Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Quadratic Expressions Important Questions Very Short Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2A Quadratic Expressions Important Questions Very Short Answer Type

Question 1.

Find the roots of the quadratic equation ax^{2} + bx + c = 0. [March ’02]

Solution:

Given quadratic equation is ax^{2} + bx + c = 0

Now, multiplying with ‘4a’ on both sides

4a (ax^{2} + bx – c) = 0

= 4ax^{2}x^{2} + 4abx + 4ac = 0

(2ax)^{2} + 2 . 2ax . b + b^{2} + b^{2} – b^{2} + 4ac = 0

(2ax + b)^{2} = b^{2} – 4ac

2ax + b = ± \(\sqrt{b^2-4 a c}\)

2ax = – b ± \(\sqrt{b^2-4 a c}\)

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

∴ The roots of the quadratic equation, ax^{2} + bx + c = 0 are \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\).

Question 2.

Fmd the roots of the following equation √3x^{2} + 10x – 8√3 = 0.

Solution:

Given quadratic equation is √3x^{2} + 10x – 8√3 = 0

Comparing this with ax^{2} + bx + c = 0,

we have a = √3, b = 10 c = – 8√3

The roots of quadratic equation are \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

= \(\frac{-10 \pm \sqrt{100+96}}{2 \sqrt{3}}=\frac{-10 \pm \sqrt{196}}{2 \sqrt{3}}\)

= \(\frac{-10 \pm 14}{2 \sqrt{3}}=\frac{-10+14}{2 \sqrt{3}} \text { (or) } \frac{-10-14}{2 \sqrt{3}}\)

= \(\frac{4}{2 \sqrt{3}} \text { (or) } \frac{-24}{2 \sqrt{3}}=\frac{2}{\sqrt{3}} \text { (or) } \frac{-12}{\sqrt{3}}\)

= \(\frac{2}{\sqrt{3}}\) (or) – 4√3

∴ The roots of the quadratic equation are \(\frac{2}{\sqrt{3}}\) (or) – 4√3

Question 3.

Form a quadratic equation whose roots are 7 ± 2√5. [TS – May ’16; March ’11, ’05, AP – March 2018]

Solution:

Let α = 7 + 2√5, β = 7 – 2√5

Now, α + β = 7 + 2√5 + 7 – 2√5 = 14

αβ = (7 + 2√5) (7 – 2√5)

= 49 – 20 = 29

The quadratic equation whose roots are α, β is

x^{2} – (α + β)x + αβ = 0

x^{2} – 14x + 29 = 0.

Question 4.

Form a quadratic equation whose roots are – 3 ± 5i.

Solution:

Let α = – 3 + 5i, β = – 3 – 5i

Now, α + β = – 3 + 5i – 3 – 5i = – 6

αβ = (- 3 + 5i) (- 3 – 5i)

= 9 – 25i^{2} = 34

The quadratic equation whose roots are α, β is x^{2} – (α + β)x + αβ = 0

⇒ x^{2} + 6x + 34 = 0

Question 5.

Form a quadratic equation whose roots are \(\frac{p-q}{p+q}\), – \(\frac{p+q}{p-q}\) (p ≠ q).

Solution:

The quadratic equation whose roots are α, β is x^{2} – (α + β)x + αβ = 0

x^{2} + \(\frac{4 p q}{p^2-q^2}\) . x – 1 = 0

⇒ (p^{2} – q^{2})x^{2} + 4pqx – (p^{2} – q^{2}) = 0

Question 6.

Find the nature of the roots of 4x^{2} – 20x + 25 = 0.

Solution:

Given quadratic equation is 4x^{2} – 20x + 25 = 0.

Comparing this with ax^{2} + bx + c = 0, we get

a = 4; b = – 20; c = 25

b^{2} – 4ac = 400 – 4 . 4 . 25 = 400 – 400 = 0

Since ^{2} – 4ac = 0 then the roots of the given equation are real and equal.

Question 7.

Find the nature of the roots of 2x^{2} – 8x + 3 = 0.

Solution:

Given quadratic equation is 2x^{2} – 8x + 3 = 0

Comparing this with ax^{2} + bx + c = 0, we get

a = 2, b = – 8, c = 3

Now, b^{2} – 4ac = 64 – 4 . 2 . 3

Since, b^{2} – 4ac > 0 then the roots of the given equation are real and distinct.

Question 8.

Find the nature of the roots of 2x^{2} – 7x + 10 = 0.

Solution:

Given quadratic equation is 2x^{2} – 7x + 10 = 0

Comparing this with ax^{2} + bx + c = 0, we get a = 2, b = – 7, c = 10

Now, b^{2} – 4ac = 49 – 80 = – 31 < 0

Since, b^{2} – 4ac < 0 then the roots of the given equation are conjugate complex numbers.

Question 9.

Find the nature of the roots of 3x^{2} + 7x + 2 = 0.

Solution:

Given quadratic equation is 3x^{2} + 7x + 2 = 0

Comparing this with ax^{2} + bx + c = 0, we get a = 3, b = 7, c = 2

Now, b^{2} – 4ac = 49 – 4 . 3 . 2

= 49 – 24 = 25 = 52 > 0

Since, b^{2} – 4ac > 0 then the roots of the given equation are rational and unequal.

Question 10.

If α, β are the roots of ax^{2} + bx + c = 0 then find \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\).

Solution:

If α, β are the roots of ax^{2} + bx + c = 0 then

α + β = \(\frac{-b}{a}\)

αβ = \(\frac{c}{a}\)

Now, \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\beta^2+\alpha^2}{\alpha^2 \beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{(\alpha \beta)^2}\)

= \(\frac{\left(\frac{-b}{a}\right)^2-2 \cdot \frac{c}{a}}{\left(\frac{c}{a}\right)^2}=\frac{b^2-2 c a}{c^2}\)

Question 11.

If α, β are the roots of ax^{2} + bx + c = 0 then find \(\frac{1}{\alpha}+\frac{1}{\beta}\). [March ’10]

Solution:

If α, β are the roots of ax^{2} + bx + c = 0

then α + β = \(\frac{-b}{a}\)

αβ = \(\frac{c}{a}\)

Now \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{\frac{-b}{a}}{\frac{c}{a}}=\frac{-b}{c}\)

Question 12.

If α, β are the roots of ax^{2} + bx + c = 0 then find α^{2} + β^{2}. [May ’13]

Solution:

If α, β are the roots of ax^{2} + bx + c = 0

then α + β = \(\frac{-b}{a}\), αβ = \(\frac{c}{a}\)

Now α^{2} + β^{2} = (α + β)^{2} – 2αβ

= \(\left(\frac{-b}{a}\right)^2-\frac{2 c}{a}=\frac{b^2}{a^2}-\frac{2 c}{a}=\frac{b^2-2 a c}{a^2}\).

Question 13.

For what values of m, the equation x^{2} – 15 – m (2x – 8) = 0 will have equal roots ?

[March ’13 ’04, May ’08, TS – Mar. 2015; AP – Mar. ’19,’17]

Solution:

Given equation is x^{2} – 15 – m (2x – 8) = 0

x^{2} – 15 – 2mx + 8m = 0

⇒ x^{2} – 2mx + 8m- 15 = 0 ……………(1)

Comparing this equation with

ax^{2} + bx + c = 0, we get

a = 1, b = – 2m, c = 8m – 15

Since (1) have equal roots

⇒ b^{2} – 4ac = 0

⇒ 4m^{2} – 4 (1) (8m – 15) = 0

⇒ 4m^{2} – 32m + 60 = 0

⇒ m^{2} – 8m + 15 = 0

⇒ m^{2} – 5m – 3m+ 15 = 0

⇒ m (m – 5) – 3 (m – 5) = 0

⇒ (m – 3) (m – 5) = 0

⇒ m = 3 (or) m = 5

∴ The values of m are 3, 5.

Question 14.

For what values of in, (m + 1) x^{2} + 2(m + 3) x + (m + 8) = 0 has equal roots? [March ‘03]

Solution:

Given quadratic equation is (m + 1) x^{2} + 2 (m + 3) x +(m + 8) = 0 …………..(1)

Comparing this with ax^{2} + bx + c = 0, we get

a = m + 1, b = 2 (m + 3), c = m + 8

Since (1) have equal roots then b^{2} – 4ac =0

⇒ [2 (m + 3)]^{2} – 4 (m + 1) (m + 8) = 0

⇒ 4 (m + 3)^{2} – 4(m + 1) (m + 8) = 0

⇒ m^{2} + 6m + 9 – m^{2} – 8m – m – 8 = 0

⇒ – 3m – 1 = 0 m = \(\frac{1}{3}\)

Question 15.

For what values of m, (3m + 1) x^{2} + 2(m + 1) x + m = 0 will have equal roots?

Solution:

Given quadratic equation is (3m + 1)x^{2} + 2(m + 1)x +m = 0 ………….(1)

Comparing this with ax^{2} + bx + c = 0, we get

a = 3m + 1, b = 2(m + 1), c = m

since (1) have equal roots then b^{2} – 4ac = 0

4 (m + 1)^{2} – 4m (3m + 1) = 0

m^{2} + 2m + 1 – 3m^{2} – m = 0

– 2m^{2} + m + 1 = 0

2m^{2} – m – 1 = 0

2m^{2} – 2m + m – 1 = 0

2m (m – 1) + 1 (m – 1) = 0

(m – 1) (2m + 1) = 0

m = 1, m = – \(\frac{1}{2}\)

Question 16.

Prove that the roots of (x – a) (x – b) = h^{2} are always real. [AP – May 2013, May ’09]

Solution:

Given quadratic equation is

(x – a) (x – b) = h^{2}

⇒ x^{2} – ax – bx + ab – h^{2} = 0

⇒ x^{2} + (- a – b)x + ab – h^{2} = 0

Comparing this with ax^{2} + bx + c =0, we get

⇒ a = 1, b = – a – b, c = ab – h^{2}

Now,

b^{2} – 4ac = (- a – b)^{2} – 4 (1) (ab – h^{2})

= a^{2} + b^{2} + 2ab – 4ab + 4h^{2}

=(a^{2} + b^{2} – 2ab) + 4h^{2}

= (a – b)^{2} + (2h)^{2} > 0

Since b^{2} – 4ac > 0 then the roots of the given equation are always real.

Question 17.

Find the quadratic equation, the sum of whose roots is 1 and sum of the sqares of the roots is 13. [May ’07]

Solution:

Let α, β be the roots of required equation.

Given that, the sum of roots, α + β = 1

Sum of squares of the roots, α^{2} + β^{2} = 13

We know that, (α + β)^{2} = α^{2} + β^{2} + 2αβ

⇒ 1 = 13 + 2αβ

⇒ 2αβ = – 12

⇒ αβ = – 6

∴ The quadratic equation whose roots are α, β is x^{2} – (α + β)x + αβ = 0

⇒ x^{2} – x – 6 = 0

Question 18.

Solve the equation 4^{x – 1} – 3 . 2^{x – 1} + 2 = 0. [March ’04]

Solution:

Given equation is 4^{x – 1} – 3 . 2^{x – 1} + 2 = 0

\(\frac{4^x}{4}-3 \cdot \frac{2^x}{2}+2\) = 0

\(\frac{\left(2^2\right)^x}{4}-3 \cdot \frac{2^x}{2}+2\) = 0

\(\frac{\left(2^x\right)^2}{4}-3 \cdot \frac{2^x}{2}+2\) = 0

Let 2^{x} = a

⇒ \(\frac{\mathrm{a}^2}{4}-\frac{3 \mathrm{a}}{2}\) + 2 = 0

⇒ a^{2} – 6a + 8 = 0

⇒ a^{2} – 4a – 2a + 8 = 0

⇒ (a – 4) (a – 2) = 0

⇒ a = 4 (or) 2.

Case – I:

lf a = 4

⇒ 2^{x} = 4

⇒ 2^{x} = 2^{2}

⇒ x = 2.

Case – 2:

If a = 2

⇒ 2^{x} = 2

⇒ x = 1

∴ The solution set of the given equation is {2, 1}.

Question 19.

Find the maximum or minimum of the expression 12x – x^{2} – 32 x varies over R.

Solution:

Given quadratic expression is – x + 12x – 32

Comparing this expression with ax + bx + c, we have a = – 1, b = 12, c = – 32

Here a = – 1 < 0.

Then – x^{2} + 12x – 32 has absolute maximum at x = \(\frac{-b}{2 a}=\frac{-12}{2(-1)}\) = 6

∴ The maximum value = \(\frac{4 a c-b^2}{4 a}\)

= \(\frac{4(-1)(-32)-144}{4(-1)}\)

= \(\frac{128-144}{-4}=\frac{-16}{-4}\) = 4.

Question 20.

If the quadratic equations ax^{2} + 2bx + c = 0 and ax^{2} + 2cx + b = 0, (b ≠ c) have a common root then show that a + 4b + 4c = 0.

Solution:

Given quadratic equations are ax^{2} + 2bx + c = 0

Comparing this equation with a_{1}x^{2} + b_{1}x + c_{1} = 0, we get

a_{1} = a, b_{1} = 2b, c_{1} = c

Now, a_{2}x + 2cx + b = 0

Comparing this with a_{2}x^{2} + b_{2}x + c_{2} = 0

we get a_{2} = a, b_{2} = 2c, c_{2} = b

The condition for two quadratic equations

a_{1}x^{2} + b_{1}x + c_{1} = 0 and a_{2}x^{2} + b_{2}x + c_{2} = 0 to have a common root is

(c_{1}a_{2} – c_{2}a_{1})^{2} = (a_{1}b_{2} – a_{2}b_{1}) (b_{1}c_{2} – b_{2}c_{1})

⇒ (ca – ba)^{2} = (a_{2}c – a_{2}b) (2bb – 2cc)

⇒ a^{2} (c – b)^{2} = 2a (c – b) 2 (b^{2} – c^{2})

⇒ a^{2} (c – b)^{2} = 4a (c – b) (b – c) (c – b)

⇒ a = – 4 (c + b) a = – 4c – 4b

a + 4b + 4c = 0.

Question 21.

If x^{2} – 6x + 5 = 0 and x^{2} – 12x + p = 0 have a common root, then find p. [TS – May 2015, Mar. 2017]

Solution:

Given quadratic equations are x^{2} – 6x + 5 = 0

Comparing this equation with

a_{1}x^{2} + b_{1}x + c_{1} = 0, we get

a_{1} = 1, b_{1} = – 6, c_{1} = 5

Now, x^{2} – 12x + p = 0,

Corn paring with a_{2}x^{2} + b_{2}x + c_{2} = 0

we get a_{2} = 1, b_{2} = – 12, c_{2} = p

The condition for two quadratic equations

a_{1}x^{2} + b_{1}x + c_{1} = 0 and a_{2}x2 + b_{2}x + c_{2} = 0 to have a common root is

(c_{1}a_{2} – c_{2}a_{1})^{2} = (a_{1}b_{2} – a_{2}b_{1}) (b_{1}c_{2} – b_{2}c_{1})

(5 (1) – p(1))^{2} = (1 (- 12) – 1 (- 6)) ((- 6)p – (- 12) 5)

⇒ (5 – p)^{2} = – 6 (- 6p + 60)

⇒ p^{2} + 25 – 10p = 36p – 360

⇒ p^{2} – 16p + 385 = 0

⇒ p^{2} – 35p – 11p + 385 = 0

⇒ p (p – 35) – 11 (p – 35) = 0

⇒ (p – 11) (p – 35) = 0

⇒ p = 11 (or) p = 35

∴ p = 11 (or) 35.

Question 22.

For what values of x, the expression x^{2} – 5x + 6 is positive?

Solution:

The given quadratic expression is x^{2} – 5x + 6

⇒ x^{2} – 3x – 2x + 6 = x (x – 3) – 2 (x – 3) = (x – 2)(x – 3)

i) If 2 < x < 3 then the expression x^{2} – 5x + 6 is ‘- ve’.

ii) If x < 2 (or) x> 3 then the expression x^{2} – 5x + 6 is ‘+ ve’.

iii) If x = 2 (or) x = 3 then the expression x^{2} – 5x + 6 = 0

Question 23.

For what values of x, the expression x^{2} – 5x – 6 is negative? [March ‘95]

Solution:

Given quadratic expression is x^{2} – 5x – 6

= x^{2} – 6x + x – 6

= x (x – 6) + 1 (x – 6)

= (x + 1)(x – 6)

i) If – 1 < x < 6 then the expression x^{2} – 5x – 6 is ‘-ve’.

ii) If x < – 1 or x > 6 then the expression x^{2} – 5x – 6 is ‘+ ve’.

iii) If x = – 1 (or) x = 6 then the expression x^{2} – 5x – 6 = 0.

Question 24.

Discuss the signs of the expression x^{2} – 5x + 4 for x ∈ R. [May ’95].

Solution:

Given quadratic expression is x^{2} – 5x + 4

= x^{2} – 4x – x + 4

= x (x – 4) + 1 (x – 4)

= (x – 1) (x – 4)

i) If 1 < x < 4 then the expression x^{2} – 5x + 4 is ’-ve’.

ii) If x < 1 (or) x > 4 then the expression x^{2} – 5x + 4 is ’+ve’.

iii) If x = 1 (or) x = 4 then the expression x^{2} – 5x + 4 = 0.

Question 25.

Find the maximum or minimum value of the expression 12x – x^{2} – 32. [May ’06]

Solution:

Given quadratic expression is – x^{2} + 12x – 32 = 0

Comparing this expression with ax^{2} + bx + c, we have a = – 1, b = 12, c = – 32

Here, a = – 1 < 0.

Then – x^{2} + 12x – 32 has absolute maximum at x = \(\frac{-b}{2 a}=\frac{-12}{2(-1)}\) = 6

The maximum value = \(\frac{4 a c-b^2}{4 a}\)

= \(\frac{4(-1)(-32)-144}{4(-1)}=\frac{128-144}{-4}\)

= \(\frac{-16}{-4}\) = 4.

Question 26.

Find the maximum or minimum value of the expression 2x – 7 – 5x^{2}. [May ’10. March ’14, ’12].

Solution:

Given quadratic expression is – 5x^{2} + 2x – 7

Comparing this expression with ax^{2} + bx + c,

we have a = – 5, b = 2, c = – 7

Here, a = – 5 < 0, – 5x^{2} + 2x – 7 has absolute maximum at

x = \(\frac{-b}{2 a}=\frac{-2}{2(-5)}=\frac{1}{5}\)

∴ Maximum value is \(\frac{4 a c-b^2}{4 a}=\frac{4(-5)(-7)-4}{4(-5)}\)

= \(\frac{140-4}{-20}=\frac{136}{-20}=\frac{-34}{5}\)

Question 27.

Find the maximum or minimum value of the expression 3x^{2} + 2x + 11. [May ’92]

Solution:

Given quadratic expression is 3x^{2} + 2x + 11

Comparing this expression with ax^{2} + bx + c, we have a = 3, b = 2, c = 11

Here, a = 3 >0, 3x^{2} + 2x + 11 has absolute minimum at

x = \(\frac{-b}{2 a}=\frac{-2}{2(3)}=\frac{-1}{3}\)

∴ The minimum value = \(\frac{4 a c-b^2}{4 a}=\frac{4(3)(11)-4}{4(3)}\)

= \(\frac{132-4}{12}=\frac{128}{12}=\frac{32}{3}\).

Question 28.

Find the maximum or minimum value of the expression x^{2} – x + 7 as x varies over R. [May ’14]

Solution:

Given quadratic expression is x^{2} – x + 7

Comparing this expression with ax^{2} + bx + c, we have a = 1, b = – 1, c = 7

Here, a = 1 > 0, x^{2} – x + 7 has absolute minimum at x = \(\frac{-b}{2 a}=\frac{-(-1)}{2 \cdot 1}=\frac{1}{2}\)

∴ The minimum value = \(\frac{4 a c-b^2}{4 a}=\frac{4(1)(7)-(-1)^2}{4(1)}\)

= \(\frac{28-1}{4}=\frac{27}{4}\)

Question 29.

Find the roots of the equation 6√5 x^{2} – 9x – 3√5 = 0.

Solution:

\(\frac{\sqrt{5}}{2}, \frac{-1}{\sqrt{5}}\)

Question 30.

Form a quadratic equation whose roots are 2√3 – 5 and – 2√3 – 5.

Solution:

x^{2} + 10x + 13 = 0.

Question 31.

Form a quadratic equation whose roots are \(\frac{\mathbf{m}}{\mathbf{n}}, \frac{-\mathbf{n}}{\mathbf{m}}\), (m ≠ 0, n ≠ 0)

Solution:

mnx^{2} – (m^{2} – n^{2})x – mn = 0

Question 32.

If α, β are the roots of ax^{2} + bx + c = 0 then find α^{3} + β^{3}.

Solution:

\(\frac{3 a b c-b^3}{a^3}\)

Question 33.

For what values of m, x^{2} + (m + 3) x + (m + 6) = 0 will have equal roots?

Solution:

m = 3, – 5.

Question 34.

For what values of m, z^{2} – 2 (1 + 3m) x + 7 (3 + 2m) = 0 will have equal roots?

Solution:

m = 2, \(\frac{-10}{9}\).

Question 35.

For what vaIueof m, (2m+ 1)x^{2} + 2(m + 3)x + (m + 5) = 0 will have equal rooti?

Solution:

m = \(\frac{-5 \pm \sqrt{41}}{2}\)

Question 36.

Find the quadratic equation, the sum of whose roots is 7 and sum of the squares of the roots is 25.

Solution:

x^{2} – 7x + 12 = 0

Question 37.

So1ve the equation 3^{1 + x} + 3^{1 – x} = 10.

Solution:

{- 1, 1}

Question 38.

Solve 7^{1 + x} + 7^{1 – x} = 50 for real x.

Solution:

{- 1, 1}

Question 39.

If x^{2} – 6x + 5 = 0 and x^{2} – 3ax + 35 = 0 have a common root, then find a.

Solution:

4 (or) 2.

Question 40.

For what values of x, the expression x^{2} – 5x + 14 is positive?

Solution:

∀ x ∈ R

Question 41.

For what values of x, the expression 3x^{2} + 4x + 4 is positive?

Solution:

∀ x ∈ R

Question 42.

For what values of x, the expression x^{2} – 7x + 10 is negative?

Solution:

2 < x < 5

Question 43.

For what values of x, the expression 15 + 4x – 3x^{2} is negative? [AP – Mar. 2015]

Solution:

x < – \(\frac{5}{3}\) (or) x > 3.

Question 44.

Find the changes in the sign of 4x – 5x^{2} + 2 for x ∈ R.

Solution:

Positive for \(\frac{2-\sqrt{14}}{5}<x<\frac{2+\sqrt{14}}{5}\)

negative for x < \(\frac{2-\sqrt{14}}{5}\) (or) x > \(\frac{2+\sqrt{14}}{5}\)