2 Marks Questions

Question 12 Marks

Very-Short-Answer Question:
If 3 is a zero of the polynomial $2x^2 + x + k$, find the value of k.

Answer

Since 3 is a zero of $f(x) = 2x^2 + x + k$, we have
$f(3) = 0$
$\Rightarrow 2(3)^2 + 3 + k = 0$
$\Rightarrow 18 + 3 + k = 0$
$\Rightarrow 21 + k = 0$
$\Rightarrow k = -21$

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Question 22 Marks

Very-Short-Answer Question:
State division algorithm for polynomials.

Answer

If f(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can always find polynomials q(x) and r(x) such that f(x) = q(x)g(x) + r(x),
where r(x) = 0 or degree r(x) < degree g(x).

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Question 32 Marks

Very-Short-Answer Question:
If $\alpha$ and $\beta$ are the zeroes of a polynomial $2x^2 + 7x + 5$, write the value of $\alpha+\beta+\alpha\beta.$

Answer

Let $\alpha,\ \beta$ are the zeros of $2x^2 + 7x + 5$
Then we have
$\alpha+\beta=-\frac{7}2{}$
$\alpha\beta=\frac{5}{2}$
Hence,
$\alpha+\beta+\alpha\beta=(\alpha+\beta)+\alpha\beta$
$=-\frac{7}{2}+\frac{5}2{}$
$=\frac{-7+5}{2}$
$=\frac{-2}{2}$
$=-1$

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Question 42 Marks

Very-Short-Answer Question:
If the sum of the zeros of the quadratic polynomial $kx^2 - 3x + 5$ is 1, write the value of k.

Answer

Let $f(x) = kx^2 - 3x + 5$
Sum of zeros $=\frac{-(\text{coefficient of }\text{x}^2)}{\text{coefficient of x}}$
$=\frac{-(-3)}{\text{k}}=\frac{3}{\text{k}}$
Given, sum of zeros = 1
$\therefore\frac{3}{\text{k}}=1$
$\Rightarrow\text{k}=3$

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Question 52 Marks

Very-Short-Answer Question:
Find the zeros of the polynomial $x^2 + x - p(p + 1).$

Answer

We have,
$f(x) = x^2 + x - p(p + 1)$
$= x^2 + (p + 1)x - px - p(p + 1)$
$= x[x + (p + 1)] - p[x + (p + 1)]$
$= (x - p)[x + (p + 1)]$
$\therefore$ $f(x) = 0$
$\Rightarrow (x - p)[x + (p + 1)] = 0$
$\Rightarrow x - p = 0 or x + (p + 1) = 0$
$\Rightarrow x = p or x = -(p + 1)$
so, the zeros of f(x) are p and -(p + 1).

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Question 62 Marks

Very-Short-Answer Question:
If -2 is a zero of the polynomial $3x^2 + 4x + 2k$ then find the value of k.

Answer

Since -2 is a zero of $f(x) = 3x^2 + 4x + 2K$, we have
$f(2) = 0$
$\Rightarrow 3(-2)^2 - 4(-2) + 2k = 0$
$\Rightarrow 12 - 8 + 2k = 0$
$\Rightarrow 4 + 2k = 0$
$\Rightarrow 2k = -4$
$\Rightarrow k = -2$

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Question 72 Marks

If one zero of the polynomial $\left(a^2+9\right) x^2+13 x+6 a$ is the reciprocal of the other, find the value of $a$.

Answer

$\left(a^2+9\right) x^2+13 x+6 a=0$
Here, $A=\left(a^2+9\right), B=13$ and $C=6 a$
Let $\alpha$ and $\frac{1}{\alpha}$ be the two zeros.
Then, product of the zeros $=\frac{ C }{ A }$
$\Rightarrow\alpha,\ \frac{1}{\alpha}=\frac{\text{6a}}{\text{a}^2+9}$
$\Rightarrow1=\frac{\text{6a}}{\text{a}^2+9}$
$\Rightarrow a^2 + 9 = 6a$
$\Rightarrow a^2 - 6a + 9 = 0$
$\Rightarrow a^2 - 2 \times a \times 3 + 3^2 = 0$
$\Rightarrow (a - 3)^2= 0$
$\Rightarrow a - 3 = 0$
$\Rightarrow a = 3$

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Question 82 Marks

Very-Short-Answer Question:
If one zero of the quadratic polynomial $k x^2+3 x+k$ is 2 then find the value of $k$.

Answer

Since 2 is a zero of $f(x)=k x^2+3 x+k$, we have
$f(2) = 0$
$\Rightarrow k(2)^2 + 3(2) + k = 0$
$\Rightarrow 4k + 6 + k = 0$
$\Rightarrow 5k + 6 = 0$
$\Rightarrow 5k = -6$
$\Rightarrow\text{k}=-\frac{6}{5}$

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Question 92 Marks

Very-Short-Answer Question:
If 1 is $a$ zero of the polynomial $a x^2-3(a-1) x-1$, then find the value of $a$.

Answer

Since 1 is a zero of $f(x)=a x^2-3(a-1) x-1$, we have
$f(1) = 0$
$\Rightarrow a(1)^2 - 3(a - 1)1 - 1 = 0$
$\Rightarrow a - 3a + 3 - 1 = 0$
$\Rightarrow -2a + 2 = 0$
$\Rightarrow 2a = 2$
$\Rightarrow a = 1$

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Question 102 Marks

Very-Short-Answer Question:
Write the zeros of the polynomial $x^2 - x - 6$

Answer

We have, $f(x) = x^2 - x - 6 = x^2- 3x + 2x - 6 = x(x - 3) + 2(x - 3) = (x + 2)(x - 3)$
$\therefore$ $f(x) = 0$
$\Rightarrow (x + 2)(x - 3) = 0$
$\Rightarrow x + 2 = 0 or x - 3 = 0$
$\Rightarrow x = -2 or x = 3$
So, the zeros of f(x) are -2 and 3

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Question 112 Marks

Very-Short-Answer Question:
If one zero of the polynomial $x^2- 4x + 1$ is $2+\sqrt3.$ Write the other zero.

Answer

Let the other zero of the polynomial $(x^2- 4x + 1)$ be $\alpha.$
Then, sum of roots $=\frac{-(-4)}{1}$
$\therefore2+\sqrt3+\alpha=4$
$\Rightarrow\alpha=2-\sqrt3$
Thus, the other zero is $\big(2-\sqrt3\big).$

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