Question

Find the sum and product of the zeroes of the following quadratic polynomials.

$48x^2-13x-1$

Answer 1

The standard form of any polynomial is $\left(\mathrm{coefficient}\mathrm{of}{x}^2\right)x^2+\left(\mathrm{coefficient}\mathrm{of}x\right)x+\mathrm{constant}\mathrm{term}$.

Here, coefficient of $x^2$ is $48$, coefficient of $x$ is $-13$, and constant term is $-1$.

Step 1: To find the sum of zeroes:

Sum of zeroes = $\frac{-\left(\mathrm{coefficient}\mathrm{of}x\right)}{\mathrm{coefficient}\mathrm{of}{x}^2}=\frac{-\left(-13\right)}{48}=\frac{13}{48}$

Step 2: To find the product of zeroes:

Product of zeroes = $\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2}=\frac{-1}{48}=-\frac{1}{48}$

Hence, the sum and product of zeroes are $\frac{13}{48}$ and $-\frac{1}{48}$respectively.