Students must practice this TS Intermediate Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(a) to find a better approach to solving the problems.
TS Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Exercise 3(a)
I.
Question 1.
Find the roots of the following equations.
i) x2 – 7x + 12 = 0
ii) – x2 + x + 2 = 0
iii) 2x2 + 3x + 2 = 0
iv) √3x2 + 10x – 8√3 = 0
v) 6√5x2 – 9x – 3√5 = 0
Solution:
x2 – 7x + 12 = 0
(x – 4) (x – 3) = 0
x = 4, 3.
ii) – x2 + x + 2 = 0
x2 – x – 2 = 0
(x – 2) (x + 1) = 0
x = 2, – 1
iii) 2x2 + 3x + 2 = 0
x = \(\frac{-3 \pm \sqrt{9-16}}{4}\)
x = \(\frac{-3 \pm \sqrt{7} i}{4}\)
iv) √3x2 + 10x – 8√3 = 0
v) 6√5x2 – 9x – 3√5 = 0
Question 2.
Form quadratic equations whose roots are
i) 2, 5
ii) \(\frac{\mathbf{m}}{\mathbf{n}}, \frac{-\mathbf{n}}{\mathbf{m}}\) (m ≠ 0, n ≠ 0)
iii) \(\frac{\mathbf{p}-\mathbf{q}}{\mathbf{p}+\mathbf{q}}, \frac{-(\mathbf{p}+\mathbf{q})}{\mathbf{p}-\mathbf{q}}\) (p ≠ ± q)
iv) 7 ± 2√5
v) – 3 ± 5i
Solution:
i) α = 2, β = 5 → roots then quadratic equation be
(x – α) (x – β) = 0
(x – 2) (x – 5) = 0
x2 – 7x + 10 = 0.
ii) α = \(\frac{\mathrm{m}}{\mathrm{n}}\), β = – \(\frac{\mathrm{n}}{\mathrm{m}}\)
(x – \(\frac{\mathrm{m}}{\mathrm{n}}\)) (x + \(\frac{\mathrm{n}}{\mathrm{m}}\)) = 0
x + x (\(\frac{\mathrm{n}}{\mathrm{m}}-\frac{\mathrm{m}}{\mathrm{n}}\)) – \(\frac{m}{n} \cdot \frac{n}{m}\) = 0
x2 + x \(\frac{\left(n^2-m^2\right)}{n m}\) – 1 = 0
mnx2 + x (n2 – m2) – nnm = 0
x2 – x(α + β) + αβ = 0.
iii) x2 – \(\left(\frac{p-q}{p+q}-\frac{p+q}{p-q}\right)\) x – \(\left(\frac{p+q}{p-q}\right)\left(\frac{p-q}{p+q}\right)\) = 0
x – \(\left(\frac{(p-q)^2-(p+q)^2}{p^2-q^2}\right)\)x – 1 = 0
x + \(\frac{4 p q}{p^2-q^2}\)x – 1 = 0
(p2 – q2)x2 + 4px – (p2 – q2) = 0.
iv) x2 – x(7 + 2√5 + 7 – 2√5)+ (7 + 2√5) (7 – 2√5) = 0
x2 – x(14) + (49 – 20) = 0
x2 – 14x + 29 = 0.
v) x2 – x (- 3 + 5i – 3 – 5i) + (- 3 + 5i) (- 3 – 5i) = 0
x2 – x (- 6) + (9 + 25) = 0
x2 + 6x + 34 = 0.
Question 3.
Find the nature of the roots of the following equations without finding the roots.
i) 2x2 – 8x + 3 = 0
ii) 9x2 – 30x + 25 = 0
iii) x2 – 12x + 32 = 0
iv) 2x22 – 7x + 10 = 0
Solution:
i) 2x2 – 8x + 3 = 0
b2 – 4ac = 64 – 24 > 0
roots are real and distinct.
ii) 9x2 – 30x + 25
b2 – 4ac = 900 – 4 × 9 × 25 = 0
Roots are real and equal.
iii) x2 – 12x + 32 = 0
(12)2 – 4 × 32 > 0
Roots are real and distinct.
iv) 2x2 – 7x 10 = 0
(- 7)2 – 4 × 2 × 10 < 0
Roots are imaginary.
Question 4.
If α, β are the roots of the equation ax2 + bx + c = 0, find the value of following expressions in terms of a, b, c.
i) \(\frac{1}{\alpha}+\frac{1}{\beta}\)
ii) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\)
iii) α4β7 + α7β4
iv) \(\left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2\), if c ≠ 0
v) \(\frac{\alpha^2+\beta^2}{\alpha^{-2}+\beta^{-2}}\), if c ≠ 0.
Solution:
i) α + β = \(\frac{-b}{a}\);
αβ = \(\frac{c}{a}\)
\(\frac{\alpha+\beta}{\alpha \beta}=\frac{\frac{-b}{a}}{\frac{c}{a}}=\frac{-b}{c}\)
ii) \(\frac{\alpha^2+\beta^2}{\alpha^2 \beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha^2 \beta^2}\)
= \(\frac{\frac{b^2}{a^2}-\frac{2 c}{a}}{\frac{c^2}{a^2}}\)
= \(\frac{b^2-2 c a}{c^2}\)
iii) α4β7 + α7β4
= α4β4 (β3 + α3)
= (αβ)4 [(α + β)3 – 3αβ (α + β)]
= \(\frac{c^4}{a^4}\left[\frac{-b^3}{a^3}+\frac{3 c}{a} \cdot \frac{b}{a}\right]\)
= \(\frac{c^4}{a^4}\left[\frac{3 a b c-b^3}{a^3}\right]\)
iv) \(\left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2\)
v) \(\frac{\alpha^2+\beta^2}{\alpha^{-2}+\beta^{-2}}\)
= α2β2
= \(\frac{c^2}{a^2}\).
Question 5.
Find the values of m for which the following equations have equal roots.
i) x2 – m(2x – 8) – 15 = 0
ii) (m + 1) x2 + 2 (m + 3) x + (m + 8) = 0
iii) x2 + (m + 3) x + (m + 6) = 0
iv) (3m + 1) x2 + 2 (m + 1) x + m = 0
v) (2m + 1) x2 + 2 (m + 3) x + m + 5 = 0
Solution:
i) x2 – 2xm + 8m – 15 = 0
b2 – 4ac = 0
(- 2m)2 – 4 (8m – 15) = 0
4m2 – 32m + 60 = 0
m2 – 8m + 15=0
(m – 5) (m – 3) = 0
m = 3, 5.
ii) (m + 1)x2 + 2(m + 3)x + (m + 8) = 0
b2 – 4ac = 0
4(m + 3)2 – 4(m + 8) (m + 1) = 0
(m + 3)2 (m2 + 9m + 8) = 0
6m + 9 – 9m – 8 = 0
– 3m + 1 = 0
m = \(\frac{1}{3}\).
iii) x2 + (m + 3) x + (m + 6) = 0
b2 – 4ac = 0
(m + 3)2 – 4 (m + 6) = 0
m2 e 6m +9-4m-24-0
m2 + 2m – 15 = 0
(m + 5) (m – 3) = 0
m = 3, – 5.
iv) (3m + 1)x2 + 2(m + 1)x + m = 0
b2 – 4ac = 0
4(m + 1)2 – 4m(3m + 1) = 0
m2 + 2m + 1 – 3m2 – m = 0
– 2m2 + m + 1 = 0
2m2 – m – 1 = 0
2m2 – 2m + m – 1 = 0
2m (m – 1) + (m – 1) = 0
(m – 1) (2m + 1)= 0
m = 1, m = – \(-\frac{1}{2}\)
v) (2m + 1)x2 + 2(m + 3)x + (m + 5) = 0.
b2 – 4ac = 0
4(m + 3)2 – 4(2m + 1) (m + 5) = 0
m2 + 6m + 9 – 2m2 – 11m – 5 = 0
– m2 – 5m + 4 = 0
m2 + 5m – 4 = 0
m = \(\frac{-5 \pm \sqrt{25+16}}{2}\)
m = \(\frac{-5}{2}\).
Question 6.
If α and β are the roots of equation x2 + px + q = 0, form a quadratic equation where roots are (α – β)2 and (α + β)2.
Solution:
α + β = \(\frac{-b}{a}\); αβ = \(\frac{c}{a}\)
x2 – [(α – β)2 + (α + β)2]x + [(α – β) (α + β)]2 = 0
x2 – [2(α2 + β2)]x + [α + β]2 [(α + β)2 – 4αβ] = o
x2 – 2[(α – β)2 – 2αβ]x + (α + β)2 [(α + β)2 – 4αβ] = o
x2 – 2 \(\left[\frac{b^2-2 a c}{a^2}\right]\) x + \(\frac{b^2}{a^2}\left[\frac{b^2-4 a c}{a^2}\right]\) = 0
Here b = p, c = q, a = 1
x2 – 2 (p2 – 2q) x + p2 (p2 – 4q) = 0
Question 7.
If x2 + bx + c = 0, x2 + cx + b = 0 (b ≠ c) have a common root, then show that b + c + 1 = 0.
Solution:
x2 + bx + c = 0
x2 + cx + b = 0
α2 is common root.
α2 + bα + c = 0
α2 + cα + b = 0
α (b – c) + c – b = 0
α (b – c) = b – c
α = 1
∴ 1 + b + c = 0.
Question 8.
Prove that roots of (x – a) (x – b) = h2 are always real.
Solution:
(x – a) (x – b) = h2
x2 – x (a + b) + ab – h2 = 0
Discriminant = (a + b)2 – 4 (ab – h2)
= (a + b)2 – 4ab + 4h2
= (a – b)2 + 4h2 > 0
∴ Roots are real.
Question 9.
Find the condition that one root of the quadratic equation ax2 + bx + c = 0 shall be n times the other, where n is positive integer.
Solution:
α + nα = – b/a
α . nα = \(\frac{c}{a}\)
α = \(\frac{-b}{a(n+1)}\)
\(\frac{n b^2}{a^2(n+1)^2}=\frac{c}{a}\)
nb2 = ac (n + 1)2
Question 10.
Find two consecutive potive even Integers, the sum of whose squares in 340.
Solution:
2n, 2n + 2
(2n)2 + (2n + 2)2 = 340
4n2 + 4n2 + 8n + 4 = 340
8n2 + 8n + 4 = 340
2n2 + 2n + 1 = 85
2n2 + 2n – 84 = 0
n2 + n – 42 = 0
(n + 7) (n – 6) = 0
n = 6
12, 14 are two numbers.
II.
Question 1.
If x1, x2 are the roots of equation ax2 + bx + c = 0 and c ≠ 0. Find the value of (ax1 + b)-2 + (ax2 + b)-2 in terms of a, b, c.
Solution:
ax12 + bx1 + c = 0
x1 (ax1 + b) + c = 0
(ax1 + b) = \(\frac{-\mathrm{c}}{\mathrm{x}_1}\)
Similarly (ax2 + b) = \(\frac{-\mathrm{c}}{\mathrm{x}_2}\)
∴ (ax1 + b)-2 + (ax2 + b)-2 = \(\)
= \(\frac{1}{c^2}\) [(x1 + x2)2 – 2x1x2]
= \(\frac{1}{c^2}\left[\frac{b^2-2 a c}{a^2}\right]\)
= \(\frac{b^2-2 a c}{a^2 c^2}\)
Question 2.
If α, β are the roots of equation ax2 + bx + c = 0, find a quadratic equation whose roots are α2 + β2 and α-2 + β-2.
Solution:
Solve the following equations:
Question 3.
2x4 + x3 – 11x2 + x + 2 = 0
Solution:
2x4 + x3 – 10x2 – x2 + x + 2 = 0
x2 (2x2 + x – 10) – (x2 – x – 2) = 0
x2 [(2x + 5) (x – 2) – [(x – 2) (x + 1)] = 0
(x – 2) [x2 (2x + 5) – x – 1] = 0
(x – 2) [2x3 + 5x2 – x – 1] = 0
(x – 2) [(2x – 1) (x2 + 3x + 1)] = 0
x = 2, \(\frac{1}{2}\); x2 + 3x + 1 = 0
x = \(\frac{-3 \pm \sqrt{9-4}}{2}\)
x = \(\frac{-3 \pm \sqrt{5}}{2}\).
Question 4.
31+x + 31-x = 10
Solution:
Let 3x = t
3. t + \(\frac{3}{t}\) = 10
3t2 + 3 – 10t = 0
3t2 – 10t + 3 = 0
3t2 – 9t – t + 3 = 0
3t (t – 3) – 1 (t – 3) = 0
(3t – 1) (t – 3) = 0
t = \(\frac{1}{3}\), t = 3
3x = 3-1, 3x = 31
x = – 1, 1.
Question 5.
4x – 1 – 3 . 2x – 1 + 2 = 0.
Solution:
\(\frac{4^x}{4}-\frac{3 \cdot 2^x}{2}\) + 2 = 0
4x – 6 . 2x + 8 = 0
2x = t
t2 – 6t + 8 = 0
(t – 4) (t – 2) = 0
2x = t
t2 – 6t + 8 = 0
(t – 4) (t – 2) = 0
2x = 22; 2x = 21
x = 1, 2.
Question 6.
\(\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}\) when x ≠ 0, x ≠ 3.
Solution:
\(\sqrt{\frac{x}{x-3}}\) = t
t + \(\frac{1}{t}=\frac{5}{2}\)
2t2 – 5t + 2 = 0
2t2 – 4t – t + 2 = 0
2t (t – 2) – 1 (t – 2) = 0
(2t – 1) (t – 2) = 0
t = \(\frac{1}{2}\); t = 2
\(\frac{x}{x-3}\) = 4
x = 4x – 12
12 = 3x
x = 4
(or)
\(\frac{x}{x-3}=\frac{1}{4}\)
4x = x – 3
3x = – 3
x = – 1.
Question 7.
\(\sqrt{\frac{3 x}{x+1}}+\sqrt{\frac{x+1}{3 x}}\) = 2 when x ≠ 0, x ≠ 1.
Solution:
Let \(\sqrt{\frac{3 x}{x+1}}\) = t
t + \(\frac{1}{t}\) = 2
t2 – 2t + 1 = 0
(t – 1)2 = 0
t = 1
\(\frac{3 x}{x+1}\) = 1
3x = x + 1
x = \(\frac{1}{2}\).
Question 8.
2 \(\left(x+\frac{1}{x}\right)^2\) – 7 \(\left(x+\frac{1}{x}\right)\) + 5 = 0, when x ≠ 0.
Solution:
x + \(\frac{1}{x}\) = t
2t2 – 7t + 5 = 0
2t2 – 5t – 2t + 5 = 0
2t (t – 1) – 5 (t – 1) = 0
t = \(\frac{5}{2}\), t = 1 but x + \(\frac{1}{x}\) ≥ 2 ∀ x ∈ R+
x + \(\frac{1}{x}\) = \(\frac{5}{2}\) only possible
2x2 – 5x + 2 = 0
(2x – 1) (x – 2) = 0
x = \(\frac{1}{2}\), 2
Now if x + \(\frac{1}{x}\) = 1
x2 – x + 1 = o
x = \(\frac{1 \pm \sqrt{1-4}}{2}\)
x = \(\frac{1 \pm \sqrt{3} i}{2}\).
Question 9.
\(\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)\) + 6 = 0 when x ≠ o.
Solution:
\(\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)\) + 6 = 0
x + \(\frac{1}{x}\) = t
t2 – 5t + 4 = 0
(t – 4) (t – 1) = 0
t = 4, 1
x + \(\frac{1}{x}\) = 4
x2 – 4x + 1 = 0
x = \(\frac{4 \pm \sqrt{16-4}}{2}\)
x = 2 ± √3
x + \(\frac{1}{x}\) = 1
x2 – x + 1 = 0
x = \(\frac{1 \pm \sqrt{1-4}}{2}\)
x = \(\frac{1 \pm \sqrt{3} i}{2}\).
Question 10.
Find a quadratic equation for which the sum of the roots is 7 and the sum of the squares of the roots is 25.
Solution:
α + β = 7, α2 + β2 = 25
(α + β)2 – 2αβ= 25
49 – 2αβ = 25
24 = 2αβ
αβ = 12
x2 – (7)x + 12 = 0.