RS Aggarwal Solutions Class 10 Ex 10H

Question 1:

Which of the following is a quadratic equation?

(a) \(x^{2}-3\sqrt{x}+2=0\)

(b) \(x+\frac{1}{x}=x^{2}\)

(c) \(x^{2}+\frac{1}{x^{2}}=5\)

(d) \(2x^{2}-5x=(x-1)^{2}\)

Solution:

(d) \(2x^{2}-5x=(x-1)^{2}\)

A quadratic equation is the equation with degree 2.

Because, 2x2 – 5x = (x – 1)2

\(\Rightarrow\) 2x2 – 5x = x2 – 2x + 1

\(\Rightarrow\) 2x2 – 5x – x2 + 2x – 1 = 0

\(\Rightarrow\) x2 – 3x – 1 = 0, which is a quadratic equation

Question 2:

Which of the following is a quadratic equation?

(a) (x2 + 1) = (2 – x)2 + 3

(b) x3 – x2 = (x – 1)3

(c) 2x2 + 3 = (5 + x)(2x – 3)

(d) none of these

Solution:

(b) x3 – x2 = (x – 1)3

Because, x3 – x2 = (x – 1)3

\(\Rightarrow\) x3 – x2 = x3 – 3x2 + 3x – 1

\(\Rightarrow\) 2x2 – 3x + 1 = 0, which is a quadratic equation

Question 3:

Which of the following is not a quadratic equation?

(a) 3x – x2 = x2 + 5

(b) (x + 2)2 = 2(x2 – 5)

(c) \((\sqrt{2}x+3)^{2}=2x^{2}+6\)

(d) (x – 1)2 = 3x2 + x – 2

Solution:

(c) \((\sqrt{2}x+3)^{2}=2x^{2}+6\)

Because, \((\sqrt{2}x+3)^{2}=2x^{2}+6\)

\(\Rightarrow\) \(2x^{2}+9+6\sqrt{2}x=2x^{2}+6\)

\(\Rightarrow\) \(6\sqrt{2}x+3=0\), which is not a quadratic equation

Question 4:

If x = 3 is a solution of the equation 3x2 + (k – 1)x + 9 = 0, then k = ?

(a) 11

(b) -11

(c) 13

(d) -13

Solution:

(b) -11

It is given that x = 3 is a solution of 3x2 + (k – 1)x + 9 = 0; therefore, we have:

3(3)2 + (k – 1) x 3 + 9 = 0

\(\Rightarrow\) 27 + 3(k – 1) + 9 = 0

\(\Rightarrow\) 3(k – 1) = -36

\(\Rightarrow\) (k – 1) = -12

\(\Rightarrow\) k = -11

Question 5:

If one root of the equation 2x2 + ax + 6 = 0 is 2 then a = ?

(a) 7

(b) -7

(c) \(\frac{7}{2}\)

(d) \(\frac{-7}{2}\)

Solution:

(b) -7

It is given that one root of the equation 2x2 + ax + 6 = 0 is 2.

Therefore, 2 x 22 + a x 2 + 6 = 0

\(\Rightarrow\) 2a + 14 = 0

\(\Rightarrow\) a = -7

Question 6:

The sum of the roots of the equation x2 – 6x + 2 = 0 is

(a) 2

(b) -2

(c) 6

(d) -6

Solution:

(c) 6

Sum of the roots of the equation x2 – 6x + 2 = 0 is

\(\alpha +\beta =\frac{-b}{a}=\frac{-(-6)}{1}=6\), where \(\alpha\) and \(\beta\) are the roots of the equation.

Question 7:

If the product of the roots of the equation x2 – 3x + k = 10 is -2 then the value of k is

(a) -2

(b) -8

(c) 8

(d) 12

Solution:

(c) 8

It is given that the product of the roots of the equation x2 – 3x + k = 10 is -2.

The equation can be rewritten as:

x2 – 3x + (k – 10) = 0

Product of the roots of a quadratic equation = \(\frac{c}{a}\)

\(\Rightarrow\) \(\frac{c}{a}\) = -2

\(\Rightarrow\) \(\frac{(k-10)}{1}\) = -2

\(\Rightarrow\) k = 8

Question 8:

The ratio of the sum and product of the roots of the equation 7x2 – 12x + 18 = 0 is

(a) 7 : 12

(b) 7 : 18

(c) 3 : 2

(d) 2 : 3

Solution:

(c) 2 : 3

Given:

7x2 – 12 x + 18 = 0

Therefore, \(\alpha +\beta =\frac{12}{7}\) and \(\alpha \beta =\frac{18}{7}\), where \(\alpha\) and \(\beta\) are the roots of the equation

Therefore, Ratio of the sum and product of the roots = \(\frac{12}{7}:\frac{18}{7}\)

= 12 : 18

= 2 : 3

Question 9:

If one root of the equation 3x2 – 10x + 3 = 0 is \(\frac{1}{3}\) then the other root is

(a) \(\frac{-1}{3}\)

(b) \(\frac{1}{3}\)

(c) -3

(d) 3

Solution:

(d) 3

Given:

3x2 – 10x + 3 = 0

One root of the equation is \(\frac{1}{3}\).

Let the other root be \(\alpha\).

We know that:

Product of the roots = \(\frac{c}{a}\)

\(\Rightarrow\) \(\frac{1}{3}\times \alpha =\frac{3}{3}\)

\(\Rightarrow\) \(\alpha\) = 3

Question 10:

If one root of 5x2 + 13x + k = 0 be the reciprocal of the other root then the value of k is

(a) 0

(b) 1

(c) 2

(d) 5

Solution:

(d) 5

Let the roots of the equation (5x2 + 13x + k = 0) be \(\alpha\) and \(\frac{1}{\alpha }\).

Therefore, Product of the roots = \(\frac{c}{a}\)

\(\Rightarrow\) \(\alpha \times \frac{1}{\alpha } =\frac{k}{5}\)

\(\Rightarrow\) 1 = \(\frac{k}{5}\)

\(\Rightarrow\) k = 5

Question 11:

If the sum of the roots of the equation kx2 + 2x + 3k = 0 is equal to their product then the value of k is

(a) \(\frac{1}{3}\)

(b) \(\frac{-1}{3}\)

(c) \(\frac{2}{3}\)

(d) \(\frac{-2}{3}\)

Solution:

(d) \(\frac{-2}{3}\)

Given:

kx2 + 2x + 3k = 0

Sum of the roots = Product of the roots

\(\Rightarrow\) \(\frac{-2}{k}=\frac{3k}{k}\)

\(\Rightarrow\) 3k = -2

\(\Rightarrow\) k = \(\frac{-2}{3}\)

Question 12:

The roots of a quadratic equation are 5 and -2. Then, the equation is

(a) x2 – 3x + 10 = 0

(b) x2 – 3x – 10 = 0

(c) x2 + 3x – 10 = 0

(d) x2 + 3x + 10 = 0

Solution:

(b) x2 – 3x – 10 = 0

It is given that the roots of the quadratic equation are 5 and -2.

Then, the equation is:

x2 – (5 – 2)x + 5 x (-2) = 0

\(\Rightarrow\) x2 – 3x – 10 = 0

Question 13:

If the sum of the roots of a quadratic equation is 6 and their product is 6, the equation is

(a) x2 – 6x + 6 = 0

(b) x2 + 6x – 6 = 0

(c) x2 – 6x – 6 = 0

(d) x2 + 6x + 6 = 0

Solution:

(a) x2 – 6x + 6 = 0

Given:

Sum of roots = 6

Product of roots = 6

Thus, the equation is:

x2 – 6x + 6 = 0

Question 14:

If \(\alpha\) and \(\beta\) are the roots of the equation 3x2 + 8x + 2 = 0 then \(\left ( \frac{1}{\alpha }+\frac{1}{\beta } \right )\) = ?

(a) \(\frac{-3}{8}\)

(b) \(\frac{2}{3}\)

(c) -4

(d) 4

Solution:

It is given that \(\alpha\) and \(\beta\) are the roots of the equation 3x2 + 8x + 2 = 0.

Therefore, \(\alpha +\beta =-\frac{8}{3}\) and \(\alpha \beta =\frac{2}{3}\)

\(\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-\frac{8}{3}}{\frac{2}{3}}=-4\)

Hence, the correct answer is option C.

Question 15:

The roots of the equation ax2 + bx + c = 0 will be reciprocal of each other if

(a) a = b

(b) b = c

(c) c = a

(d) none of these

Solution:

(c) c = a

Let the roots of the equation (ax2 + bx + c = 0) be \(\alpha\) and \(\frac{1}{\alpha }\).

Therefore, Product of the roots = \(\alpha\) x \(\frac{1}{\alpha }\) = 1

\(\Rightarrow\) \(\frac{c}{a}=1\)

\(\Rightarrow\) c = a

Question 16:

If the roots of the equation ax2 + bx + c = 0 are equal then c = ?

(a) \(\frac{-b}{2a}\)

(b) \(\frac{b}{2a}\)

(c) \(\frac{-b^{2}}{4a}\)

(d) \(\frac{b^{2}}{4a}\)

Solution:

(d) \(\frac{b^{2}}{4a}\)

It is given that the roots of the equation (ax2 + bx + c = 0) are equal.

Therefore, (b2 – 4ac) = 0

\(\Rightarrow\) b2 = 4ac

\(\Rightarrow\) c = \(\frac{b^{2}}{4a}\)

Question 17:

If the equation 9x2 + 6kx + 4 = 0 has equal roots then k = ?

(a) 2 or 0

(b) -2 or 0

(c) 2 or -2

(d) 0 only

Solution:

(c) 2 or -2

It is given that the roots of the equation (9x2 + 6kx + 4 = 0) are equal.

Therefore, (b2 – 4ac) = 0

\(\Rightarrow\) (6k)2 – 4 x 9 x 4 = 0

\(\Rightarrow\) 36k2 = 144

\(\Rightarrow\) k2 = 4

\(\Rightarrow\) k = \(\pm\)2

Question 18:

If the equation x2 + 2(k + 2)x + 9k = 0 has equal roots then k = ?

(a) 1 or 4

(b) -1 or 4

(c) 1 or -4

(d) -1 or -4

Solution:

(a) 1 or 4

It is given that the roots of the equation (x2 + 2(k + 2)x + 9k = 0) are equal.

Therefore, (b2 – 4ac) = 0

\(\Rightarrow\) {2(k + 2)}2 – 4 x 1 x 9k = 0

\(\Rightarrow\) 4(k2 + 4k + 4) – 36k = 0

\(\Rightarrow\) 4k2 + 16k + 16 – 36k = 0

\(\Rightarrow\) 4k2 – 20k + 16 = 0

\(\Rightarrow\) k2 – 5k + 4 = 0

\(\Rightarrow\) k2 – 4k – k + 4 = 0

\(\Rightarrow\) k(k – 4) – (k – 4) = 0

\(\Rightarrow\) (k – 4)(k – 1) = 0

\(\Rightarrow\) k = 4 or k = 1

Question 19:

If the equation 4x2 – 3kx + 1 = 0 has equal roots then k = ?

(a) \(\pm \frac{2}{3}\)

(b) \(\pm \frac{1}{3}\)

(c) \(\pm \frac{3}{4}\)

(d) \(\pm \frac{4}{3}\)

Solution:

(d) \(\pm \frac{4}{3}\)

It is given that the roots of the equation (4x2 – 3kx + 1 = 0) are equal.

Therefore, (b2 – 4ac) = 0

\(\Rightarrow\) (3k2) – 4 x 4 x 1 = 0

\(\Rightarrow\) 9k2 = 16

\(\Rightarrow\) k2 = \(\frac{16}{9}\)

\(\Rightarrow\) k = \(\pm \frac{4}{3}\)

Question 20:

The roots of ax2 + bx + c = 0, a \(\neq\) 0 are real and unequal, if (b2 – 4ac) is

(a) > 0

(b) = 0

(c) < 0

(d) none of these

Solution:

(a) > 0

The roots of the equation are real and unequal when (b2 – 4ac) > 0.

Question 21:

In the equation ax2 + bx + c = 0, it is given that D = (b2 – 4ac) > 0. Then, the roots of the equation are

(a) real and equal

(b) real and unequal

(c) imaginary

(d) none of these

Solution:

We know that when discriminant, D > 0, the roots of the given quadratic equation are real and unequal.

Hence, the correct answer is option B.

Question 22:

The roots of the equation 2x2 – 6x + 7 = 0 are

(a) real, unequal and rational

(b) real, unequal and irrational

(c) real and equal

(d) imaginary

Solution:

(d) imaginary

Because, D = (b2 – 4ac)

= (-6)2 – 4 x 2 x 7

= 36 – 56

= -20 < 0

Thus, the roots of the equation are imaginary.

Question 23:

The roots of the equation 2x2 – 6x + 3 = 0 are

(a) real, unequal and rational

(b) real, unequal and irrational

(c) real and equal

(d) imaginary

Solution:

(b) real, unequal and irrational

Because, D = (b2 – 4ac)

= (-6)2 – 4 x 2 x 3

= 36 – 24

= 12

12 is greater than 0 and it is not a perfect square; therefore, the roots of the equation are real, unequal and irrational.

Question 24:

If the roots of 5x2 – kx + 1 = 0 are rwal and distinct then

(a) \(-2\sqrt{5}\) < k < \(2\sqrt{5}\)

(b) k > \(2\sqrt{5}\) only

(c) k < \(-2\sqrt{5}\) only

(d) either k > \(2\sqrt{5}\) or k < \(-2\sqrt{5}\)

Solution:

(d) either k > \(2\sqrt{5}\) or k < \(-2\sqrt{5}\)

It is given that the roots of the equation (5x2 – kx + 1 = 0) are real and distinct.

Therefore, (b2 – 4ac) > 0

\(\Rightarrow\) (-k)2 – 4 x 5 x 1 > 0

\(\Rightarrow\) k2 – 20 > 0

\(\Rightarrow\) k2> 20

\(\Rightarrow\) k > \(\sqrt{20}\) or k < \(-\sqrt{20}\)

\(\Rightarrow\) k > \(2\sqrt{5}\) or k < \(-2\sqrt{5}\)

Question 25:

If the equation x2 + 5kx + 16 = 0 has no real roots then

(a) k > \(\frac{8}{5}\)

(b) k < \(\frac{-8}{5}\)

(c) \(\frac{-8}{5}\) < k < \(\frac{8}{5}\)

(d) none of these

Solution:

(c) \(\frac{-8}{5}\) < k < \(\frac{8}{5}\)

It is given that the equation (x2 + 5kx + 16 = 0) has no real roots.

Therefore, (b2 – 4ac) < 0

\(\Rightarrow\) (5k)2 – 4 x 1 x 16 < 0

\(\Rightarrow\) 25k2 – 64 < 0

\(\Rightarrow\) k2< \(\frac{64}{25}\)

\(\Rightarrow\) \(\frac{-8}{5}\) < k < \(\frac{8}{5}\)

Question 26:

If the equation x2 – kx + 1 = 0 had no real roots then

(a) k < -2

(b) k > 2

(c) -2 < k < 2

(d) none of these

Solution:

(c) -2 < k < 2

It is given that the equation x2 – kx + 1 = 0 has no real roots.

Therefore, (b2 – 4ac) < 0

\(\Rightarrow\) (-k)2 – 4 x 1 x 1 < 0

\(\Rightarrow\) k2< 4

\(\Rightarrow\) -2 < k < 2

Question 27:

For what values of k, the equation kx2 – 6x – 2 = 0 has real roots?

(a) k \(\leq \frac{-9}{2}\)

(b) k \(\geq \frac{-9}{2}\)

(c) k \(\leq\) -2

(d) none of these

Solution:

(b) k \(\geq \frac{-9}{2}\)

It is given that the roots of the equation (kx2 – 6x – 2 = 0) are real.

Therefore, D \(\geq\) 0

\(\Rightarrow\) (b2 – 4ac) \(\geq\) 0

\(\Rightarrow\) (-6)2 – 4 x k x (-2) \(\geq\) 0

\(\Rightarrow\) 36 + 8k \(\geq\) 0

\(\Rightarrow\) k \(\geq\) \(\frac{-36}{8}\)

\(\Rightarrow\) k \(\geq\) \(\frac{-9}{2}\)

Question 28:

The sum of a number and its reciprocal is \(2\frac{1}{10}\). The number is

(a) \(\frac{5}{4}\) or \(\frac{4}{5}\)

(b) \(\frac{4}{3}\) or \(\frac{3}{4}\)

(c) \(\frac{5}{6}\) or \(\frac{6}{5}\)

(d) \(\frac{1}{6}\) or 6

Solution:

(a) \(\frac{5}{4}\) or \(\frac{4}{5}\)

Let the required number be x.

According to the question:

x + \(\frac{1}{x}\) = \(\frac{41}{20}\)

\(\Rightarrow\) \(\frac{x^{2}+1}{x}=\frac{41}{20}\)

\(\Rightarrow\) 20x2 – 41x + 20 = 0

\(\Rightarrow\) 20x2 – 25x – 16x + 20 = 0

\(\Rightarrow\) 5x(4x – 5) – 4(4x – 5) = 0

\(\Rightarrow\) (4x – 5)(5x – 4) = 0

\(\Rightarrow\) x = \(\frac{5}{4}\) or x = \(\frac{4}{5}\)

Question 29:

The perimeter of a rectangle is 82 m and its area is 400 m2. The breadth of the rectangle is

(a) 25 m

(b) 20 m

(c) 16 m

(d) 9 m

Solution:

(c) 16 m

Let the length and breadth of the rectangle be l and b.

Perimeter of the rectangle = 82 m

\(\Rightarrow\) 2 x (l + b) = 82

\(\Rightarrow\) l + b = 41

\(\Rightarrow\) l = (41 – b) – – – – – – – (i)

Area of the rectangle = 400 m2

\(\Rightarrow\) l x b = 400 m2

\(\Rightarrow\) (41 – b)b = 400 (using (i))

\(\Rightarrow\) 41b – b2 = 400

\(\Rightarrow\) b2 – 41b + 400 = 0

\(\Rightarrow\) b2 – 25b – 16b + 400 = 0

\(\Rightarrow\) b(b – 25) – 16(b – 25) = 0

\(\Rightarrow\) (b – 25)(b – 16) = 0

\(\Rightarrow\) b = 25 or b = 16

If b = 25, we have:

l = 41 – 25 = 16

Since, l cannot be less than b,

Therefore, b = 16 m

Question 30:

The length of a rectangular field exceeds its breadth by 8 m and the area of the field is 240 m2. The breadth of the field is

(a) 20 m

(b) 30 m

(c) 12 m

(d) 16 m

Solution:

Let the breadth of the rectangular filed be x m.

Therefore, Length of the rectangular filed = (x + 8) m

Area of the rectangular filed = 240 m2 (Given)

Therefore, \((x+8)\times x=240\) (Area = Length x Breadth)

\(\Rightarrow\) x2 + 8x – 240 = 0

\(\Rightarrow\) x2 + 20x – 12x – 240 = 0

\(\Rightarrow\) x(x + 20) – 12(x + 20) = 0

\(\Rightarrow\) (x + 20)(x – 12) = 0

\(\Rightarrow\) x + 20 = 0 or x -12 = 0

\(\Rightarrow\) x = -20 or x = 12

Therefore, x = 12 (Breadth cannot be negative)

Thus, the breadth of the field is 12 m.

Hence, the correct answer is option C.

Question 31:

The roots of the quadratic equation 2x2 – x – 6 = 0 are

(a) -2, \(\frac{3}{2}\)

(b) 2, \(\frac{-3}{2}\)

(c) -2, \(\frac{-3}{2}\)

(d) 2, \(\frac{3}{2}\)

Solution:

The given quadratic equation is 2x2 – x – 6 = 0

2x2 – x – 6 = 0

\(\Rightarrow\) 2x2 – 4x + 3x – 6 = 0

\(\Rightarrow\) 2x(x – 2) + 3(x – 2) = 0

\(\Rightarrow\) (x – 2)(2x + 3) = 0

\(\Rightarrow\) x – 2 = 0 or 2x + 3 = 0

\(\Rightarrow\) x = 2 or x = \(\frac{-3}{2}\)

Thus, the roots of the given equation are 2 and \(\frac{-3}{2}\).

Hence, the correct answer is option B.

Question 32:

The sum of two natural numbers is 8 and their product is 15. Find the numbers.

Solution:

Let the required numbers be x and (8 – x).

It is given that the product of the two numbers is 15.

Therefore, x(8 – x) = 15

\(\Rightarrow\) 8x – x2 = 15

\(\Rightarrow\) x2 – 8x + 15 = 0

\(\Rightarrow\) x2 – 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 required numbers are 3 and 5.

Question 33:

Show that x = -3 is a solution of x2 + 6x + 9 = 0.

Solution:

The given equation is x2 + 6x + 9 = 0.

Putting x = -3 in the given equation, we get

L.H.S. = (-3)2 + 6 x (-3) + 9 = 9 – 18 + 9 = 0 = R.H.S.

Therefore, x = -3 is a solution of the given equation.

Question 34:

Show that x = -2 is a solution of 3x2 + 13x + 14 = 0.

Solution:

The given equation is 3x2 + 13x + 14 = 0.

Putting x = -2 in the given equation, we get

L.H.S. = 3 x (-2)2 + 13 x (-2) + 14 = 12 – 26 + 14 = 0 = R.H.S.

Therefore, x = -2 is a solution of the given equation.

Question 35:

If x = \(\frac{-1}{2}\) is a solution of the quadratic equation 3x2 + 2kx – 3 = 0, find the value of k.

Solution:

It is given that x = \(\frac{-1}{2}\) is a solution of the quadratic equation 3x2 + 2kx – 3 = 0.

Therefore, \(3\times \left ( \frac{-1}{2} \right )^{2}+2k\times \left ( \frac{-1}{2} \right )-3=0\)

\(\Rightarrow\) \(\frac{3}{4}-k-3=0\)

\(\Rightarrow\) k = \(\frac{3-12}{4}=-\frac{9}{4}\)

Hence, the value of k is \(-\frac{9}{4}\).

Question 36:

Find the roots of the quadratic equation 2x2 – x – 6 = 0.

Solution:

The given quadratic equation is 2x2 – x – 6 = 0.

2x2 – x – 6 = 0

\(\Rightarrow\) 2x2 – 4x + 3x – 6 = 0

\(\Rightarrow\) 2x(x – 2) + 3(x – 2) = 0

\(\Rightarrow\) (x – 2)(2x + 3) = 0

\(\Rightarrow\) x – 2 = 0 or 2x + 3 = 0

\(\Rightarrow\) x = 2 or x = \(-\frac{3}{2}\)

Hence, the roots of the given equation are 2 and \(-\frac{3}{2}\).

Question 37:

Find the solution of the quadratic equation \(3\sqrt{3}x^{2}+10x+\sqrt{3}=0\).

Solution:

The given quadratic equation is \(3\sqrt{3}x^{2}+10x+\sqrt{3}=0\)

\(3\sqrt{3}x^{2}+10x+\sqrt{3}=0\)

\(\Rightarrow\) \(3\sqrt{3}x^{2}+9x+x+\sqrt{3}=0\)

\(\Rightarrow\) \(3\sqrt{3}x(x+\sqrt{3})+1(x+\sqrt{3})=0\)

\(\Rightarrow\) \((x+\sqrt{3})(3\sqrt{3}x+1)=0\)

\(\Rightarrow\) \(x+\sqrt{3}=0\) or \(3\sqrt{3}x+1=0\)

\(\Rightarrow\) x = \(-\sqrt{3}\) or x = \(-\frac{1}{3\sqrt{3}}=-\frac{\sqrt{3}}{9}\)

Hence, \(-\sqrt{3}\) and \(-\frac{\sqrt{3}}{9}\) are the solutions of the given equation.

Question 38:

If the roots of the quadratic equation 2x2 + 8x + k = 0 are equal then find the value of k.

Solution:

It is given that the roots of the quadratic equation 2x2 + 8k + k = 0 are equal.

Therefore, D = 0

\(\Rightarrow\) 82 – 4 x 2 x k = 0

\(\Rightarrow\) 64 – 8k = 0

\(\Rightarrow\) k = 8

Hence, the value of k is 8.

Question 39:

If the quadratic equation \(px^{2}-2\sqrt{5}px+15=0\) has two equal roots then find the value of p.

Solution:

It is given that quadratic equation \(px^{2}-2\sqrt{5}px+15=0\) has two equal roots.

Therefore, D = 0

\(\Rightarrow\) \((-2\sqrt{5}p)^{2}-4\times p\times 15=0\)

\(\Rightarrow\) 20p2 – 60p = 0

\(\Rightarrow\) 20p(p – 3) = 0

\(\Rightarrow\) p = 0 or p – 3 = 0

\(\Rightarrow\) p = 0 or p = 3

For p = 0, we get 15 = 0, which is not true.

Therefore, p \(\neq\) 0

Hence, the value of p is 3.

Question 40:

If 1 is a root of the equation ay2 + ay + 3 = 0 and y2 + y + b = 0 then find the value of ab.

Solution:

It is given that y = 1 is a root of the equation ay2 + ay + 3 = 0.

Therefore, a x (1)2 + a x 1 + 3 = 0

\(\Rightarrow\) a + a + 3 = 0

\(\Rightarrow\) 2a + 3 = 0

\(\Rightarrow\) a = \(-\frac{3}{2}\)

Also, y = 1 is a root of the equation y2 + y + b = 0.

Therefore, (1)2 + 1 + b = 0

\(\Rightarrow\) 1 + 1 + b = 0

\(\Rightarrow\) b + 2 = 0

\(\Rightarrow\) b = -2

Therefore, ab = \(\left ( -\frac{3}{2} \right )\times (-2)=3\)

Hence, the value of ab is 3.

Question 41:

If one zero of the polynomial x2 – 4x + 1 is \((2+\sqrt{3})\), write the other zero.

Solution:

Let the other zero of the given polynomial be \(\alpha\).

Now,

Sum of the zeroes of the given polynomial = \(\frac{-(-4)}{1}=4\)

Therefore, \(\alpha +(2+\sqrt{3})=4\)

\(\Rightarrow\) \(\alpha =4-2-\sqrt{3}=2-\sqrt{3}\)

Hence, the other zero of the given polynomial is \((2-\sqrt{3})\).

Question 42:

If one root of the quadratic equation 3x2 – 10x + k = 0 is reciprocal of the other, find the value of k.

Solution:

Let \(\alpha\) and \(\beta\) be the roots of the equation 3x2 – 10x + k = 0.

Therefore, \(\alpha =\frac{1}{\beta }\) (Given)

\(\Rightarrow\) \(\alpha \beta =1\)

\(\Rightarrow\) \(\frac{k}{3}=1\) (Product of the roots = \(\frac{c}{a}\))

\(\Rightarrow\) k = 3

Hence, the value of k is 3.

Question 43:

If the roots of the quadratic equation px(x – 2) + 6 = 0 are equal, find the value of p.

Solution:

It is given that the roots of the quadratic equation px2 – 2px + 6 = 0 are equal.

Therefore, D = 0

\(\Rightarrow\) (-2p)2 – 4 x p x 6 = 0

\(\Rightarrow\) 4p2 – 24p = 0

\(\Rightarrow\) 4p(p – 6) = 0

\(\Rightarrow\) p = 0 or p – 6 = 0

\(\Rightarrow\) p = 0 or p = 6

For p = 0, we get 6 = 0, which is not true.

Therefore, p \(\neq\) 0

Hence, the value of p is 6.

Question 44:

Find the values of k so that the quadratic equation x2 – 4kx + k = 0 has equal roots.

Solution:

It is given that the quadratic equation x2 – 4kx + k = 0 has equal roots.

Therefore, D = 0

\(\Rightarrow\) (-4k)2 – 4 x 1 x k = 0

\(\Rightarrow\) 16k2 – 4k = 0

\(\Rightarrow\) 4k(4k – 1) = 0

\(\Rightarrow\) k = 0 or 4k – 1 = 0

\(\Rightarrow\) k = 0 or k = \(\frac{1}{4}\)

Hence, 0 and \(\frac{1}{4}\) are the required values of k.

Question 45:

Find the values of k for which the quadratic equation 9x2 – 3kx + k = 0 has equal roots.

Solution:

It is given that the quadratic equation 9x2 – 3kx + k = 0 has equal roots.

Therefore, D = 0

\(\Rightarrow\) (-3k)2 – 4 x 9 x k = 0

\(\Rightarrow\) 9k2 – 36k = 0

\(\Rightarrow\) 9k(k – 4) = 0

\(\Rightarrow\) k = 0 or k – 4 = 0

\(\Rightarrow\) k = 0 or k = 4

Hence, 0 and 4 are the required values of k.

Question 46:

Solve: \(x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\).

Solution:

\(x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\)

\(\Rightarrow\) \(x^{2}-\sqrt{3}x-x+\sqrt{3}=0\)

\(\Rightarrow\) \(x(x-\sqrt{3})-1(x-\sqrt{3})=0\)

\(\Rightarrow\) \((x-\sqrt{3})(x-1)=0\)

\(\Rightarrow\) \(x-\sqrt{3}=0\) or x – 1 = 0

\(\Rightarrow\) x = \(\sqrt{3}\) or x = 1

Hence, 1 and \(\sqrt{3}\) are the roots of the given equation.

Question 47:

Solve: 2x2 + ax – a2 = 0.

Solution:

2x2 + ax – a2 = 0

\(\Rightarrow\) 2x2 + 2ax – ax – a2 = 0

\(\Rightarrow\) 2x(x + a) – a(x + a) = 0

\(\Rightarrow\) (x + a)(2x – a) = 0

\(\Rightarrow\) x + a = 0 or 2x – a = 0

\(\Rightarrow\) x = -a or x = \(\frac{a}{2}\)

Hence, -a and \(\frac{a}{2}\) are the roots of the given equation.

Question 48:

Solve: \(3x^{2}+5\sqrt{5}x-10=0\)

Solution:

\(3x^{2}+5\sqrt{5}x-10=0\)

\(\Rightarrow\) \(3x^{2}+6\sqrt{5}x-\sqrt{5}x-10=0\)

\(\Rightarrow\) \(3x(x+2\sqrt{5})-\sqrt{5}(x+2\sqrt{5})=0\)

\(\Rightarrow\) \((x+2\sqrt{5})(3x-\sqrt{5})=0\)

\(\Rightarrow\) \(x+2\sqrt{5}=0\) or \(3x-\sqrt{5}=0\)

\(\Rightarrow\) x = \(-2\sqrt{5}\) or x = \(\frac{\sqrt{5}}{3}\)

Hence, \(-2\sqrt{5}\) and \(\frac{\sqrt{5}}{3}\) are the roots of the given equation.

Question 49:

Solve: \(\sqrt{3}x^{2}+10x-8\sqrt{3}=0\)

Solution:

\(\sqrt{3}x^{2}+10x-8\sqrt{3}=0\)

\(\Rightarrow\) \(\sqrt{3}x^{2}+12x-2x-8\sqrt{3}=0\)

\(\Rightarrow\) \(\sqrt{3}x(x+4\sqrt{3})-2(x+4\sqrt{3})=0\)

\(\Rightarrow\) \((x+4\sqrt{3})(\sqrt{3}x-2)=0\)

\(\Rightarrow\) \(x+4\sqrt{3}=0\) or \(\sqrt{3}x-2=0\)

\(\Rightarrow\) x = \(-4\sqrt{3}\) or x = \(\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\)

Hence, \(-4\sqrt{3}\) and \(\frac{2\sqrt{3}}{3}\) are the roots of the given equation.

Question 50:

Solve: \(\sqrt{3}x^{2}-2\sqrt{2}x-2\sqrt{3}=0\)

Solution:

\(\sqrt{3}x^{2}-2\sqrt{2}x-2\sqrt{3}=0\)

\(\Rightarrow\) \(\sqrt{3}x^{2}-3\sqrt{2}x+2\sqrt{2}x-2\sqrt{3}=0\)

\(\Rightarrow\) \(\sqrt{3}x(x-\sqrt{6})+\sqrt{2}(x-\sqrt{6})\)

\(\Rightarrow\) \((x-\sqrt{6})(\sqrt{3}x+\sqrt{2})=0\)

\(\Rightarrow\) \(x-\sqrt{6}=0\) or \(\sqrt{3}x+\sqrt{2}=0\)

\(\Rightarrow\) x = \(\sqrt{6}\) or x = \(-\frac{\sqrt{2}}{\sqrt{3}}=-\frac{\sqrt{6}}{3}\)

Hence, \(\sqrt{6}\) and \(-\frac{\sqrt{6}}{3}\) are the roots of the given equation.

Question 51:

Solve: \(4\sqrt{3}x^{2}+5x-2\sqrt{3}=0\)

Solution:

\(4\sqrt{3}x^{2}+5x-2\sqrt{3}=0\)

\(\Rightarrow\) \(4\sqrt{3}x^{2}+8x-3x-2\sqrt{3}=0\)

\(\Rightarrow\) \(4x(\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2)=0\)

\(\Rightarrow\) \((\sqrt{3}x+2)(4x-\sqrt{3})=0\)

\(\Rightarrow\) \(\sqrt{3}x+2=0\) or \(4x-\sqrt{3}=0\)

\(\Rightarrow\) x = \(-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}\) or x = \(\frac{\sqrt{3}}{4}\)

Hence, \(-\frac{2\sqrt{3}}{3}\) and \(\frac{\sqrt{3}}{4}\) are the roots of the given equation.

Question 52:

Solve: 4x2 + 4bx – (a2 – b2) = 0

Solution:

4x2 + 4bx – (a2 – b2) = 0

\(\Rightarrow\) 4x2 + 4bx – (a – b)(a + b) = 0

\(\Rightarrow\) 4x2 + 2[(a + b) – (a – b)]x – (a – b)(a + b) = 0

\(\Rightarrow\) 4x2 + 2(a + b)x – 2(a – b)x – (a – b)(a + b) = 0

\(\Rightarrow\) 2x[2x + (a + b)] – (a – b)[2x + (a + b)] = 0

\(\Rightarrow\) [2x + (a + b)][2x – (a – b)] = 0

\(\Rightarrow\) 2x + (a + b) = 0 or 2x – (a – b) = 0

\(\Rightarrow\) x = \(-\frac{a+b}{2}\) or x = \(\frac{a-b}{2}\)

Hence, \(-\frac{a+b}{2}\) and \(\frac{a-b}{2}\) are the roots of the given equation.

Question 53:

Solve: x2 + 5x – (a2 + a – 6) = 0

Solution:

x2 + 5x – (a2 + a – 6) = 0

\(\Rightarrow\) x2 + 5x – (a + 3)(a – 2) = 0

\(\Rightarrow\) x2 + [(a + 3) – (a – 2)]x – (a + 3)(a – 2) = 0

\(\Rightarrow\) x2 + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0

\(\Rightarrow\) x[x + (a + 3)] – (a – 2)[x + (a + 3)] = 0

\(\Rightarrow\) [x + (a + 3)][x – (a – 2)] = 0

\(\Rightarrow\) x + (a + 3) = 0 or x – (a – 2) = 0

\(\Rightarrow\) x = -(a + 3) or x = (a – 2)

Hence, -(a + 3) and (a – 2) are the roots of the given equation.

Question 54:

x2 + 6x – (a2 + 2a – 8) = 0

Solution:

x2 + 6x – (a2 + 2a – 8) = 0

\(\Rightarrow\) x2 + 6x – (a2 + 2a – 8) = 0

\(\Rightarrow\) x2 + [(a + 4) – (a – 2)]x – (a + 4)(a – 2) = 0

\(\Rightarrow\) x2 + (a + 4)x – (a – 2)x – (a + 4)(a – 2) = 0

\(\Rightarrow\) x[x + (a + 4) – (a – 2)[x + (a + 4)] = 0

\(\Rightarrow\) [x + (a + 4)][x – (a – 2)] = 0

\(\Rightarrow\) x + (a + 4) = 0 or x – (a – 2) = 0

\(\Rightarrow\) x = -(a + 4) or x = (a – 2)

Hence, -(a + 4) and (a – 2) are the roots of the given equation.

Question 55:

x2 – 4ax + 4a2 – b2 = 0

Solution:

x2 – 4ax + 4a2 – b2 = 0

\(\Rightarrow\) x2 – 4ax + (2a + b)(2a – b) = 0

\(\Rightarrow\) x2 – [(2a + b) (2a – b)]x + (2a + b)(2a – b) = 0

\(\Rightarrow\) x2 – (2a + b)x – (2a – b)x + (2a + b)(2a – b) = 0

\(\Rightarrow\) x[x – (2a + b)] – (2a – b)[x – (2a + b)] = 0

\(\Rightarrow\) [x – (2a + b)][x – (2a – b) = 0

\(\Rightarrow\) x – (2a + b) = 0 or x – (2a – b) = 0

\(\Rightarrow\) x = (2a + b) or x = (2a – b)

Hence, (2a + b) and (2a – b) are the roots of the given equation.


Practise This Question

A 10 cm tall candle when viewed using a convex lens appears to be 20 cm tall. The focal length is 5 cm. What is the object distance?