# Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Quadratic Expressions Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 1.
Form quadratic equation whose root 7 ± 2$$\sqrt{5}$$ (Mar. ’11, ’05)
Solution:
α + β = 7 + 2$$\sqrt{5}$$ + 7 – 2$$\sqrt{5}$$ = 14
αβ = (7 + 2$$\sqrt{5}$$) (7 – 2$$\sqrt{5}$$) = 49 – 20 = 29
The required equation is
x2 – (α + β)x + αβ = 0
x2 – 14x + 29 = 0

Question 2.
Form quadratic equation whose root -3 ± 5i. (Mar. ’07)
Solution:
α + β = -3 + 5i – 3 – 51 = -6
αβ = (-3 + 5i)(-3 – 5i)
= 9 + 25 = 34
The required equation is
x2 – (α + β)x + αβ = 0
x2 + 6x + 34 = 0

Question 3.
For what values of x, 15 + 4x – 3x2 expressions are negative? (AP Mar. ’15)
Solution:
The roots of 15 + 4x – 3x2 = 0 are
$$\frac{-4 \pm \sqrt{16+180}}{-6}$$ i.e., $$\frac{-5}{3}$$, 3
∴ The expression 15 + 4x – 3x2 is negative if
x < $$\frac{-5}{3}$$ or x > 31 ∵ a = -3 < 0

Question 4.
If α, β are the roots of the equation ax2 + bx + c = 0, find the value $$\frac{1}{\alpha^{2}}$$ + $$\frac{1}{\beta^{2}}$$ expressions in terms of a, b, c. (AP & TS Mar. ‘16, 08)
Solution:

Question 5.
$$\frac{p-q}{p+q}$$, $$\frac{-p+q}{p-q}$$, (p ≠ ±q) (Mar. ’06)
Solution:
α + β = $$\frac{p-q}{p+q}$$ – $$\frac{p+q}{p-q}$$

Question 6.
Find the values of m for which the following equations have equal roots?
i) x2 – 15 – m(2x – 8) = 0. (AP Mar. ’17) (TS Mar. ’15 13)
Solution:
Given equation is x2 – 15 – m(2x – 8) = 0
x2 – 2mx + 8m – 15 = 0
a = 1, b = -2m, c = 8m – 15
b2 – 4ac = (-2m)2 – 4(1) (8m – 15)
= 4m2 – 32m + 60
= 4(m2 – 8m + 15)
= 4(m – 3)(m – 5)
Hint: If the equation ax2 + bx + c = 0 has equal roots then its discriminant is zero.
∵ The roots are equal b2 – 4ac = 0
⇒ 4(m – 3) (m – 5) = 0
⇒ m – 3 = 0 or m – 5 = 0
∴ m = 3 or 5

Question 7.
(m + 1)x2 + 2(m + 3)x + (m + 8) = 0.
Solution:
Given equation is
(m + 1)x2 + 2(m + 3)x + (m + 8) = 0
a = m + 1, b = 2(m + 3), c = m + 8
b2 – 4ac = (2(m + 3)]2 – 4(m + 1) (m + 8)]
= 4(m2 + 6m + 9) – 4(m2 + 8m + m + 8)
= 4m2 + 24m + 36 – 4m2 – 36 m – 32
= -12m + 4
= -4(3m – 1)
∵ The roots are equal ⇒ b2 – 4ac = 0
⇒ -4(3m – 1) = 0
⇒ 3m – 1 = 0
⇒ 3m = 1
∴ m = $$\frac{1}{3}$$

Question 8.
If x is real, prove that $$\frac{x}{x^{2}-5 x+9}$$ lies between 1 and $$\frac{-1}{11}$$. (Mar. ‘14, 13, ‘08, ‘02; May 11, ‘07)
Solution:
Let y = $$\frac{x}{x^{2}-5 x+9}$$ ⇒ yx2 – 5yx + 9y = x
⇒ yx2 + (-5y – 1)x + 9y = 0
x ∈ R ⇒ (-5y – 1)2 – 4y(9y) ≥ 0
⇒ 25y2 + 1 + 10y – 36y2 ≥ 0
⇒ -11y2 + 10y + 1 ≥ 0 —— (1)
⇒ -11y2 + 10y + 1 = 0 ⇒ -11y2 + 11y – y + 1 = 0
⇒ 11y(-y + 1) + 1(-y + 1) = 0
⇒ (-y + 1)(11y + 1) = 0 ⇒ y = 1, $$\frac{-1}{11}$$
-11y2 + 10y + 1 ≥ 0
∴ y2 coeff is be, but the exp is ≥ 0 from (1)
⇒ $$\frac{-1}{11}$$ ≤ y ≤ 1 ⇒ y lies between 1 and $$\frac{-1}{11}$$

Question 9.
Theorem : The roots of ax2 + bx + c = 0 are

(Mar. ’02)
Proof:
Given quadratic equation is ax2 + bx + c = 0
⇒ 4a(ax2 + bx + c) = 0
⇒ 4a2x2 + 4abx + 4ac = 0
⇒ (2ax)2 + 2(2ax) (b) + b2 – b2 + 4ac = 0
⇒ (2ax + b)2 = b2 – 4ac

Question 10.
Find the maximum value of the function $$\frac{x^{2}+14 x+9}{x^{2}+2 x+3}$$ over R.
Solution:
Let y = $$\frac{x^{2}+14 x+9}{x^{2}+2 x+3}$$
⇒ yx2 + 2yx + 3y = x2 + 14x + 9
⇒ (y – 1)x2 + 2(y – 7)x + 3y – 9 = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 7)]2 – 4(y – 1) (3y – 9) ≥ 0
⇒ 4[(y2 – 14y + 49) [(3y2 – 12y + 9)] ≥ 0
⇒ -2y2 – 2y + 40 ≥ 0
⇒ y2 + y – 20 ≤ 0
⇒ (y + 5) (y -4) ≤ 0
⇒ -5 ≤ y ≤ 4
⇒ y ∈ [-5, 4]
⇒ Maximum value of y = 4
∴ Maximum value of the function
$$\frac{x^{2}+14 x+9}{x^{2}+2 x+3}$$ over R is 4.

Question 11.
If x2 – 6x + 5 = 0 and x2 – 12x + p = 0 have a common root, then find p. (TS Mar. ’17)
Solution:
Given x2 – 6x + 5 = 0, x2 – 12x + p = 0 have a common root.
If α is the common root then
α2 -6α + 5 = 0, α2 – 12α + p = 0
α2 – 6α + 5 = 0 ⇒ (α – 1) (α – 5) = 0
⇒ α = 1 or 5
If α = 1 then α2 – 12α + p = 0
⇒ 1 – 12 + p = 0 ⇒ p = 11
If α = 5 then α2 – 12α + p = 0
⇒ 25 – 60 + p = 0 ⇒ p = 35
∴ p = 11 or 35

Question 12.
If x1, x2 are the roots of the quadratic equation ax2 + bx + c = 0 and c ≠ 0, find the value of (ax1 + b)-2 + (ax2 + b)-2 in terms of a, b, c. (TS Mar. ’17)
Solution:
x1, x2 are the roots of the equation
ax2 + bx + c = 0

Question 13.
Prove that $$\frac{1}{3 x+1}$$ + $$\frac{1}{x+1}$$ – $$\frac{1}{(3 x+1)(x+1)}$$ does not lie between 1 and 4, if x is real. (AP & TS Mar. ’16, AP Mar. 15, ’11) (AP Mar. ‘17)
Solution:

⇒ 3yx2 + 4yx + y = 4x + 1
⇒ 3yx2 + (4y – 4) x + (y – 1) = 0
x ∈ R ⇒ (4y – 4)2 – 4(3y)(y – 1) ≥ 0
⇒ 16y2 + 16 – 32y – 12y2 + 12y ≥ 0
⇒ 4y2 – 20y + 16 ≥ 0
4y2 – 20y + 16 = 0
⇒ y2 – 5y + 4 = 0
⇒ (y – 1)(y – 4) = 0
⇒ y = 1, 4
⇒ 4y2 – 20y + 16 ≥ 0.
⇒ y ≤ 1 or y ≥ 4
⇒ y does not lie between 1 and 4
Since y2 coeff is the and exp ≥ 0.

Question 14.
Solve the following equations: (T.S Mar. ‘15)
2x4 + x3 – 11x2 + x + 2 = 0
Solution:
Dividing by x2

Substituting in (1)
2(a2 – 2) + a – 11 = 0
⇒ 2a2 – 4 + a – 11 = 0
⇒ 2a2 + a – 15 = 0
⇒ (a + 3) (2a – 5) = 0
⇒ a + 3 = 0 or 2a – 5 = 0
a = -3 or 2a = 5 a = $$\frac{5}{2}$$

Case(i) : a = -3
x + $$\frac{1}{x}$$ = -3
x2 + 1 = -3
x2 + 3x + 1 = 0
x = $$\frac{-3 \pm \sqrt{9-4}}{2}$$ = $$\frac{-3 \pm \sqrt{5}}{2}$$

Case (ii) : a = $$\frac{5}{2}$$
x + $$\frac{1}{x}$$ = $$\frac{5}{2}$$
⇒ $$\frac{x^{2}+1}{x}$$ = $$\frac{5}{2}$$
⇒ 2x2 + 2 = 5x
⇒ 2x2 – 5x + 2 = 0
⇒ (2x – 1) (x – 2) = 0
⇒ 2x – 1 = 0 or x – 2 = 0
⇒ x = $$\frac{1}{2}$$, 2
∴ The roots are $$\frac{1}{2}$$, 2, $$\frac{-3 \pm \sqrt{5}}{2}$$

Question 15.
For what values of x, the following expressions are positive? (May ’11)
i) x2 – 5x + 6
Solution:
x2 – 5x + 6 = (x – 2) (x – 3)
Roots of x2 – 5x + 6 = 0 are 2, 3 which are real.
The expression x2 – 5x + 6 is positive if x < 2 or x > 3, ∴ a = 1 > 0.

ii) 3x2 + 4x + 4
Solution:
Here a = 3, b = 4, c = 4,
Δ = b2 – 4ac
= 16 – 48 = -32 < 0
∴ 3x2 + 4x + 4 is positive ∀ x ∈ R,
∵ a = 3 >0 and Δ < 0
Hint: ax2 + bx + c and ‘a’ have same sign ∀ x ∈ R, if Δ < 0

iii) 4x – 5x2 + 2
Solution:
Roots of 4x – 5x2 + 2 = 0 are

iv) x2 – 5x + 14
Solution:
Here a = 1, b = -5, c = 14,
Δ = b2 – 4ac
= 25 – 56 = -31 < 0
∴ Δ < 0 ∵ a = 1 > 0 and Δ < 0
⇒ x2 – 5x + 14 is positive ∀ x ∈ R.

Question 16.
Determine the range of the $$\frac{x^{2}+x+1}{x^{2}-x+1}$$ expressions. (Mar ’04)
Solution:
Let y = $$\frac{x^{2}+x+1}{x^{2}-x+1}$$
⇒ x2y – xy + y = x2 + x + 1
⇒ x2y – xy + y – x2 – x – 1 = 0
⇒ x2(y – 1) – x(y + 1) + (y – 1) = 0
x is real ⇒ b2 – 4ac ≥ 0
⇒ (y + 1)2 – 4(y – 1)2 ≥ 0
⇒ (y + 1)2 – (2y – 2)2 ≥ 0
⇒ (y + 1 + 2y – 2)(y + 1 – 2y + 2) ≥ 0
⇒ (3y – 1)(-y + 3) ≥ 0
⇒ -(3y – 1) (y – 3) ≥ 0
a = coeff. of y2 = -3 < 0., But
The expression ≥ 0
⇒ y lies between $$\frac{1}{3}$$ and 3
∴ The range of $$\frac{x^{2}+x+1}{x^{2}-x+1}$$ is $$\left[\frac{1}{3}, 3\right]$$

Question 17.
Theorem : Let α, β be the real roots of ax2 + bx + c = 0 and α < β. Then
i) x ∈ R, α < x < β ⇒ ax2 + bx + c and ’a’ have opposite signs.
ii) x ∈ R, x < α or x > β ⇒ ax2 + bx + c and ‘a’ have the same sign. (Apr. ’96, ’93)
Proof:
α, β are the roots of ax2 + bx + c = 0
⇒ ax2 + bx + c = a(x – α) (x – β)
⇒ $$\frac{a x^{2}+b x+c}{a}$$ = (x – α) (x – β)

i) Suppose x ∈ R, α < x < β
α < x < β ⇒ x – α > 0, x – β < 0
⇒ (x – α) (x – β) < 0
⇒ ax2 + bx + c, a have opposite signs.

ii) Suppose x ∈ R, x < α
x < α < 13
⇒ x – α < 0, x – β < 0 ⇒ (x – α) (x – β) > 0
⇒ $$\frac{a x^{2}+b x+c}{a}$$ > 0
⇒ ax2 + bx + c and a have the same sign.
Suppose x ∈ R, x > β
x > β > α
⇒ x – α > 0, x – β > 0
⇒ (x – α)(x – β) > 0
⇒ $$\frac{a x^{2}+b x+c}{a}$$ > 0
⇒ ax2 + bx + c and a have the same sign,
∴ x ∈ R, x < α or x > β
⇒ ax2 + bx + c and a have the same sign.

Question 18.
Theorem: Let f(x) = ax2 + bx + c be a quadratic function. (Apr. ‘01)
i) If a > 0 then f(x) has minimum value at x = $$\frac{-b}{2 a}$$ and the minimum value = $$\frac{4 a c-b^{2}}{4 a}$$
ii) If a < 0 then f(x) has maximum value at x = $$\frac{-b}{2 a}$$ and the maximum value $$\frac{4 a c-b^{2}}{4 a}$$
Proof:
ax2 + bx + c =

≤ $$\frac{4 a c-b^{2}}{4 a}$$,
when a < 0
∴ If a < 0, then $$\frac{4 a c-b^{2}}{4 a}$$ is the maximum fór f when x = $$\frac{-b}{2 a}$$
Second Proof : f(x) = ax2 + bx + c
⇒ f'(x) = 2ax + b
⇒ f”(x) = 2a
If f'(x) = 0, then 2ax + b = 0 and hence x = $$-\frac{b}{2 a}$$
If a > 0, then f”(x) > 0 and hence ‘f’ has minimum value at x = $$-\frac{b}{2 a}$$.
Minimum of

If a < 0,
then f”(x) < 0 and hence ‘f’ has maximum value at x = $$-\frac{b}{2 a}$$ maximum of

Question 19.
Theorem : The roots of ax2 + bx + c = 0 are $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$.
Proof:
Given quadratic equation is ax2 + bx + c = 0
⇒ 4a (ax2 + bx + c) = 0
⇒ 4a2x2 + 4abx + 4ac = 0
⇒ (2ax)2 + 2(2ax) (b) + b2 – b2 + 4ac = 0
⇒ (2ax + b)2 = b2 – 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 ax2 + bx + c = 0 are
$$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$

Question 20.
Theorem : Let a, b, c ∈ R and a ≠ 0. Then the roots of ax2 + bx + c = 0 are non-real complex numbers if and only if ax2 + bx + c and a have the same sign for all x ∈ R.
Proof :
The condition for the equation
ax2 + bx + c = o to have non-real complex roots is b2 – 4ac < 0, i.e., 4ac – b2 > 0.

Hence 4ac – b2 > 0, so that b2 – 4ac < 0. Thus b2 – 4ac < 0 if and only if ax2 + bx + c and a have the same sign for all real x.

Question 21.
Theorem: If the roots of ax2 + bx + c = 0 are real and equal to α = $$\frac{-b}{2 a}$$; then for α ≠ x ∈ R, ax2 + bx + c and ‘a’ have the same sign.
Proof:
The roots are equal ⇒ b2 – 4ac = 0
⇒ 4ac – b2 = 0

∴ For α ≠ x ∈ R, ax2 + bx + c and a have the same sign.

Question 22.
Theorem : Let α, β be the real roots of ax2 + bx + c = 0 and α < β. Then
i) x ∈ R, α < x < 13 ⇒ ax2 + bx + c and ’a’ have opposite signs.
ii) x ∈ R, x < α or x > β ⇒ ax2 + bx + c and ‘a’ have the same sign.
Proof.
α, β are the roots of ax2 + bx + c = 0
⇒ ax2 + bx + c = a(x – α)(x – β)
⇒ $$\frac{a x^{2}+b x+c}{a}$$ = (x – α)(x – β)

i) Suppose x ∈ R, α < x < β
α 0, x – β < 0
⇒ (x – α)(x – β) < 0 ⇒ $$\frac{a x^{2}+b x+c}{a}$$ < 0
⇒ ax2 + bx + c, a have opposite signs

ii) Suppose x ∈ R, x < α
x < α < β ⇒ x – α < 0, x – β < 0 ⇒ (x – α) (x – β) > 0 ⇒ $$\frac{a x^{2}+b x+c}{a}$$ > 0
⇒ ax2 + bx + c and a have the same sign.
Suppose x ∈ R, x > 3
x > β > α ⇒ x – α > 0, x – β > 0
⇒ (x – α) (x – β) > 0
⇒ $$\frac{a x^{2}+b x+c}{a}$$ > 0
⇒ ax2 + bx + c and a have the same sign,
x ∈ R, x < α or x > β
⇒ ax2 + bx + c and a have the same sign

Question 23.
Theorem : Let f(x) = ax2 + bx+ c be a quadratic function.
i) If a > 0 then f(x) has minimum value at x = $$\frac{-\mathbf{b}}{2 a}$$ and the minimum value = $$\frac{4 a c-b^{2}}{4 a}$$
ii) If a < 0 then f(x) has maximum value at x = $$\frac{-\mathbf{b}}{2 a}$$ and the maximum value = $$\frac{4 a c-b^{2}}{4 a}$$ (Apr. ’01)
Proof.
ax2 + bx+ c =

Second Proof: f(x) = ax2 + bx + c
⇒ f'(x) = 2ax + b ⇒ f”(x) = 2a
If f'(x) = 0, then 2ax + b = 0 and hence x = $$-\frac{b}{2 a}$$
If a > 0, then f”(x) > 0 and hence ‘f’ has minimum value at x = $$-\frac{b}{2 a}$$
Minimum of

If a < 0 then f”(x) < 0 and hence ‘f’ has maximum value at x = –$$\frac{b}{2 a}$$
maximum of

Question 24.
Theorem : A necessary and sufficient condition for the quadratic equations
a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 to have a common root is (c1a2 – c2a1)2 = (a1b2 – a2b1) (b1c2 – b2c1).
Proof:
Necessity
Let α be a common root of the given equations.
Then a1α2 + b1α + c1 = 0 ——(1)
a2α2 + b2α + c2 = 0 —— (2)
On multiplying euqation (1) by a2, equation (2) by a1 and then subtracting the latter from the former, we get

On multiplying euqation (1) by b2, equation (2) by b1 and then subtracting the latter from the former, we get
α2 (a1b2 – a2b1) = b1c2 – b2c1 —— (4)
On squaring both sides of equation (3) and using (4) we obtain

Therefore the given equations have the same roots.

Case (ii): a1b2 – a2b1 ≠ 0

Similarly we can prove that a2α2 + b2α + c2 = 0
Thus α is a common root of the given equations.

Question 25.
Find the roots of the equation 3x2 + 2x – 5 = 0.
Solution:
The roots of the quadratic equation
ax2 + bx + c = 0 are $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$
Here a = 3, b = 2 and c = -5.
Therefore the roots of the given equation are

Hence 1 and –[/latex]\frac{5}{3}[/latex] are the roots of the given equation.
Another method
We can also obtain these roots in thè following way.
3x2 + 2x – 5 = 3x2 + 5x – 3x – 5
= x(3x + 5) -1 (3x + 5)
= (x – 1) (3x + 5)
= 3(x – 1) $$\left(x+\frac{5}{3}\right)$$
Since 1 and –[/latex]\frac{5}{3}[/latex] are the zeros of 3x2 + 2x – 5, they are the roots of 3x2 + 2x – 5 = 0.

Question 26.
Find the roots of the equation 4x2 – 4x + 17 = 3x2 – 10x – 17.
Solution:
Given equation can be rewritten as x2 + 6x + 34 = 0. The roots of the quadratic equation
ax2 + bx + c = 0 are $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$
Here a = 1, b = 6 and c = 34
Therefore the roots of the given equation are

Hence the roots of the given equation are -3 + 5i and -3 – 5i

Question 27.
Find the roots of the equation
$$\sqrt{3}$$x2 + 10x – 8$$\sqrt{3}$$ = 0.
Solution:
Here a = $$\sqrt{3}$$, b = 10, c = -8$$\sqrt{3}$$

Question 28.
Find the nature of the roots of 4x2 – 20x + 25 = 0
Solution:
Here a =4, b = -20, c = 25
Hence Δ = b2 – 4ac
= (-20)2 – 4(4) (25) = 0
Since Δ is zero and a, b, c are real, the roots of the given equation are real and equal.

Question 29.
Find the nature of the roots of 3x2 + 7x + 2 = 0
Solution:
Here a = 3, b = 7, c = 2
Hence Δ = b2 – 4ac
= 49 – 4(3) (2)
= 49 – 24
= 25 = (5)2 > 0
Since a, b, c are rational numbers and Δ = 52 is a perfect square and positive, the roots of the given equation are rational and unequal.

Question 30.
For what values of m, the equation x2 – 2(1 + 3m)x + 7(3 + 2m) = 0 will have equal roots?
Solution:
The given equation will have equal roots iff its discriminant is 0.
Here Δ = [(-2(1 + 3m)]2 – 4(1) [7 + (3 + 2m)]
= 4(1 + 9m2 + 6m) – 28(3 + 28m)
= 9m2 – 8m – 20
= (m – 2)(9m + 10)
Hence Δ = 0 ⇔ m = 2, $$\frac{-10}{9} .$$

Question 31.
If α and β are the roots of ax2 + bx + c = 0, find the value of α2 + β2 and α3 + β3 in terms of a, b, c.
Solution:
α, β are the roots of ax2 + bx + c = 0

Question 32.
Show that the roots of the equation x2 – 2px + p2 – q2 + 2qr – r2 = 0 are rational, given that p, q, r are rational.
Solution:
Here a = 1, b = -2p, c = p2 – q2 + 2qr – r2
∆ = b2 – 4ac
= (-2p)2 – 4(1) (p2 – q2 + 2qr – r2)
= 4p2 – 4p2 + 4q2 + 8qr + 4r2
= 4(q – r)2.
∵ p, q, r are rational, the coefficient of the given equation are rational and is a square of a rational number 2(q – r).
∴ The roots of the given equation are rational.

Question 33.
Form a quadratic equation whose roots are 2$$\sqrt{3}$$ – 5 and -2$$\sqrt{3}$$ – 5.
Solution:
Let α = 2$$\sqrt{3}$$ – 5 and β = -2$$\sqrt{3}$$ – 5
Then α + β = (2$$\sqrt{3}$$ – 5) + (-2$$\sqrt{3}$$ – 5) = -10
and αβ = (2$$\sqrt{3}$$ – 5) + (-2$$\sqrt{5}$$ – 5)
= (-5)2 – (-2$$\sqrt{3}$$)2
= 25 – 4 × 3
= 13
x2 – (α + β)x + αβ = 0
⇒ x2 – (-10)x + 13 = 0
⇒ x2 + 10x + 13 = 0

Question 34.
Find the quadratic equation, the sum of whose roots is 1 and the sum of squares of the roots is 13.
Solution:
Let α, β be the roots of a required quadratic equation.
Then α + β = 1 and α2 + β2 = 13
Now αβ = $$\frac{1}{2}$$[(α + β)2 – (α2 + β2)]
= $$\frac{1}{2}$$[(1)2 – (13)]
Therefore x2 – (α + β)x + αβ = 0
⇒ x2 – (1)x + (-6) = 0
⇒ x2 – x – 6 = 0

Question 35.
Let α and β be the roots of the quadratic equation ax2 + bx + c = 0, c ≠ 0, then form the quadratic equation whose roots are $$\frac{1-\alpha}{\alpha}$$ and $$\frac{1-\beta}{\beta}$$.
Solution:
From the given equation

Question 36.
Solve x2/3 + x1/3 – 2 = 0
Solution:
(x1/3)2 + (x1/3) – 2 = 0
Let x1/3 = a, a2 + a – 2 = 0
⇒ (a +2)(a – 1) = 0 ⇒ a = 1, -2
Now, x1/3 = 1 ⇒ x = (1)3 = 1
x1/3 = -2 ⇒ x = (-2)3 = -8
∴ roots are 1, -8.

Question 37.
Solve 71 + x + 71 – x = 50 for real x.
Solution:
The given equation can be written as,
7.7x + $$\frac{7}{7^{x}}$$ – 50 = 0
Let 7x = a
7a + $$\frac{7}{a}$$ – 50 = 0
⇒ 7a2 + 7 – 50a = 0
⇒ 7a2 – 49a – a + 7 = 0
⇒ 7a(a – 7) – 1(a – 7) = 0
⇒ (a – 7)(7a – 1) = 0
∴ a = 7, $$\frac{1}{7}$$
Now, if a = 7 then 7x = 71 ⇒ x = 1
a = $$\frac{1}{7}$$ then 7x = $$\frac{1}{7}$$ = 7-1
x = -1
∴ x = -1, 1

Question 38.
Solve

Solution:
On taking $$\sqrt{\frac{x}{1-x}}$$ = a
a + $$\frac{1}{a}$$ = $$\frac{13}{6}$$
⇒ $$\frac{a^{2}+1}{a}$$ = $$\frac{13}{6}$$
⇒ $$\frac{a^{2}+1}{a}$$ = $$\frac{13}{6}$$
⇒ 6a2 + 6 = 13a
⇒ 6a2 – 13a + 6 = 0
⇒ 6a2 – 9a – 4a + 6 = 0
⇒ 3a(2a – 3) – 2(2a – 3) = 0
⇒ (2a – 3)(3a – 2) = 0
a = $$\frac{3}{2}$$, a = $$\frac{2}{3}$$
If a = $$\frac{3}{2}$$ then $$\sqrt{\frac{x}{1-x}}$$ = $$\frac{3}{2}$$
⇒ $$\frac{x}{1-x}$$ = $$\frac{9}{4}$$
⇒ 4x = 9 – 9x
⇒ 13x = 9
⇒ x = $$\frac{4}{13}$$
∴ x = $$\frac{9}{13}$$, $$\frac{4}{13}$$

Question 39.
Find all number which exceed their square root by 12.
Solution:
Let the required number be ‘x’
given, x = $$\sqrt{x}$$ + 12
⇒ x – 12= $$\sqrt{x}$$
Squaring on both sides
(x – 12)2 = ($$\sqrt{x}$$)2
⇒ x2 – 24x + 144 = x
⇒ x2 – 25x + 144 = 0
⇒ (x – 9)(x – 16) = 0
⇒ x = 9, 16
But x = 9 does not satisfy the given condition
x = 16
∴ The required numbér = 16.

Question 40.
If x2 + 4ax + 3 = 0 and 2x2 + 3ax – 9 = 0 have a common root, then find the values
of a and the common root.
Solution:

Substitute in the above equation
[(3) (2) – (-9) (1)]2 = [(1) (3a) – (2) (4a)] [(4a) (-9) – (3a) (3)]
(15)2 = (-5a) (-45a)
⇒ 225 = 225a2 ⇒ a2 = 1 ⇒ a = ±1

Case (i) : If a = 1, the given equations become x2 + 4x + 3 = 0 and 2x2 + 3x – 9 = 0
⇒ (x + 1)(x + 3) = 0 and (2x + 3)(x + 3) = 0
⇒ x = 3, -1 and -3, $$\frac{3}{2}$$
In this case, the common root of the given equations is -3

Case (ii) : If a = -1, the given equations become x2 – 4x + 3 = 0 and 2x2 – 3x – 9 = 0
⇒ (x – 1) (x – 3) = 0 and (2x + 3) (x – 3) = 0
⇒ x = 1, 3 and x = 3, –$$\frac{3}{2}$$
In this case, the common root of the given equation is 3.

Question 41.
Prove that there is unique pair of consecutive positive odd integers such that the sum of their squares is. 290 and find it.
Solution:
The difference of two consecutive odd integers is 2.
So, we have to prove that there is a unique positive odd integer ‘x’ such that
x2 + (x + 2)2 = 290 —– (1)
x2 + (x + 2)2 = 290
⇒ x2 + x2 + 4x + 4 = 290
⇒ 2x2 + 4x – 286 = 0
⇒ x2 + 2x – 143 = 0
⇒ (x + 13) (x – 11) = 0
⇒ x = -13, 11
Hence 11 is the only positive odd integer satisfying equation (1).
∴ 11, 13 is the unique pair of integers which satisfies the given condition.

Question 42.
The cost of a piece of cable wire is Rs. 35/-, If the length of the piece of wire is 4 meters more and each meter costs, Rs. 1/— less, the cost would remain unchanged. What is the length of the wire?
Solution:
Let the length of the piece of the wire be ‘l’ meters and the cost of each meter be Rs. x.
given lx. = 35 —— (1)
and (l + 4) (x – 1) = 35
⇒ lx – l + 4x – 4 = 35
⇒ 35 – l + 4x – 4 = 35
⇒ 4x = l + 4
⇒ x = $$\frac{l+4}{4}$$
Substitute x in (1), we get l$\frac{l+4}{4}$ = 35
⇒ l2 + 4l = 140
⇒ l2 + 4l – 140 = 0
⇒ l2 + 14l – 10l – 140 = 0
⇒ l(l + 14) – 10(l + 14) = 0
⇒ (l – 10)(l + 14) = 0
⇒ l = -14, 10
Since length cannot be negative, l = 10
∴ The length of the piece of wire is 10 meters.

Question 43.
One fourth of a herd of goats was seen in the forest. Twice the square root of the number in the herd had gone up the hill and the remaining 15 goats were on the bank of the river. Find the total number of goats.
Solution:
Let the number of goats in the herd be ‘x’.
The number of goats seen in the forest = $$\frac{x}{4}$$
The number of goats gone upto the hill = $$2 \sqrt{x}$$
The number of goats on the bank of a river = 15
∴ $$\frac{x}{4}$$ + 2$$\sqrt{x}$$ + 15 = x
⇒ x + 8$$\sqrt{x}$$ + 60 = 4x
⇒ 3x – 8$$\sqrt{x}$$ – 60 = 0
Put $$\sqrt{x}$$ = y
⇒ 3y2 – 8y – 60 = 0
⇒ 3y2 – 18y + 10y – 60 = 0
⇒ 3y(y – 6) + 10(y – 6) = 0
⇒ (3y + 10)(y – 6) = 0
⇒ y = 6, –$$\frac{10}{3}$$
Since y cannot be negative, y = 6
⇒ $$\sqrt{x}$$ = 36
∴ x = 36
∴ Total number of goats = 36

Question 44.
In a cricket match Anil took one wicket less than twice the number of wickets taken by Ravi. If the product of the
number of wickets taken by them is 15, find the number of wickets taken by each of them.
Solution:
Let the number of wickets taken by Ravi be ‘x’ and the number of wickets taken by Anil is 2x – 1.
Given x(2x – 1) = 15
⇒ 2x2 – x – 15 = 0
⇒ 2x2 – 6x + 5x – 15 = 0
⇒ 2x(x – 3) + 5(x – 15) = 0
⇒ (x – 3)(2x + 5) = 0
⇒ x = 3, –$$\frac{5}{2}$$
Since the number of wickets be integer,
x = 3 and 2x – 1 = 2(3) – 1 = 5
∴ The number of wickets taken by Anil and Ravi are 5 and 3 respectively.

Question 45.
Some points on a plane are marked and they are connected pairwise by line segments. If the total number of line
segments formed is 10, find the number of marked points on the plane.
Solution:
Let the number of points marked on the plane be ’x’.
The total number of line segments actually formed is

Given $$\frac{x(x-1)}{2}$$ = 10
⇒ x2 – x = 20
⇒ x2 – x – 20 = 0
⇒ (x – 5)(x + 4) = 0
Since x cannot be negatives x = 5
∴ The number of points marked on the plane is 5.

Question 46.
Suppose that the quadratic equations ax2 + bx + c = 0 and bx2 + cx + a = 0 have a common root. Then show that a3 + b3 + c3 = 3abc.
Solution:
Let α be the common root of the equations

Question 47.
For what values of x1 the expression x2 – 5x – 14 is positive?
Solution:
Since x2 – 5x – 14 = (x + 2) (x – 7), the roots of the equation -2 and 7
Here the coefficient of x2 is positive.
Hence x2 – 5x – 14 is positive when x < -2 or x > 7.

Question 48.
For what values ofx. the expression -6x2 + 2x3 is negative?
Solution:
-6x2 + 2x – 3 = 0 can be written as 6x2 – 2x + 3 = 0

∴ The roots of -6x2 + 2x – 3 = 0 are non-real complex numbers.
Hence coefficient of x2 is -6, which is negative.
∴ -6x2 + 2x – 3 < 0 for all x ∈ R.

Question 49.
Find the value of x at which the following expressions have maximum or minimum.

i) x2 + 5x + 6
Solution:
Here a = 1 > 0, the expression has minimum value at x = $$\frac{-b}{2 a}$$
= $$\frac{-5}{2(1)}$$
= $$\frac{-5}{2}$$

ii) 2x – x2 + 7
Solution:
Here a = -1 < 0, the expression has maximum value at x
= $$\frac{-b}{2 a}$$
= $$\frac{-2}{2 x-1}$$ = 1

Question 50.
Find the maximum or minimum value of the quadratic expression
(i) 2x – 7 – 5x2
(ii) 3x2 + 2x + 17 (Mar. ’14)
Solution:
i) Compare the given equation with ax2 + bx + c, we get
a = -5, b = 2, c = -7
Here a = -5 < 0, the given expression has maximum value at x = $$\frac{-b}{2 a}$$ = $$\frac{-2}{2(-5)}$$
= $$\frac{1}{5}$$
and maximum value = $$\frac{4 a c-b^{2}}{4 a}$$
= $$\frac{4(-5) \cdot(-7)-(?)^{2}}{4(-5)}$$
= $$\frac{140-4}{-20}$$ = $$\frac{-34}{5}$$

(ii) Compare 3x2 + 2x + 11 with
ax2 + bx + c, we get a = 3, b = 2, c = 11
∵ a = 3 > 0, the given expression has minimum value at

Question 51.
Find the changes in the sign of 4x – 5x2 + 2 for x ∈ R and find the extreme value.
Solution:
The roots of 4x – 5x2 + 2 = 0
⇒ 5x2 – 4x – 2 = 0
roots are = $$\frac{2 \pm \sqrt{14}}{5}$$
∴ $$\frac{2-\sqrt{14}}{5}$$ < x < $$\frac{2+\sqrt{14}}{5}$$ the sign of 4x – 5x2 + 2 is positive.
x < $$\frac{2-\sqrt{14}}{5}$$ or x > $$\frac{2+\sqrt{14}}{5}$$, the sign of 4x – 5x2 + 2 is negative.
Since a = -5 < 0, then maximum value = $$\frac{4 a c-b^{2}}{4 a}$$
= $$\frac{4(-5)(2)-(4)^{2}}{4(-5)}$$
= $$\frac{-56}{-20}$$ = $$\frac{14}{5}$$
Hence extreme value = $$\frac{14}{5}$$

Question 52.
Find the solution set of x2 + x – 12 ≤ 0 by both algebraic and graphical methods.
Solution:
Algebraic Method : We have x2 + x – 12 = (x + 4) (x – 3).
Hence -4 and 3 are the roots of the equation x2 + x – 12 = 0.
Since the coefficient of x2 in the quadratic expression x2 + x – 12 = 0 is positive, x2 + x – 12 is negative if -4 < x < 3 and positive if either x < -4 or x > 3.
Hence x2 + x – 12 ≤ 0 ⇔ -4 ≤ x ≤ 3.
Therefore the solution set is
{x ∈ R : -4 ≤ x ≤ 3}.
Graphical Method: Let y = f(x) = x2 + x – 12.
The values of y at some selected values of s are given in the following table:

The graph of the function y = f(x) is drawn using the above tabulated values. This is shown in Fig.

Therefore the graph of y = f(x) we observe that
y = x2 – x – 12 < 0 if f -4 ≤ x ≤ 3.
Hence the solution set is {x ∈ R : -4 ≤ x ≤ 3}.

Question 53.
Find the set of values of x for which the inequalities x2 – 3x – 10 < 0, 10x – x2 – 16
> 0 hold simultaneously.
Solution:
∵ x2 – 3x – 10 < 0
⇒ x2 – 5x + 2x – 10 < 0
⇒ x(x – 5) + 2(x – 5) < 0
⇒ (x – 5)(x + 2) < 0
∴ x2 coeff. is the and expression is <0
⇒ x ∈ (-2, 5) ——- (1)
Now 10x – x2 – 16 > 0
⇒ x2 – 10x + 16 < 0
⇒ (x – 2) (x – 8) < 0
∴ x2 coeff is the ana expression is < 0
⇒ x ∈ (2, 8) ——- (2)
Hence x2 – 3x – 10 < 0 and 10x – x2 – 16 > 0
⇔ x ∈ (-2, 5) ∩ (2, 8)
∴ The solution set is {x/x ∈ R : 2 < x < 5}

Question 54.
Solve the inequation $$\sqrt{x+2}$$ > $$\sqrt{8-x^{2}}$$.
Solution:
When a, b ∈ R and a ≥ 0, b ≥ 0 then $$\sqrt{a}$$ > $$\sqrt{b}$$ ⇔ a > b ≥ 0
∴ $$\sqrt{x+2}$$ = $$\sqrt{8-x^{2}}$$
⇔ x + 2 > 8 – x2 ≥ 0 and x > -2, |x| < 2$$\sqrt{2}$$
We have (x + 2) > 8 – x2
⇔ x2 + x – 6 > 0
⇔ (x + 3)(x – 2) > 0
⇔ x ∈ (-∞, -3)(2, ∞) .
We have 8 – x2 ≥ 0
⇔ x2 ≤ 8 ⇔ |x| < 2 $$\sqrt{2}$$ ⇔ x ∈ [-2 $$\sqrt{2}$$, 2 $$\sqrt{2}$$] Also x + 2 > 0 ⇔ x > -2
Hence x + 2 > 8 – x2 ≥ 0
⇔ x ∈ ((-∞, -3) ∪ (2, ∞)) ∩ [-2 $$\sqrt{2}$$, 2 $$\sqrt{2}$$] and x > -2
⇔ x ∈ (2, 2 $$\sqrt{2}$$)
∴ The solution set is [x ∈ R : -2 < x ≤ 2 $$\sqrt{2}$$]

Question 55.
Solve the equation
$$\sqrt{(x-3)(2-x)}$$ < $$\sqrt{4 x^{2}+12 x+11}$$.
Solution:
The given inequation is equivalent to the following two inequalities.
(x – 3)(2 – x) ≥ 0 and
(x – 3)(2 – x) < 4x2 + 12x + 11
(x – 3)(2 – x) ≥ 0 ⇔ (x – 2)(x – 3) ≤ 0
⇔ 2 ≤ x ≤ 3
-x2 + 5x – 6 < 4x2 + 12x + 11
⇔ 5x2 + 7x + 17 > 0
The discriminant = b2 – 4ac = 49 – 340 < 0 ⇒ x ∈ R : a = 5 > 0 and expressive is the
Hence the solution set of the given inequation is {x ∈ R : 2 ≤ x ≤ 3}

Question 56.
Solve the inequation

Solution:

⇔ either 6 + x – x2 = 0, 2x + 5 ≠ 0 and x + 4 ≠ 0 or 6 + x – x2 > 0
and $$\frac{1}{2 x+5}$$ ≥ $$\frac{1}{x+4}$$
We have 6 + x – x2 = -(x2 – x – 6)
= -(x + 2) (x – 3)
Hence 6 + x + x2 = 0 ⇔ x = -2 or x = 3
2x + 5 = 0 ⇔ x = –$$\frac{5}{2}$$
x + 4 = 0 ⇔ x = -4
∴ 6 + x – 2x2 > 0 ⇔ -2 < x < 3 x ∈ (-2, 3), 2x + 5 > -4 + 5 = 1 > 0 and x + 4 > -2 + 4 = 2 > 0
x + 4 > -2 + 4 = 2 > 0
x ∈ (-2, 3), $$\frac{1}{2 x+5}$$ ≥ $$\frac{1}{x+4}$$
⇔ 2x + 5 ≤ x + 4,
2x + 5 ≤ x + 4 ⇔ x ≤ 1
Hence 6 + x – x2 > 0 and $$\frac{1}{2 x+5}$$ ≥ $$\frac{1}{x+4}$$
⇔ -2 < x ≤ -1
∴ The solution set is {-2, 3} ∪ (-2, -1)
= [-2, -1] ∪ {3}

Question 57.
Find the maximum value of the function
$$\frac{x^{2}+14 x+9}{x^{2}+2 x+3}$$ over R. (May ’05)
Solution:
Let y = $$\frac{x^{2}+14 x+9}{x^{2}+2 x+3}$$
⇒ yx2 + 2yx + 3y = x2 + 14x + 9
⇒ (y – 1)x2 + 2(y – 7)x + 3y – 9 = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 7)2] – 4(y – 1) (3y – 9) ≥ 0
⇒ 4[(y2 – 14y + 49) [(3y2 – 12y + 9)] ≥ 0
⇒ -2y2 – 2y + 40 ≥ 0
⇒ y2 + y – 20 ≤ 0
⇒ (y + 5) (y – 4) ≤ 0
⇒ -5 ≤ y ≤ 4
⇒ y ∈ [-5, 4]
∴ Maximum value of y = 4
∴ Maximum value of the function
$$\frac{x^{2}+14 x+9}{x^{2}+2 x+3}$$ over R is 4.

Question 58.
Show that none of the values of the function over $$\frac{x^{2}+34 x-71}{x^{2}+2 x-7}$$ over R lies between 5 and 9. (Mar. 2005)
Solution:
Let y = $$\frac{x^{2}+34 x-7}{x^{2}+2 x-7}$$
⇒ x2 + 2x – 7 = x2 + 34x – 71
⇒ (y – 1)x2 + 2(y – 17)x + (-7y + 71) = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 17)]2 – 4(y – 1)(-7y + 71)] ≥ 0
⇒ 4[(y2 – 34y + 289) – (-7y2 + 78y – 71)] ≥ 0
⇒ 8y2 – 112y + 360 ≥ 0
⇒ y2 – 14y + 45 ≥ 0
⇒ (y – 5) (y – 9) ≥ 0
⇒ y ≤ 5 or y ≥ 9
⇒ y does not lies between 5 and 9, since expression is ≥ 0 and y2 coeff is the
∴ None of the values of the function
$$\frac{x^{2}+34 x-7}{x^{2}+2 x-7}$$ over R lies between 5 and 9.

Question 59.
Solve the inequation
$$\sqrt{x^{2}-3 x-10}$$ > (8 – x)
Solution:
(i) $$\sqrt{x^{2}-3 x-10}$$ > (8 – x)
⇒ x2 – 3x – 10 ≥ 0 and (i) (8 – x < 0)
or (ii) x2 – 3x – 10 > (8 – x)2 and (8 – x ≥ 0)
Now x2 – 3x – 10 = (x – 5) (x + 2)
Hence x2 – 3x – 10 ≥ 0
⇒ x ∈ (-∞, -2] ∪ [5, ∞)
8 – x < 0 ⇒ x ∈ (8, ∞)
∴ x2 – 3x – 10 ≥ 0 and 8 – x < 0
⇔ x ∈ (-∞, -2] ∪ [5, ∞) and x ∈ (8, ∞)
⇔ x ∈ (8, ∞)

(ii) x2 – 3x – 10 > (8 – x)2
⇔ x2 – 3x – 10 > 64 + x2 – 16x
⇔ 13x > 74