Question

$\text{A quadratic polynomial whose zeroes}\text{are}\frac{3}{4}\text{and}\frac{1}{2}\text{is}$

• A. $\frac{8x^2-10x+3}{8}$
• B. $\frac{10x^2-8x+3}{4}$
• C. $\frac{8x^2+10x+3}{8}$
• D. $\frac{10x^2+8x+3}{4}$

Answer 1

The correct option is A $\frac{8x^2-10x+3}{8}$

$\text{Given:}\alpha =\frac{3}{4}\text{and}\beta =\frac{1}{2}$

$\therefore \alpha +\beta =\frac{3}{4}+\frac{1}{2}=\frac{3+2}{4}=\frac{5}{4}\text{and}\alpha \beta =\frac{3}{4}\times \frac{1}{2}=\frac{3}{8}$

$\text{We know that, if}\alpha ,\beta \text{are the zeros of}\text{the quadratic polynomial, the}\text{quadratic polynomial is given by}{x}^2-(\alpha +\beta )x+\alpha \beta \Rightarrow x^2-\frac{5}{4}x+\frac{3}{8}\Rightarrow \frac{8x^2-10x+3}{8}$