iii) ` x=4" is not its solution."` |
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Answer» The given equation is `2x^(2)-5x-3=0.` (i) On substituting x = 3 in the given equation, we get `LHS" "=2xx3^(2)-5xx3-3=(18-15-3)=0=RHS.` `:." "x=3" is a solution of "2x^(2)-5x-3=0.` (ii) On substituting `x=(-1)/(2)` in the given equation, we get `LHS" "=2xx((-1)/(2))^(2)-5xx((-1)/(2))-3` `={2xx(1)/(4)+5xx(1)/(2)-3}` `={(1)/(2)+(5)/(2)-3}=0=RHS.` `:." "x=(-1)/(2)` is a solution of `2x^(2)-5x-3=0.` (iii) On substituting x = 4 in the given equation, we get `LHS=2xx4^(2)-5xx4-3=(32-20-3)=9ne0.` Thus, `LHSneRHS`. `:." "x=4` is not a solution of `2x^(2)-5x-3=0.` |
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| 188. |
Find the nature of the roots of the quadratic equation `3x^(2)-4sqrt(3)x+4=0` and hence solve it. |
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Answer» This is of the form `ax^(2)+bx+c=0,` where a=3, `b=-4sqrt(3)` and c=4. `:." "D=(b^(2)-4ac)={(-4sqrt(3))^(2)-4xx3xx4}=(48-48)=0.` This shows that the given quadratic equation has real and equal roots. Each root `=(-b)/(2a)=(4sqrt(3))/(6)=(2sqrt(3))/(3).` Hence, `(2sqrt(3))/(3)" and "(2sqrt(3))/(2)` are the roots of the given equation. |
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| 189. |
Find the roots of the following quadratic equations, if they exist, using the quadratic formula:(i) `3x^2-5x+2=0` (ii) `x^2+4x+5=0`(iii) `2x^2-2sqrt(2)x+1=0` |
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Answer» (i) `3x^2-5x+2=0` `= 3x^2 -3x - 2x +2=0` `= 3x(x-1)-2(x-1)=0` `=(3x-2)(x-1)=0` `x = 2/3 , 1 ` (ii) `x^2+4x+5=0` `D= b^2 - 4ac = (4)^2 -(4)(1)(5)` ` = 16- 20 =-4` `x= (-b+- sqrt D)/(2a)` `= (-4 +- 2i)/2 = -2+-i` (iii) `2x^2 - 2 sqrt2 x +1 = 0` `D= b^2 - 4ac` `= (-2 sqrt2)^2 - 4(2)(1)` `= 8-8 = 0` `x = -b/(2a) = -(-2 sqrt2)/(2(2)) ` `=1/sqrt2` answer |
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| 190. |
Show thatthe equation `x^2+a x-4=0`has realand distinct roots for all real values of `a`. |
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Answer» The given equation is `x^(2)-ax-4=0.` This is of the form `Ax^(2)+Bx+C=0,` where A=1, B=a and C=-4. `:." "D=(B^(2)-4AC)={a^(2)-4xx1xx(-4)}=(a^(2)+16)bt0` fro all real values of a. Thus, `Dgt0` for all real values of a. Hence, the given equation has real and distinct roots for all real values of a. |
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| 191. |
Check whether the following are quadratic equations: (i) `(2x-1)(x-3)=(x+4)(x-2)" "(ii)" "(x+2)^(3)=2x(x^(2)-1)` (iii) `(x+1)^(3)=x^(3)+x+6" "(iv)" "x(x+3)+6=(x+2)(x-2)` |
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Answer» We have `(i)" "(2x-1)(x-3)=(x+4)(x-2)` `implies" "2x^(2)-7x+3=x^(2)+2x-8impliesx^(2)-9x+11=0.` This is of the form `ax^(2)+bx+c=0`, where `a=1,b=-9` and `c=11`. Hence, the given equation is a quadratic equation. (ii) `(x+2)^(3)=2x(x^(2)-1)` `implies" "x^(3)+8+6x(x+2)=2x^(3)-2x` `implies" "x^(3)+6x^(2)+12x+8=2x^(3)-2x` `implies" "x^(3)-6x^(2)-14x-8=0.` This is not of the form `ax^(2)+bx+c=0.` Hence, the given equation is not a quadratic equation. (iii) `(x+1)^(3)=x^(3)+x+6` `implies" "x^(3)+1+3x(x+1)=x^(3)+x+6` `implies" "3x^(2)+2x-5=0.` This is of the form `ax^(2)+bx+c=0`, where `a=3,b=2` and `c=-5.` Hence, the given equation is a quadratic equation. (iv)` x(x+3)+6=(x+2)(x-2)` `implies" "x^(2)+3x+6=x^(2)-4` `implies" "3x+10=0.` This is not of the form `ax^(2)+bx+c=0.` Hence, the given equation is not a quadratic equation. |
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| 192. |
Solve for :`x :(x-1)/(x-2)+(x-3)/(x-4)=3 1/3,x!=2,4` |
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Answer» The given equation is `(x-1)/(x-2)+(x-3)/(x-4)=(10)/(3)` `implies" "((x-1)(x-4)+(x-3)(x-2))/((x-2)(x-4))=(10)/(3)` `implies" "((x^(2)-5x+4)(x^(2)-5x+6))/((x^(2)-6x+8))=(10)/(3)implies(2x^(2)-10x+10)/(x^(2)-6x+8)=(10)/(3)` `implies" "3(2x^(2)-10x+10)=10(x^(2)-6x+8)` `implies" "6x^(2)-30x+30=10x^(2)-60x+80` `implies" "4x^(2)-30x+50=0implies2x^(2)-15x+25=0." "...(i)` This equation is of the form `ax^(2)+bx+c=0,` where a=2, b=-15 and c=25. `:." "D=(b^(2)-4ac)={(-15)^(2)-4xx2xx25}=(225-200)=25gt0.` So, the given equation has real roots. Now, `sqrt(D)=sqrt(25)=5.` `alpha=(-b+sqrt(D))/(2a)=((15+5))/(2xx2)=(20)/(4)=5,` `beta=(-b-sqrt(D))/(2a)=((15-5))/(2xx2)=(10)/(4)=(5)/(2).` Hence, the required values of x are 5 and `(5)/(2).` |
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| 193. |
Show that the equation `2x^(2)-6x+3=0` has real roots and find these roots. |
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Answer» The given equation is `2x^(2)-6x+3=0.` This is of the form `ax^(2)+bx+c=0,` where a=2, b=-6 and c=3. `:." "D=(b^(2)-4ac)={(-6)^(2)-4xx2xx3}=(36-24)=12gt0.` So, the given equation has real unequal roots. Solving `2x^(2)-6x+3=0` by quadratic formula, we have `x=(6+-sqrt(36-4xx2xx3))/((2xx2))=(6+-sqrt(36-24))/(4)=(6+-sqrt(12))/(4)` `implies" "x=(6+-2sqrt(3))/(4)impliesx=(3+-sqrt(3))/(2).` So, `((3+sqrt(3)))/(2)" and "((3-sqrt(3)))/(2)` are the roots of the given equation. |
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| 194. |
Solve each of the following quadratic equations: `x^(2)+3sqrt(3)x-30=0` |
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Answer» Correct Answer - `x=2sqrt(3)" or "x=-5sqer(3)` `x^(2)+3sqrt(3)x-30=0impliesx^(2)+5sqrt(3)x-2sqrt(3)x-30=0impliesx(x+5sqrt(3))-2sqrt(3)(x+5sqrt(3))=0.` |
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| 195. |
Which of the following are quadratic equations? `(i) x^(2)-5x+3=0`(ii) `2x^(2)-3sqrt(2)x+6=0` (iii) `3x^(2)-2sqrtx+8=0`(iv) `2x^(2)-3=0` (v) `x+(1)/(x)=x^(2)`(vi) `x^(2)+(1)/(x^(2))=4(1)/(4)` |
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Answer» (i) Clearly, `(x^(2)-5x+3)` is a quadratic polynomial. `:." "x^(2)-5x+3=0` isa quadratic equation. (ii) Clearly, `(2x^(2)-3sqrt(2)x+6)` is a quadratic polynomial. `:." "2x^(2)-3sqrt(2)x+6=0` is a quadratic equation. (iii) `3x^(2)-2sqrt(x)+8` is not of the form `ax^(2)+bx+c=0.` `:." "3x^(2)-2sqrt(x)+8=0` is not a quadratic equation. (iv) `2x^(2)-3=0` is of the form `ax^(2)+bx+c=0,` where `a=2,b=0` and `C=-3`. `:." "2x^(2)-3=0` is a quadratic equation. (v) `x+(1)/(x)=x^(2)impliesx^(2)+1=x^(3)impliesx^(3)-x^(2)-1=0.` And, `(x^(3)-x^(2)-1)` being a polynomial of degree 3, it is not quadratic. Hence, `x+(1)/(x)=x^(2)` is not a quadratic equation. (vi) `x^(2)+(1)/(x^(2))=(17)/(4)implies4x^(2)+4=17x^(2)implies4x^(4)-17x^(2)+4=0.` C,ear,y.`4x^(4)-17x^(2)+4` is a polynomial of degree 4. `:." "x^(2)+(1)/(x^(2))=(17)/(4)` is not a quadratic equation. |
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| 196. |
Solve each of the following quadratic equations: `x^(2)-(sqrt(3)+1)x+sqrt(3)=0` |
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Answer» Correct Answer - `x=sqrt(3)" or "x=1` `x^(2)-(sqrt(3)+1)x+sqrt(3)=0impliesx^(2)-sqrt(3)x-x+sqrt(3)=0impliesx(x-sqrt(3))-(x-sqrt(3))=0.` |
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| 197. |
Find the nature of roots of the quadratic equation `4x^(2)-5x+3=0`. |
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Answer» The given equation is `4x^(2)-5x+3=0.` This is of the form `ax^(2)+bx+c=0,` where a=4, b=-5 and c=3. `:." "D=(b^(2)-4ac)={(-5)^(2)-4xx4xx3}=(25-48)=23lt0.` Hence, the given equation has no real roots. |
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| 198. |
Solve each of the following quadratic equations: `x^(2)-3sqrt(5)x+10=0` |
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Answer» Correct Answer - `x=sqrt(5)" or "x=2sqrt(5)` `x^(2)-3sqrt(5)x+10=0impliesx^(2)-2sqrt(5)x-sqrt(5)x+10=0impliesx(x-2sqrt(5))-sqrt(5)(x-2sqrt(5))=0.` |
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| 199. |
One of the roots of the quadratic equation `kx^(2) - 14x - 5 = 0 ` is 5. Complete the following activity to find the value of k . |
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Answer» 5 is the root of the quadratic equation `kx^(2) - 14x - 5 = 0 ` `:.` substitute x=5 in the equation `k xx square^(2)-square xx5-5=0` `:. 25k - 70 -5 = 0 ` `:. 25k = square ` `:. k = "-" " ":. k = 3 ` Activity 5 is the root of the quadratic equation `kx^(2) - 14x - 5 = 0 ` `:.` substitute `x=5` in the equation. `k xx 5^(2) - 14 xx5 -5=0` `:. 25k -70-5=0` `:. 25k = 75` `:. k = ( 75)/(25)` `:. k = 3 ` |
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| 200. |
Complete the following activity to determine the nature of the roots of the quadratic equation `2x^(2) - 5x + 3 = 0 ` |
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Answer» `2x^(2) - 5 x + 3 = 0` Here, `a=2, b = - 5 , c = 3 ` `b^(2) =4 ac = square ^(2)- 4(2)(3)` `= square -24` `= square ` `:. b^(2) - 4ac gt 0` The roots are `square` Activity `:` `a=2,b= -5,c = 3 ` `b^(2) - 4ac = (-5)^(2)- 4(2) (3)` `= 25 -24` `= 1 ` ` :. b^(2) - 4ac gt 0` The roots are real and unequal. |
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