Question

Find a quadratic polynomial with $\frac{1}{8}$ as the sum and 2 as product of its zeroes.

• A. $8x^2+x+16$
• B. $8x^2-x+16$
• C. $8x^2-x-16$
• D. $8x^2+x-16$

Answer 1

The correct option is B

$8x^2-x+16$

Let, $\alpha \&\beta$ be zeroes.
Given, $\alpha +\beta =\frac{1}{8};\alpha \beta =2$
Therefore required polynomial is $p\left(x\right)=k[x^2-(\alpha +\beta )x+\alpha \beta ]$ where k is a real number.

Now, consider $p\left(x\right)=k[x^2-\frac{1}{8}x+2]$

$p\left(x\right)=k\left[\frac{8x^2-x+16}{8}\right]$

$p\left(x\right)=k\times \frac{1}{8}[8x^2-x+16]$

Let's take k = 8

(Note: multiplying a polynomial with a non-zero constant doesn't change the zeroes of the polynomial).

$\Rightarrow p\left(x\right)=8x^2-x+16$

Therefore, $(8x^2-x+16)$ is a polynomial with $\frac{1}{8}$ as the sum and 2 as product of its zeroes.