Vardaan Learning Institute
Answer Key: Polynomials
FORMULA
Relationship between Zeroes and Coefficients:
For a quadratic polynomial $ax^2 + bx + c$:
Sum of zeroes $(\alpha + \beta) = -\frac{b}{a}$
Product of zeroes $(\alpha\beta) = \frac{c}{a}$
SECTION A: OBJECTIVE TYPE (Answers)
1. Number of zeroes for 3 intersections:
Ans: (d) 3
Reason: The number of points where graph cuts X axis = Number of Zeroes.
2. Value of $\alpha + \beta$:
Ans: (a) 5
Sum = -(-5)/1 = 5.
3. Zeroes -3 and 4:
Ans: (c) $x^2 - x - 12$
Sum=1, Prod=-12. $x^2 - (1)x + (-12)$.
4. Finding k if 2 is zero:
Ans: (b) -10
$4 + 6 + k = 0 \Rightarrow k = -10$.
5. Shape of graph:
Ans: (c) Parabola
6. Zeroes 2 and -3. find a, b:
Ans: (d) a = 0, b = -6
7. Sum of zeroes is 3, find k:
Ans: (a) 6
Sum = k/2. k/2 = 3. k = 6.
8. Value of $1/\alpha + 1/\beta$:
Ans: (b) -1
9. Polynomials with zeroes -2 and 5:
Ans: (d) More than 3
Reason: Any polynomial $k(x+2)(x-5)$ works for any non-zero k. There are infinite such polynomials.
10. Zeroes of $x^2 + 99x + 127$:
Ans: (b) both negative
Reason: Sum = -99 (negative), Product = 127 (positive). For product to be positive, both signs must be
same. Must be negative for Sum to be negative.
SECTION B: SHORT ANSWER (Solutions)
11. Zeroes of $6x^2 - 3 - 7x$:
Zeroes: $-1/3$ and $3/2$. Verification done in standard process.
12. Quadratic poly with sum 1/4, prod -1:
Ans: $4x^2 - x - 4$.
13. Zeroes double of $2x^2 - 5x - 3$:
Old zeroes: $-1/2, 3$. New: $-1, 6$.
New Poly: $x^2 - 5x - 6$.
So $p = -5, q = -6$.
14. Find k if -3 is zero:
k = 4/3.
SECTION C: LONG ANSWER (Solutions)
15. $\alpha - \beta = 1$:
$k = 6$. Derived using $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$.
16. Zeroes $1/\alpha, 1/\beta$:
New Sum = 4/5. New Prod = 2/5.
Ans: $5x^2 - 4x + 2$.
17. $(\alpha- \beta)^2 = 144$:
Ans: $p = \pm 18$.
18. Find a, b (Sum 4, Prod 3):
Ans: $a = 5, b = 3.5$.
19. Evaluate $\alpha^2 + \beta^2$:
Ans: $b^2 - 2c$.
20. New poly with zeroes $\frac{\alpha^2}{\beta}, \frac{\beta^2}{\alpha}$:
Given $x^2 - 4x + 1$: $\alpha+\beta=4, \alpha\beta=1$.
Sum of new zeroes $= \frac{\alpha^3+\beta^3}{\alpha\beta} = \frac{(\alpha+\beta)^3 -
3\alpha\beta(\alpha+\beta)}{\alpha\beta}$.
$= \frac{64 - 3(1)(4)}{1} = 52$.
Product $= \alpha\beta = 1$.
Ans: $x^2 - 52x + 1$.
SECTION D: LONG ANSWER (Solutions)
21. Value of Exp for $3s^2 - 6s + 4$:
Sum $\alpha+\beta = 2$, Prod $\alpha\beta = 4/3$.
Expression: $\left(\frac{\alpha^2+\beta^2}{\alpha\beta}\right) +
2\left(\frac{\alpha+\beta}{\alpha\beta}\right) + 3\alpha\beta$.
$\alpha^2+\beta^2 = (2)^2 - 2(4/3) = 4 - 8/3 = 4/3$.
Term 1: $\frac{4/3}{4/3} = 1$.
Term 2: $2 \times \frac{2}{4/3} = 2 \times \frac{6}{4} = 3$.
Term 3: $3 \times (4/3) = 4$.
Total $= 1 + 3 + 4 = 8$.
Ans: 8
22. Evaluate for $ax^2 + bx + c$:
(i) $\alpha^2 - \beta^2$:
$= (\alpha+\beta)(\alpha-\beta) = -\frac{b}{a} \times \frac{\pm\sqrt{b^2-4ac}}{a} = \mp
\frac{b\sqrt{b^2-4ac}}{a^2}$.
(ii) $\alpha^3 + \beta^3$:
$= (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = (-\frac{b}{a})^3 - 3(\frac{c}{a})(-\frac{b}{a})$
$= -\frac{b^3}{a^3} + \frac{3bc}{a^2} = \frac{3abc - b^3}{a^3}$.
(iii) $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$:
$= \frac{\alpha^3+\beta^3}{\alpha\beta} = \left( \frac{3abc - b^3}{a^3} \right) \div \left( \frac{c}{a}
\right)$
$= \frac{3abc - b^3}{a^3} \times \frac{a}{c} = \frac{3abc - b^3}{a^2c}$.
SECTION E: CASE STUDY (Solutions)
23. Case Study: Bridge $P(x) = -x^2 + 2x + 8$
(i) Downwards (coeff of $x^2$ is negative).
(ii) Zeroes: 4, -2.
(iii) $P(0) = 8$.
(iv) Yes, $P(4) = 0$.