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Nature of Roots. A quadratic equation ax² + bx + c = 0, a ≠ 0, has two roots which by the quadratic formula are as under :

\(\frac{-b+\sqrt{b^2-4ac}}{2a}\) and \(\frac{-b-\sqrt{b^2-4ac}}{2a}\)

The expression b² – 4ac is called the discriminant. 

Examining the nature of the roots means to see what type of roots the equation has, that is, whether they are real or imaginary, real or irrational, equal or unequal. The nature of the roots depends entirely on the value of the discriminant D = b² – 4ac 

Thus, if a, b, c are rational, then

I. If D = b² – 4ac > 0 (i.e, positive), the roots are real and unequal. 

Also,

(a) If D = b² – 4ac is a perfect square, the roots are rational. 

(b) If D = b² – 4ac is not a perfect square, the roots are irrational. 

(c) If D = b² – 4ac = 0, the roots are equal, each being equal to \(\frac{-b}{2a}\). 

So, ax² + bx + c = 0 is a perfect square if D = 0. 

II. If D = b² – 4ac < 0 (i.e., negative), the roots are imaginary (complex).

Given, \(\frac{1}{a+b+x}\) = \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{x}\) ⇒ \(\frac{1}{a+b+x}\)- \(\frac{1}{x}\) = \(\frac{1}{a}\) + \(\frac{1}{b}\)

⇒ \(\frac{x-(a+b+x)}{x(a+b+x)}\) = \(\frac{a+b}{ab}\) ⇒ \(\frac{-(a+b)}{x(a+b+x)}\) = \(\frac{a+b}{ab}\) ⇒ – ab = x² + (a + b)\(x\)

⇒ x² + (a + b)\(x\) + ab = 0 ⇒ (\(x\) + a) (\(x\) + b) = 0 ⇒ \(x\) = – a, – b.

(i) 2x² + 2x + 3 = 0                   (Here, a = 2, b = 2, c = 3)

∴ D = b² – 4ac = (2)² – 4 × 2 × 3 = 4 – 24 = – 20 < 0 

Hence, roots are imaginary. 

(ii) 2x² – 7x + 3 = 0                       (Here, a = 2, b = – 7, c = 3)

∴ D = b² – 4ac = 49 – 24 = 25 > 0 and a perfect square 

Hence, roots are real and rational. 

(iii) x² – 5x – 2 = 0.                  (Here, a = 1, b = – 5, c = – 2)

∴ D = b² – 4ac = 25 + 8 = 33 > 0 and not a perfect square 

Hence, roots are real and irrational. 

(iv) 4x² – 4x + 1 = 0                 (Here, a = 4, b = – 4, c = 1)

∴ D = b² – 4ac = 16 – 16 = 0. Hence, roots are real and equal.

If the two roots of the quadratic equation ax² + bx + c = 0 obtained by the quadratic formula be denoted by a and b, then we have

α = \(\frac{-b+\sqrt{b^2-4ac}}{2a},\) β = \(\frac{-b-\sqrt{b^2-4ac}}{2a}\)

∴ Sum of roots = α + β = \(\frac{-b+\sqrt{b^2-4ac}}{2a}\) + \(\frac{-b-\sqrt{b^2-4ac}}{2a}\) = \(\frac{-2b}{2a}=\frac{-b}{a}\)

Product of roots = αβ = \(\bigg(\frac{-b+\sqrt{b^2-4ac}}{2a}\bigg)\) x \(\bigg(\frac{-b-\sqrt{b^2-4ac}}{2a}\bigg)\)

= \(\frac{(-b)^2-(\sqrt{b^2-4ac})^2}{4a^2}\) = \(\frac{b^2-(b^2-4ac)}{4a^2}\) = \(\frac{4ac}{4a^2}\) = \(\frac{c}{a}.\)

Thus, 

Sum of roots = \(\frac{\text{-Coeff. of x}}{Coeff.\,of\,x^2}\); Product of roots = \(\frac{\text{Constant term}}{Coeff.\,of\,x^2}\)

Thus, if α, β be the roots of the equation 6x² – 5x + 7 = 0, then α + β = – (– \(\frac{5}{6}\)) = \(\frac{5}{6}\), ab = \(\frac{7}{6}\).

Correct option is (B) 3b² = 16ac

Let \(\alpha\;\&\;3\alpha\) are roots of the quadratic equation \(ax^2+bx+c=0.\)

\(\therefore\) Sum of roots \(=\frac{-b}a\)

\(\Rightarrow\) \(\alpha+3\alpha\) \(=\frac{-b}a\)

\(\Rightarrow\) \(\alpha\) \(=\frac{-b}{4a}\)          _______________(1)

Product of roots \(=\frac ca\)

\(\Rightarrow\) \(\alpha.3\alpha\) \(=\frac ca\)

\(\Rightarrow\) \(3\alpha^2\) \(=\frac ca\)

\(\Rightarrow\) \(3(\frac{-b}{4a})^2\) \(=\frac ca\)            (From (1))

\(\Rightarrow\) \(\frac{3b^2}{16a^2}\) \(=\frac ca\)

\(\Rightarrow\) \(3b^2=\frac{16a^2c}a=16ac\)

Correct option is B) 3b² = 16ac

Let the roots be α and 4α. Then,

α + 4α = 5α = \(-\frac{b}{a}\)

α . 4α = 4α² = \(\frac{c}{a}\)

From (i) α =  \(\frac{b}{5a}\)

∴ From (ii) 4. \(\big(-\frac{b}{5a}\big)^2\) = \(\frac{c}{a}\) ⇒ \(\frac{4b^2}{25a^2}\) = \(\frac{c}{a}\) ⇒ 4b² = 25ac.

True, because in this case discriminant is always positive.

For example, in ax²+ bx + c = 0, as a and c have opposite sign, ac < 0

⟹ Discriminant = b² – 4ac > 0.

False. For example, equation x² + 4 = 0 has no real root.

True, because every quadratic polynomial has almost two roots.

False. For example, a quadratic equation x² – 4x + 4 = 0 has only one root which is 2.

Let’s consider the length of smaller side of rectangle as x metres

Then, the larger side will be (x + 30) metres and diagonal will be = (x + 60) metre

[From given relation]

Now, by using Pythagoras theorem we have,

x² + (x + 30)² = (x + 60)²

x² + x² + 60x + 900 = x² + 120x + 3600

2x² + 60x + 900 – x² – 120x – 3600 = 0

x² – 60x – 2700 = 0

x² – 90x + 30x – 2700 = 0 [By factorisation method]

x(x – 90) + 30(x – 90) = 0

(x – 90)(x + 30) = 0

x = 90   or x = -30 (this is neglected as the side of a rectangle can never be negative)

Therefore, we only take x = 90,

⇒ x + 30 = 90 + 30 = 120

Thus, the length of smaller side of rectangle is 90 metres and the larger side is 120 metres.

(i) Quadratic, because it is in the form of ax² + bx² + c = 0

(ii) n = 2, and also a≠0, as it will make the polynomial 0.

(iii) Putting the value of x = -1, in x² - 5x + 4

(-1)² - 5(-1) + 4

1 + 5 + 4

10

The value is 10.

(iv) The degree is the highest power of the term in the expression, so it is 3.

(v) The zeroes of a polynomial are α & β.

∴ To be zero, αx²+b α+c=0 & βx²+b β+c=0

-x² + 9 = 0

-x² = -9

x² = 9

∴ x = ±3

∴ The zeroes of the given polynomial are 3 & -3.

4x² - 1 = 0

4x² = 1

x² = 1/4

x = ± 1/2

The zeroes of the given polynomial are 1/2 & -1/2.

9 - 4x² = 0

4x² = 9

x² = 9/4

∴ x = ± 3/2

The zeroes of the given polynomial are 3/2 & -3/2 and the option (c) is correct.

Putting the value of -2, in the given polynomial,

3(-2)² + (-2) – 10

3(4) – 2 – 10

12 – 12

0

∵ the value comes out to be 0.

∴ -2 is one of the zeroes and, yes 3x² + x - 10 is a quadratic polynomial.

In the above expressions, only the second one has a positive power, unlike others.

∴ It is the only polynomial.

Equating the expression with 0,

x² + 2x = 0

x ( x+2) = 0

∴ x = 0 or x + 2 = 0

x = 0 or x = -2

The zeroes of the polynomial x² +2x are 2 & -2, having a degree of 2, being the highest power of the terms in the same expression.

Putting the value of -1, in the given polynomial,

(-1)² + 2(-1) – 3

1 – 2 – 3

3 – 3

0

∵ the value comes out to be 0.

∴ -1 is one of the zeroes of the given polynomial.

Correct option is (D) 24, 25

Let x & x+1 are two consecutive natural numbers whose product is 600.

\(\therefore\) x (x+1) = 600

\(\Rightarrow x^2+x-600=0\)

\(\Rightarrow x^2+25x-24x-600=0\)

\(\Rightarrow x(x+25)-24(x+25)=0\)

\(\Rightarrow(x+25)(x-24)=0\)

\(\Rightarrow\) x+25 = 0 or x - 24 = 0

\(\Rightarrow\) x = -25 or x = 24

\(\therefore\) x = 24      \((\because\) x is a natural number, therefore \(x\neq-25)\)

\(\Rightarrow\) x+1 = 25

Hence, required natural numbers are 24, 25.

Correct option is D) 24, 25

The correct option is: (A) 1, -8

Explanation:

Given equation is  x^2/3 + x^1/3 - 2 = 0

 Let y = x^1/3 =>  y² + y - 2 = 0

=> y² + 2y - y - 2 = 0 => (y - 1)(y + 2) = 0

=> y = 1 1 or y = - 2 => x^1/3 = 1 or x^1/3 = -2

 => x = 1 or x = -8

The correct option is: (B) x² - 12x + 32 = 0

Explanation:

Let the two roots be a and b, then

a + b = 12...(1) and a - b = 4 ...(2)

=> a = 8 and b = 4 (from (1)and (2)) 

.'. Required equation is x² - 12x + 32 = 0

Given,

kx(x – 2) + 6 = 0

It can be rewritten as,

kx² – 2kx + 6 = 0

It’s of the form of ax² + bx + c = 0

Where, a =k, b = -2k, c = 6

For the given quadratic equation to have real roots D = b² – 4ac ≥ 0

D = (-2k)² – 4(k)(6) ≥ 0

⇒ 4k² – 24k ≥ 0

⇒ 4k(k – 6) ≥ 0

⇒ k ≥ 0 and k ≥ 6

⇒ k ≥ 6

The value of k should be greater than or equal to 6 to have real roots.

A quadratic equation has two real roots if discriminant = 0 

For the given equation, we have: 

d = b² – 4 a c 

d = (2)² – 4 (1) (a² + 1) 

d = 4 – 4(a² + 1) 

d = 4(1 – a² – 1) 

d = – 4a²

Now, D = 0 when a = 0. So, the equation will have real and equal roots if a = 0. And for all other values of a, the equation will have no real roots.

No, there is no real value of ‘a’ for which the given equation has real roots.

Consider x² + ax – 1 = 0,For the quadratic equation to have real roots D ≥ 0

Here a = 1, b = a and c = 1

In the given equation D = a² – 4 ≥ o

⇒ a² ≥ 4

So for all the real values of ‘a’ which are greater than or equal to 2 and – 2 the equation will have the real roots.

First of all, the equation is x² + |x| − 6 = 0

CASE 1: x>0 then |x| = x

⇒ x² + x− 6 = 0

⇒ x² + 3x – 2x – 6 = 0

⇒ (x + 3)(x−2) = 0

⇒ x = 2, – 3

CASE 2: x<0 |x| = – x

⇒ x² − x − 6 = 0

⇒ x² − 3x + 2x−6 = 0 

⇒ (x − 3)(x + 2) = 0 

⇒ x = −2, 3

So the sum of the roots is 2 + (– 3) + (– 2) + 3

= 0

If ax² + bx + c = 0 then x \(=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

If D = b² − 4ac ≥ 0 then the values of x are real

If D = b² − 4ac < 0 then the values of x are complex

|z| is always a positive real number regardless of x being a real number or complex number.

Given eqn. is |x|² + 3|x| + 2 = 0 and a = 1, b = 3 and c = 2

|x|² + 3|x| + 2 = 0

⇒ |x| = \(\frac{-3\pm\sqrt{9-8}}{2}\)

⇒ |x| = 2 or −1

But |x| cannot be negative 

No real root for the equation.

given x²- a(x+1) - c = 0 

x²- ax - a - c = 0 

x²- ax - (a + c) = 0 

as α, β are roots of equation, we get 

sum of the roots α + β = a 

Product of roots αβ = - (a + c) 

Given (1+ α) (1+ β) 

= 1 + (α + β) + (α β) 

= 1 + a – a – c 

= 1 – c.

In the given equation x² – (k + 6)x + 2 (2k – 1) = 0 

a = 1, b = – (k + 6), c = 2 (2k – 1) 

Sum of the roots = 1/2 (product of roots) (given) 

K + 6 = 2k – 1 

K = 7

Given a and b are roots of the equation x² + a x + b = 0 

Here a = 1, b = a, c = b 

Sum of the roots = a + b = – a/1 

b = – 2a 

Product of the roots 

ab = b 

a = 1 

∴ b = – 2 x 1 = – 2 

Now a + b = 1 + (– 2) = – 1

Correct option is (C) -3 or -10

When coefficient of x is 17 then -2 and -15 are roots of equation \(x^2+px + q = 0\)      _____________(1)

i.e., when p = 17, x = -2 is a root of equation (1)

\(\therefore(-2)^2+17\times-2+q=0\)

\(\Rightarrow4-34+q=0\)

\(\Rightarrow q=34-4=30\)

Now, let p = 13 then original quadratic equation becomes

\(x^2+13x+q=0\)

\(\Rightarrow x^2+13x+30=0\)    \((\because q=30)\)

\(\Rightarrow x^2+10x+3x+30=0\)

\(\Rightarrow x(x+10)+3(x+10)=0\)

\(\Rightarrow(x+10)(x+3)=0\)

\(\Rightarrow x+10=0\;or\;x+3=0\)

\(\Rightarrow x=-10\;or\;x=-3\)

Hence, -10 and -3 are roots of original equation.

Correct option is C) -3 or -10

Correct option is (C) -5 and 3

Given equation is

\(\frac{15}{x^2-4}-\frac{2}{x-2}=1\)

\(\Rightarrow\frac{15-2(x+2)}{x^2-4}=1\)

\(\Rightarrow15-2x-4=x^2-4\)

\(\Rightarrow x^2+2x-4+4-15=0\)

\(\Rightarrow x^2+2x-15=0\)

\(\Rightarrow x^2+5x-3x-15=0\)

\(\Rightarrow x(x+5)-3(x+5)=0\)

\(\Rightarrow(x+5)(x-3)=0\)

\(\Rightarrow x+5=0\) \(or\;x-3=0\)

\(\Rightarrow x=-5\;or\;x=3\)

Hence, the roots of given equation are -5 and 3.

Correct option is C) -5 and 3

Correct option is (A) 8 and -3

Given equation is

\(2x^2-hx + 2k = 0\)

\(\therefore\) Sum of roots \(=\frac{-(-h)}2=\frac{h}2\)

\(\Rightarrow\frac h2=4\)      \((\because\) Sum of roots = 4 (given))

\(\Rightarrow\) h = 8

And product of roots \(=\frac{2k}2=k\)

\(\Rightarrow\) k = -3      \((\because\) Product of roots = -3 (given))

Hence, h = 8 & k = -3

Correct option is A) 8 and -3

(c) x² – 13x – 30 = 0.

Sum of roots = 2 + (– 15) = – 13; Product of roots = 2 × (– 15) = – 30. 

∴ Required equation is x² – (sum of roots) x + product of roots = 0 

⇒ Reqd. equation = x² – 13x – 30 = 0.

Let 3 and α be the roots of the equation x² + ax + 3 = 0 

Then, sum of roots = 3 + α = – a              ..(i), Product of roots = 3α = 3                 ...(ii) 

From (ii) α = 1.          ∴ Substituting a = 1 in (i), we get a = – 4.

∴ The second equation x² + ax + b = 0 becomes x² – 4x + b = 0. 

Let β and 3β be the roots of this equation. Then, sum of roots = β + 3β = 4 ⇒ 4β = 4 ⇒ β = 1 and Product of roots = β × 3β = b ⇒ 3β² = b ⇒ b = 3.

For being the perfect square, the roots are equal

So, d = b² – 4ac = 0

So, d = b² – 4ac = 0

⇒ d = 16 – 4 λ = 0

⇒ λ = 4

Given,

(2x + 3) (3x – 7) = 0.

So, either 2x + 3 = 0, ⇒ x = \(\frac{– 3}{2}\)

Or, 3x -7 = 0, ⇒ x = \(\frac{7}{3}\)

Thus, the roots of the given quadratic equation are x = \(\frac{– 3}{2}\) and x = \(\frac{7}{3}\) respectively.

Given.

3x² – 14x – 5 = 0

⇒ 3x² – 14x – 5 = 0

⇒ 3x² – 15x + x – 5 = 0

⇒ 3x(x – 5) + 1(x – 5) = 0

⇒ (3x + 1)(x – 5) = 0

Now, either 3x + 1 = 0 ⇒ x =\(\frac{ -1}{3}\)

Or, x – 5 = 0 ⇒ x = 5

Thus, the roots of the given quadratic equation are 5 and x = \(\frac{ -1}{3}\) respectively.

Let the sister and her sister’s age be ‘X’ yrs and ’(X+5)’ yrs respectively.

As per the question,

X(X + 5) = 104

X² + 5X – 104 = 0

X² + 13X – 8X + 132 = 0

X(X + 13) – 8(X + 13) = 0

(X + 13)(X – 8) = 0

∴ X = 8 or X = –13

The ages of sisters are 8 years & 13 years respectively, as the age can’t be negative.

Let the father and his son’s age be ‘X’ yrs and ‘Y’ yrs respectively.

X + Y = 35

X × Y = 150

X = 150/Y

150/Y + Y = 35

150 + Y² = 35Y

Y² – 35Y + 150 = 0

(Y–5) (Y–30)=0

Y=5

The son's age (Y) = 5 yrs and father's age (X) = 30 yrs.

Let the present age of Pragati be x years. 

∴ 2 years ago, 

Age of Pragati = (x – 2) years 

After 3 years, 

Age of Pragati = (x + 3) years 

According to the given condition,

(x – 2) (x + 3) = 84 

∴ x(x + 3) – 2(x + 3) = 84 

∴ x² + 3x – 2x – 6 = 84 

∴ x² + x – 6 – 84 = 0 

∴ x² + x – 90 = 0 

x² + 10x – 9x – 90 = 0 

∴ x(x + 10) – 9(x + 10) = 0 

∴ (x + 10)(x – 9) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get 

∴ x + 10 = 0 or x – 9 = 0 

∴ x = -10 or x = 9 But, age cannot be negative. 

∴ x = 9 

∴ Present age of Pragati is 9 years.

Let the first even natural number be x. 

∴ the next consecutive even natural number will be (x + 2). 

According to the given condition, 

x² + (x + 2)² = 244 

∴ x² + x² + 4x + 4 = 244 

∴ 2x² + 4x + 4 – 244 = 0

∴ 2x² + 4x – 240 = 0 

∴ x² + 2x – 120 = 0 …[Dividing both sides by 2] 

∴ x² + 12x – 10x – 120 = 0 

∴ x(x + 12) – 10 (x + 12) = 0 

∴ (x + 12) (x – 10) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get 

∴ x + 12 = 0 or x – 10 = 0 

∴ x = -12 or x = 10 

But, natural number cannot be negative. 

∴ x = 10 and x + 2 = 10 + 2 = 12 

∴ The two consecutive even natural numbers are 10 and 12.

Assume the integer be x. Then its square will be x².

And given, their sum is 90

⇒ x + x² = 90

⇒ x² + x – 90 = 0

Solving for x by factorization method, we have

x² + 10x – 9x – 90 = 0

⇒ x(x + 10) – 9(x + 10) = 0

⇒ (x + 10)(x – 9) = 0

Now, either x + 10 = 0 ⇒ x = –10

Or, x – 9 = 0 ⇒ x = 9

Thus, the values of the integer are 9 and -10 respectively.

Given, the difference between two digits is 6 and the ten's digit is bigger than the unit's digit.

So, let the unit's digit be x and ten's digit be (x + 6).

From the given condition, we have:

x(x + 6) = 27

x² + 6x − 27 = 0

x² + 9x − 3x − 27 = 0

x(x + 9) − 3(x + 9) = 0

(x + 9) (x − 3) = 0

x = −9, 3

Since, the digits of a number cannot be negative. So, x = 3.

Unit's digit = 3

Ten's digit = 9

Thus, the number is 93.

Let ‘x’ be the number of books bought by the shopkeeper.

∴ Cost of one book = Rs. 80 ÷ x.

[∵ the total cost of books is Rs. 80]

∴ Cost of one book when he buys 4 more books for same rate = Rs. 80 ÷ (x + 4).

When he buys four more books for the same amount; it will cost Re. 1 per book less than the previous.

∴ 80 ÷ (x + 4) = (80 ÷ x) − 1

⇒ 80/(x + 4) = 80/x - 1

⇒ 80x = (x + 4) (80 − x)

⇒ 80x = 80x − x² + 320 − 4x

⇒ x² + 4x + 320 = 0

Let the consecutive positive odd numbers be x and x + 2.

From the given information,

x² + (x + 2)² = 74

2x² + 4x + 4 = 74

2x² + 4x − 70 = 0

x² + 2x − 35 = 0

(x + 7)(x − 5) = 0

x = −7, 5

Since, the numbers are positive, so, x = 5.

Thus, the numbers are 5 and 7.

Let the larger diameter pipe fill it in ‘a’ hours

The smaller diameter pipe fills it in ‘a + 10’ hours

In 1 hour, larger diameter pipe fills 1/a part of the pool.

In 1 hour, smaller diameter pipe fills 1/(a + 10) part of the pool.

Given, the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled.

⇒ 4 × 1/a + 9 × 1/(a+10) = 1/2

⇒ 2(4a + 40 + 9a) = a² + 10a

⇒ 26a + 80 = a² + 10a

⇒ a² – 16a – 80 = 0

⇒ a² – 20a + 4a – 80 = 0

⇒ a(a – 20) + 4(a – 20) = 0

⇒ (a + 4)(a – 20) = 0

⇒ a = 20 hours

Time in which smaller diameter pipe fills the pool = 20 + 10 = 30 hours

Equation ax² + bx + c = 0 has cos a and cos a as two roots

sin α + cos α = \(-\frac{b}{a}\)

sin α × cos α = c/a …eq(1)

\((sin\,a\,+\,cosa)^2=\frac{b^2}{a^2}\)

\(sin^2a\,+cos^2a\,+2\,sina\,cos=\frac{b^2}{a^2}....eq(2)\)

But sin2 α + cos2 α = 1 

∴ a² (1 + 2 sinα.cos α) = b² 

Putting sin α × cos α = c/a, we get,

⇒ b² = a² + 2ac.

2x² – 5√2x + 4 = 0

Compare given equation with the general form of quadratic equation, which is ax² + bx + c = 0

Here, a = 2, b = – 5√2, c = 4

Discriminant formula: D = b² – 4ac

= (– 5√2)² – 4.2.4

= 50 – 32

= 18

Since the roots of the given equations are equal, so discriminant will be equal to zero.

=> b²(c - a)² - 4a(b - c) . c(a - b) = 0

=> b²(c² + a² - 2ac) - 4ac(ba - ca - b² + bc) = 0,

=> a²b² + b²c² + 4a²c² + 2b²ac - 4ac²bc - 4abc² = 0 

=> (ab + bc - 2ac)² = 0

=> ab + bc - 2ac = 0 

=> ab + bc = 2ac

=> 1/c+1/a=2/b

=> 2/b=1/a+1/c.

Hence Proved.

If the roots of the given equation are equal, then discriminant is zero i.e.

(c - a)² - 4(b - c) (a - b) = 0 

=> c² + a² - 2ac + 4b² - 4ab + 4ac - 4bc = 0

=>c² + a² + 4b² + 2ac - 4ab - 4bc = 0

=>(c + a - 2b)² = 0

=>c + a = 2b Hence Proved.