Solutions: Polynomials - Test 3

Q1. (b) 2/3

Sum $\alpha + \beta = -(-2)/1 = 2$. Product $\alpha\beta = 3k/1 = 3k$.
Given $\alpha + \beta = \alpha\beta \Rightarrow 2 = 3k \Rightarrow k = 2/3$.

Q2. (c) 3

Let zeroes be $\alpha$ and $1/\alpha$. Product = $\alpha \times (1/\alpha) = 1$.
Product = $c/a = k/3$. So, $k/3 = 1 \Rightarrow k = 3$.

Q3. (a) -1

$\alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$.
Sum = -5, Product = 5. Value = $-5/5 = -1$.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Sum = $2 + (-3) = -1$. Product = $2(-3) = -6$.
Poly = $k[x^2 - (-1)x + (-6)] = x^2 + x - 6$.

Q5. $4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3}$
$= 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = (4x - \sqrt{3})(\sqrt{3}x + 2)$.
Zeroes: $\frac{\sqrt{3}}{4}, -\frac{2}{\sqrt{3}}$. [1]
Sum = $\frac{3 - 8}{4\sqrt{3}} = \frac{-5}{4\sqrt{3}} = -\frac{b}{a}$. Verified. [1]

Q6. For $x^2 - x - 2$: $\alpha + \beta = 1, \alpha\beta = -2$.
New Sum $S = (2\alpha + 1) + (2\beta + 1) = 2(\alpha + \beta) + 2 = 2(1) + 2 = 4$.
New Product $P = (2\alpha + 1)(2\beta + 1) = 4\alpha\beta + 2(\alpha + \beta) + 1 = 4(-2) + 2(1) + 1 = -8 + 2 + 1 = -5$.
Poly: $x^2 - 4x - 5$. [2]

Q7. $\alpha + \beta = 8, \alpha\beta = k$.
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 64 - 2k$.
Given $64 - 2k = 40 \Rightarrow 2k = 24 \Rightarrow k = 12$. [2]

Q8. $\alpha + \beta = 5, \alpha\beta = k$. Given $\alpha - \beta = 1$.
Using $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$:
$(1)^2 = (5)^2 - 4k \Rightarrow 1 = 25 - 4k \Rightarrow 4k = 24 \Rightarrow k = 6$. [3]

Q9. Original zeroes $\alpha, \beta$. New zeroes $1/\alpha, 1/\beta$.
New Sum $S = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}$.
New Product $P = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c}$.
Poly: $k[x^2 - (-\frac{b}{c})x + \frac{a}{c}] = k[\frac{cx^2 + bx + a}{c}]$.
Required polynomial: $cx^2 + bx + a$. [3]

Q10. For $x^2 - 3x - 2$: $\alpha + \beta = 3, \alpha\beta = -2$.
Let new zeroes be $A = 2\alpha + 3\beta$ and $B = 3\alpha + 2\beta$.
Sum $S = A + B = 5\alpha + 5\beta = 5(\alpha + \beta) = 5(3) = 15$. [1]
Product $P = AB = (2\alpha + 3\beta)(3\alpha + 2\beta) = 6\alpha^2 + 4\alpha\beta + 9\alpha\beta + 6\beta^2$
$= 6(\alpha^2 + \beta^2) + 13\alpha\beta$. [1]
Now, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (3)^2 - 2(-2) = 9 + 4 = 13$. [1]
$P = 6(13) + 13(-2) = 78 - 26 = 52$. [1]
Polynomial: $x^2 - Sx + P \Rightarrow x^2 - 15x + 52$. [1]

(i) $p(x) = x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = (x-4)(x+2)$.
Zeroes are 4 and -2. [1]

(ii) Sum of zeroes = $4 + (-2) = 2$.
$-b/a = -(-2)/1 = 2$. Hence verified. [1]

(iii) New zeroes: 0 and 4.
Sum = 4, Product = 0.
Polynomial = $k(x^2 - 4x + 0) = x^2 - 4x$. [2]