MCQ
Choose the correct answer from the given four options in the following questions:The number of polynomials having zeroes as -2 and 5 is:
- A1.
- B2.
- C3.
- ✓More than 3.
✓
Answer
Correct option: D.
More than 3.
Let $p(x) = ax^2 + bx + c$ be the required polynimial whose zeroes are -2 and 5.$\therefore\ \text{Sum of zeroes}=\frac{-\text{b}}{\text{a}}$
$\Rightarrow\ \frac{-\text{b}}{\text{c}}=-2+5=\frac{3}{1}=\frac{-(-3)}{1}\ .....(\text{i})$
and Procudt of zeroes $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\ \frac{\text{c}}{\text{a}}=-2\times5=\frac{-10}{1}\ .....(\text{ii})$
From Eqs. (i) and (ii)
a = 1, b = -3 and c = -10
$\therefore p(x)=a x^2+b x+c=1 \cdot x^2-3 x-10$
$=x^2-3 x-10$
But we know that. if we multiply/divide any polynomial by any arbitraru constant. Then, the zeroes of polynomial never change.
$\therefore$ $p(x) = kx^2 - 3kx - 10k$ [where, k is a real number]
$\therefore\ \text{p}(\text{x})=\frac{\text{x}^2}{\text{k}}-\frac{3}{\text{k}}\text{x}-\frac{10}{\text{k}},$ [where, k is a nonzero real number].
$\Rightarrow\ \frac{-\text{b}}{\text{c}}=-2+5=\frac{3}{1}=\frac{-(-3)}{1}\ .....(\text{i})$
and Procudt of zeroes $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\ \frac{\text{c}}{\text{a}}=-2\times5=\frac{-10}{1}\ .....(\text{ii})$
From Eqs. (i) and (ii)
a = 1, b = -3 and c = -10
$\therefore p(x)=a x^2+b x+c=1 \cdot x^2-3 x-10$
$=x^2-3 x-10$
But we know that. if we multiply/divide any polynomial by any arbitraru constant. Then, the zeroes of polynomial never change.
$\therefore$ $p(x) = kx^2 - 3kx - 10k$ [where, k is a real number]
$\therefore\ \text{p}(\text{x})=\frac{\text{x}^2}{\text{k}}-\frac{3}{\text{k}}\text{x}-\frac{10}{\text{k}},$ [where, k is a nonzero real number].
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