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The correct option is: (C) x² - 14x + 48 = 0

Explanation:

For 1^st one,

Let the equation be x² + ax + b = 0

Since roots are 5 and 9

.'. a = -14 and b = 45

For 2^nd one,

Let the equation be x² + px + q - 0

Since roots are 12 and 4. 

. ^.. p = -16 and q = 48

Now, according to the question, b and p both are wrong.

Therefore, the correct equation would be

x² -14x + 48 = 0

The correct option is: (B) 999 

Explanation:

Let the number of bangles in a side of square = x

According to the question,

x² + 38 = Total no. of bangles                   ....(1)

Also, (x + 1)² - 25 = Total no. of bangles       ...(2)

From (1) and (2), we have

x²+38 = (x + 1)² - 25

=> 38 + 24 = 2x => x = 31

.'. Total no. of bangles = (31)² + 38 = 999

(a) 5 km/hr

Let the speed of the stream be x km/hr. 

Speed downstream = (15 + x) km/hr 

Speed upstream = (15 – x) km/hr

∴ \(\frac{30}{(15+x)}\) + \(\frac{30}{(15-x)}\) = \(4\frac{30}{60}\)

⇒ \(\frac{30(15-x)+30(15+x)}{(15+x)(15-x)}\) = \(\frac92\) 

⇒ 900 x 2 = 9 (225 - x²) 

⇒ x² - 225 = 200 ⇒ x² = 25 ⇒ x = 5.

Time = distance/speed 

Let the speed of the train be ‘a’ km/hr. 

Given, passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from the usual speed.

\(\Rightarrow \frac{150}{a}-\frac{150}{a+5}=1\)

⇒ 150(a + 5 – a) = a² + 5a 

⇒ a² + 5a – 750 = 0 

⇒ a² + 30a – 25a – 750 = 0 

⇒ a(a + 30) – 25(a + 30) = 0 

⇒ (a – 25)(a + 30) = 0 

⇒ a = 25 km/hr

Let the marks scored in maths be ‘X’.

Marks in English is ‘(40–X)’.

As, per the question,

If he got 3 marks in maths & 4 marks less in English,

Marks in Maths =X+3

Marks in English = 40–X–4 = 36–X

Product = 360

(36 – X)(X + 3) = 360

(36X + 108 – X² – 3X) = 360

(33X + 108 – X²) = 360

X² – 33X + 360 – 108 = 0

X² – 33X + 252 = 0

X² – 21X – 12X + 252 = 0

X(X – 21) – 12(X – 21) = 0

(X– 12)(X – 21) = 0

X = 12 or 21

∴ If marks in Maths = 12 then marks in English = 40 – 12 = 28

If marks in Maths = 21 then marks in English = 40 – 21 = 19

Sum of the marks in Mathematics and English = 30 

Let Moulika’s marks in Mathematics be x Then her marks in English = 30 – x 

If she got 2 more marks in Mathematics, then her marks would be x + 2. 

If she got 3 marks less in English then her marks would be 30 – x – 3 = 27 – x 

By problem (x + 2) (27 – x) = 210 

⇒ x(27 – x) + 2(27 – x) = 210

⇒ 27x – x² + 54 – 2x = 210 

⇒ -x² + 25x + 54 = 210 

⇒ x² – 25x – 54 + 210 = 0 

⇒ x² – 25x + 156 = 0 

⇒ x² – 12x – 13x + 156 = 0 

⇒ x(x – 12) – 13(x 12) = 0 

⇒ (x – 12) (x – 13) = 0 

⇒ x – 12 = 0 or x – 13 = 0 

⇒ x = 12 or x = 13 

If x = 12, then marks in Mathematics = 12

English = 30 – 12 = 18 

If x = 13, then marks in Mathematics = 13 

English = 30 – 13 = 17

Let the number of students who planned the picnic be ‘a’

Budget for the food was Rs. 480

Cost of food for each member = 480/a

Given, eight of these failed to go and thus the cost of food for each member increased by Rs. 10

\(\Rightarrow (a-8)\times(\frac{480}{a}+10)=480\)

⇒ (a – 8)(480 + 10a) = 480a 

⇒ 480a + 10a² – 3840 – 80a = 480a 

⇒ a² – 8a – 384 = 0 

⇒ a² – 24a + 16a – 384 = 0 

⇒ a(a – 24) + 16(a – 24) = 0 ⇒

 (a + 16)(a – 24) = 0 

⇒ a = 24

Number of students who attended the picnic = 24 – 8 = 16

Let ‘x’ be the length of cloth.

∵ Cost of x metre is Rs. 200.

∴ Cost per metre = Rs. 200 ÷ x

∴ Cost per metre when total size is x + 5 = Rs. 200 ÷ (x + 5)

∵ Cost of 5 metre longer cloth is Rs. 2 less for each metre.

200/x - 200/(x + 5) = 2

Let the original number of people be ‘a’.

Given, Rs. 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less.

Amount which each receives= 9000/a

\(\Rightarrow \frac{9000}{a\,+\,20}=\frac{9000}{a}-160\)

⇒ 9000a = (9000 – 160a)(a + 20) 

⇒ 9000a = 9000a + 180000 – 160a² – 3200a 

⇒ a² + 20a – 1125 = 0 

⇒ a² + 45a – 25a – 1125 = 0 

⇒ a(a + 45) – 25(a + 45) = 0 

⇒ (a – 25)(a + 45) = 0 

⇒ a = 25

x² − 3x + 2 = 0

Put x = − 1 in L.H.S.

L.H.S. = (− 1)² − 3(− 1) + 2

= 1 + 3 + 2 = 6 ≠ R.H.S.

Then x = − 1 is not the solution of the given equation.

We have to make the quadratic equation a perfect square if possible or sum of perfect square with a constant.

\((a+b)^2=a^2+2ab+b^2\)

\(2x^2+x-4=0\)

⇒ x² + x/2 – 2 = 0 

⇒ x² + 2 × 1/4 × x + (1/4)² - (1/4)² – 2 = 0

⇒ (x + 1/4)² = 33/16

⇒ x + 1/4 =  √33/4

\(\Rightarrow x=\frac{\sqrt{33}-1}{4},\frac{-\sqrt{33}-1{}}{4}\)

The correct option is: (B) > 0

Explanation:

Given equation is ax² + bx + c = 0

Roots are real and unequal, if b² - 4ac > 0

Given; x (x + 1) + 8 = (x + 2) (x − 2)

⇒ x² + x + 8 = x² − 2²

⇒ x² + x + 8 − x² + 4 = 0

⇒ x + 12 = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given; (x − 2)² + 1 = 2x − 3

⇒ x² − 2x + 4 + 1 = 2x − 3

⇒ x² − 2x + 5 − 2x + 3 = 0

⇒ x² − 4x + 8 = 0

∵ The highest power of x in the equation is 2;

∴ It is a quadratic equation.

The correct option is: (C) Statement - I is false but Statement - II is true.

Explanation:

Statement - I is false, since the quadratic equation ax² + bx + c = 0 has two distinct real roots, if b² - 4ac > 0.

Also, given equation is

2(a² + b²)x² + 2(a + b)x + 1 = 0

D = b² - 4ac = (2(a + b))² - 4(2a² + 2b²)(1)

= 4a² + 4b² + 8ab - 8a² - 8b²

= 4a² - 4b² + 8ab= 4(a - b)² < 0

.'. Given equation has no real roots. Hence, statement - II is true.

x² – (m + 1)x + 6 = 0

Put x = 3 in the given equation

(3)² – (m + 1) (3) + 6 = 0

⟹ 9 – 3m – 3 + 6 = 0

⟹ – 3m = – 12

⟹ m = 4

Put this value of m in the given equation, we get

x² – 5x + 6 = 0

⟹ x² – 3x – 2x + 6 = 0

⟹ x(x – 3) – 2(x – 3) = 0

⟹ (x – 3) (x – 2) = 0

If x – 3 = 0 Or x – 2 = 0

Then x = 3 Or x = 2

∴ 2 is the other root of the given equation

Given,

x + \(\frac{1}{x}\) = 1

On multiplying by x on both sides we have,

x² + 1 = x

⇒ x² – x + 1 = 0

It’s clearly seen that x² – x + 1 is a quadratic polynomial. Thus, the given equation is a quadratic equation.

Given; (x − 1)² = (x + 1)²

⇒ x² − x + 1² = x² + x + 1²

⇒ x² − x + 1 − x² − x − 1 = 0

⇒ −2x = 0

∵ The highest power of x in the equation is 1;

∴ It is not a quadratic equation.

Given,

3x² – 5x + 9 = x² – 7x + 3

On simplifying the equation, we have

2x² + 2x + 6 = 0

⇒ x² + x + 3 = 0 (dividing by 2 on both sides)

Now, it’s clearly seen that x² + x + 3 is a quadratic polynomial. Thus, the given equation is a quadratic equation.

2x– 2x – 3x + 3 = 0

2x (x– 1) – 3(x – 1) = 0

(2x – 3) (x – 1) = 0

2x – 3 = 0

x = 3/2

x – 1 = 0

x =1

Therefore, the roots of the equation are 3/2, 1.

 \(\cfrac{4}{x}-3\) = \(\cfrac{5}{2x+3}\),x ≠ 0, - \(\cfrac{3}{2}\)

⇒ \(\cfrac{4}{x}-\)\(\cfrac{5}{2x+3}\) = 3

⇒ \(\cfrac{8x + 12-5x}{x(2x+3)}\) = 3

⇒ \(\cfrac{3x+12}{2x^2+3x}\)= 3

⇒  \(\cfrac{x+4}{2x^2+3x}\) = 1

⇒  2x² + 3x = x + 4   (cross multiplication)

⇒ 2x² + 2x - 4 = 0

⇒ x² + x - 2 = 0

⇒ x² + 2x - x - 2 = 0

⇒ x(x + 2) - 1(x + 2) = 0

⇒ (x + 2) (x - 1) = 0

⇒ x + 2 = 0 or x - 1 = 0

⇒ x = - 2 or x = 1

Hence, - 2 and 1 are the roots of the given equation.

d = b² – 4ac

d = (7)² – 4 (4) (2)

d = 49 – 32

d = 17

Since, d > 0, roots are unique and real.

d = b² – 4ac

d = (–6)² – 4 (2) (3)

d = 36 – 24

d = 12

∴ Roots are real and unique.

Let the numbers be ‘a’ and ‘b’ 

Given, sum of two numbers is 15. The sum of their reciprocals is 1/4 

⇒ a + b = 15 

⇒ b = 15 – a 

Also, 1/a + 1/b = 3/10 

⇒ 1/a + 1/(15 – a) = 3/10 

⇒ 15 × 10 = 45a – 3a² 

⇒ a² – 15a + 50 = 0 

⇒ a² – 15a – 5a + 50 = 0 

⇒ a(a – 10) – 5(a – 10) = 0 

⇒ (a – 5)(a – 10) = 0 

⇒ a = 5, 10 

Numbers are are 5,10 or 10, 5

Let the numbers be ‘a’ and ‘b’ 

Given, sum of two numbers is 18. The sum of their reciprocals is 1/4 

⇒ a + b = 18 

⇒ b = 18 – a 

Also, 1/a + 1/b = 1/4 

⇒ 1/a + 1/(18 – a) = 1/4 

⇒ 18 × 4 = 18a – a² 

⇒ a² – 18a + 72 = 0 

⇒ a² – 12a – 6a + 72 = 0 

⇒ a(a – 12) – 6(a – 12) = 0 

⇒ (a – 6)(a – 12) = 0 

⇒ a = 6, 12 

Numbers are are 6,12 or 12, 6

Let the numbers be ‘a’ and ‘b’ 

Given, sum of two numbers is 18. The sum of their reciprocals is 1/4 

⇒ a + b = 18 

⇒ b = 18 – a 

Also, 1/a + 1/b = 1/4 

⇒ 1/a + 1/(18 – a) = 1/4 

⇒ 18 × 4 = 18a – a² 

⇒ a² – 18a + 72 = 0 

⇒ a² – 12a – 6a + 72 = 0 

⇒ a(a – 12) – 6(a – 12) = 0 

⇒ (a – 6)(a – 12) = 0 

⇒ a = 6, 12 

Numbers are are 6,12 or 12, 6

Let a positive integer be x. 

Then the second integer = x + 1 

Sum of the squares of the above integers = x² + (x + 1)² 

= x² + x² + 2x + 1 

= 2x² + 2x + 1 

By problem 2x² + 2x + 1 = 613 

⇒ 2x² + 2x – 612 = 0

⇒ x² + x – 306 = 0 

⇒ x² + 18x – 17x – 306 = 0 

⇒ x(x + 18) – 17(x + 18) = 0 

⇒ (x – 17) (x + 18) = 0 

⇒ x – 17 = 0 (or) x + 18 = 0 

⇒ x = 17 (or) -18,

we do not consider -18 

Then the numbers are (17, 17 + 1) 

i.e., 17, 18 are the required two consecutive positive integers.

Let the three consecutive numbers be a, a + 1, a + 2 

Given, there are three consecutive integers such that the sum of square of the first and the product of the other two is 46. 

⇒ a² + (a + 1)(a + 2) = 46 

⇒ 2a² + 3a + 2 = 46 

⇒ 2a² + 3a – 44 = 0 

⇒ 2a² + 11a – 8a – 44 = 0 

⇒ a(2a + 11) – 4(2a + 11) = 0 

⇒ (a – 4)(2a + 11) = 0 

Thus, a = 4 

Numbers are 4, 5, 6

Let the positive integers be ‘a’ and ‘b’. 

Given, difference of the squares of two positive integers is 180. 

⇒ a² – b² = 180 

Also, square of the smaller number is 8 times the larger. 

⇒ b² = 8a 

Thus, a² – 8a – 180 = 0 

⇒ a² – 18a + 10a – 180 = 0 

⇒ a(a – 18) + 10(a – 18) = 0 

⇒ (a + 10)(a – 18) = 0 

⇒ a = -10, 18 

Thus, the other number is 324 – 180 = b² 

⇒ b = 12 

Numbers are 12, 18

Let ‘x’ and ‘y’ be the smaller and larger integer respectively.

∴ x² + y² = 117.

[∵ The sum of the squares of two positive integers is 117]

∴ x² = 4y

[∵ the square of the smaller number equals four times the larger number.]

∴ y² + 4y – 117 = 0

Let the ones digit be ‘a’ and tens digit be ‘b’. 

Given, two-digit number is such that the product of its digits is 16. 

⇒ ab = 16 --- (1) 

Also, when 54 is subtracted from the number, the digits interchange their places 

⇒ 10b + a – 54 = 10a + b 

⇒ 9b – 9a = 54 

⇒ b – a = 6 

⇒ b = 6 + a 

Substituting in 1 

⇒ a × (6 + a) = 16 

⇒ a² + 6a – 16 = 0 

⇒ a² + 8a – 2a – 16 = 0 

⇒ a(a + 8) – 2(a + 8) = 0 

⇒ (a – 2)(a + 8) = 0 

⇒ a = 2 

Thus, b = 8 

Number is 82

Let the ones digit be ‘a’ and tens digit be ‘b’. 

Given, two-digit number is such that the product of its digits is 12. 

⇒ ab = 12 --- (1) 

Also, when 36 is added to the number, the digits interchange their places 

⇒ 10b + a + 36 = 10a + b 

⇒ 9a – 9b = 36 

⇒ a – b = 4 

⇒ a = 4 + b 

Substituting in 1 

⇒ b × (4 + b) = 12 

⇒ b² + 4b – 12 = 0 

⇒ b² + 6b – 2b – 12 = 0 

⇒ b(b + 6) – 2(b + 6) = 0 

⇒ (b – 2)(b + 4) = 0 

⇒ b = 2 

Thus, a = 6 

Number is 26

In given equation let x = \(\sqrt{6+\sqrt+{6+\sqrt{6+...}}}\)

So, 

x = √(6 + x)

Now squaring both side

x² = 6 + x

X² – x – 6 = 0 

X² – 3x + 2x – 6 = 0 

x(x – 3) + 2(x – 3) = 0 

(x – 3) (x + 2) = 0 

x = 3 or – 2

x cannot be equal to – 2 as root can never be negative.

x = 3

multiplying first equation and subtracting both the equations we get,

2x² + 4x +6λ  -2x² - 3x -5λ  = 0 

x + λ = 0 

x = -λ

Substituting  it in first equation we get,

(-λ)² +2(-λ) + 3λ = 0

λ² + λ = 0

λ = -1

Correct option is (A) 3/16

Given equation is \(4x^2-\sqrt{3}x-5=0\)

\(\Rightarrow(2x)^2-2\times2x\times\frac{\sqrt{3}}4-5=0\)     \((\because2\times2x\times\frac{\sqrt{3}}4=\sqrt3x)\)

\(\Rightarrow(2x)^2-2\times2x\times\frac{\sqrt3}4\) \(+(\frac{\sqrt3}4)^2-5-(\frac{\sqrt3}4)^2=0\)

\(\Rightarrow(2x-\frac{\sqrt3}4)^2-(5+\frac3{16})=0\)

Thus, we have to add or subtract \(\frac{3}{16}\) to solve the given quadratic equation by the method of completing the square.

Correct option is A) 3/16

(2x – 3)(3x + 1) = 0

Either 2x – 3 = 0 or 3x + 1 = 0

x = 3/2 or x = -1/3

We write, - 3x = 3x - 6x as 9x² \(\times\)(-2) = -18x² = 3x \(\times\)(-6x)

∴ 9x² - 3x - 2 = 0

⇒ 9x² + 3x - 6x - 2 = 0

⇒ 3x(3x + 1) - 2(3x + 1) = 0

⇒ (3x + 1) (3x - 2) = 0

⇒ 3x + 1 = 0 or 3x  - 2 = 0

⇒ x = - 1/3 or x = 2/3

Hence, the roots of the given equation are - 1/3 and 2/3.

Correct option is (A) 2/3, -1/2

Given quadratic equation is

 \((3x-2)\) \((2x+1)\) = 0

\(\Rightarrow\) 3x - 2 = 0 or 2x+1 = 0

\(\Rightarrow x=\frac23\;or\;x=\frac{-1}2\)

Hence, roots of given quadratic equation are \(\frac23,\frac{-1}2.\)

Correct option is A) 2/3, -1/2

Given

\((\frac{4x-3}{2x+1})\) - \(10(\frac{2x+1}{4x-3})\) = 3

putting \(\frac{4x-3}{2x+1}\) = y, we get

y - \(\frac{10}y\) = 3

⇒ \(\frac{y^2-10}y\) = 3

⇒ y² - 10 = 3y  [On cross multiplying]

⇒ y² - 3y - 10 = 0

⇒ y² - (5 - 2)y - 10 = 0

⇒  y² - 5y + 2y - 10 = 0

 ⇒ y(y - 5) + 2(y - 5) = 0

⇒ (y - 5) (y + 2) = 0

⇒ y - 5 = 0 or y + 2 = 0

⇒ y = 5 or y = -2

Case I: 

If y = 5, we get:

\(\frac{4x-3}{2x+1}\) = 5

⇒ 4x - 3 = 5(2x + 1)  [On cross multiplying]

⇒ 4x - 3 = 10x + 5

⇒ - 6x = 8

⇒ x = \(\frac{8}6\)

⇒ x = - \(\frac{4}3\)

Case II: 

If y = -2, we get:

\(\frac{4x-3}{2x+1}\) = - 2

⇒ 4x - 3 = - 2(2x + 1)

⇒ 4x - 3 = - 4x -2

⇒ 8x = 1

⇒ x = \(\frac{1}8\)

Hence, the roots of the equation are - \(\frac{4}3\) and \(\frac{1}8\).

x² – (√3 + 1) x + √3 = 0

x² – (√3 + 1) x + √3 = 0

x² – √3x – x + √3 = 0

x(x – √3) – (x – √3) = 0

(x – 1)(x – √3) = 0

either (x – 1) = 0 or (x – √3) = 0

x = 1 or √3

4x² + 5x = 0

Or x (4x + 5) = 0

Either x = 0 or 4x + 5 = 0, then

x = -5/4 or 0

x² + 6x + 5 = 0

x² + x + 5x + 5 = 0

x(x + 1) + 5(x + 1) = 0

(x + 5)(x + 1) = 0

either x +5 = 0 or x + 1 = 0

x = -5 or -1

It is given that (x = 1) is a root of (x² + kx + 3 = 0) . 

Therefore,(x =1) must satisfy the equation. 

⇒ (1)² + k x 1 + 3 =0

⇒ k + 4 = 0

⇒ k - 4

Hence, the required value of k is - 4.

3x² – 243 = 0

or x² – 81 = 0

(x)² – (9)² = 0

(x + 9) (x – 9) = 0

Either, x + 9 = 0 or x – 9 = 0

x = -9 or 9

We write, x = 4x - 3x as 2x² x (-6) = - 12x² = 4x x (-3x)

∴ 2x² + x - 6 = 0

⇒ 2x² + 4x - 3x - 6 = 0

⇒ 2x(x + 2) - 3(x + 2) = 0

⇒ (x + 2) (2x - 3) = 0

⇒ x + 2 = 0 or 2x - 3 = 0

⇒ x = - 2 or x = 3/2

Hence, the roots of the given equation are - 2 and 3/2.

Given:

3x² - 243 = 0

⇒ 3(x² - 81) = 0

⇒ (x)² - (9)² = 0

⇒ (x +9)(x - 9) = 0

⇒ x + 9 = 0 or x - 9 = 0

⇒ x = - 9 or x = 9

Hence, - 9 and 9 are the roots of the equation 3x² - 243 = 0

(2x - 3)(3x+1) = 0

⇒ 2x - 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = -1

⇒ x = 3/2 or x = -1/3

Hence the roots of the given equation are 3/2 and 1/3.

Correct option is (D) 2, 3

\(x^2-5x+6=0\)

\(\Rightarrow\) \(x^2-2x-3x+6=0\)

\(\Rightarrow\) x (x - 2) - 3 (x - 2) = 0

\(\Rightarrow\) (x - 2) (x - 3) = 0

\(\Rightarrow\) x - 2 = 0 or x - 3 = 0

\(\Rightarrow\) x = 2 or x = 3

Hence, roots of the quadratic given equation are 2 and 3.

Correct option is D) 2, 3

4x² + 5x = 0

⇒ x(4x + 5) = 0

⇒ x = 0 or 4x + 5 = 0

⇒ x = 0 or x = - 5/4

Hence, the roots of the given equation are 0 and - 5/4

Comparing x² + 10x - 7 = 0 with ax² + bx + c = 0

a = 1, b = 10 , c = -7,

 b² - 4 ac = 10 ² - 4 x 1 x-7 

= 100 + 28 

= 128