Chapter 4.5, Problem 71SR

To determine

To find: the quadratic equation from given two roots in standard form.

Answer to Problem 71SR

The quadratic equation is ${\text{5x}}^{\text{2}}\text{+28x-12=0}$ .

Explanation of Solution

Given:

Two roots are:

  $-\frac{2}{5},6$ .

Concept used:

The equation for the quadratic equation whose roots $\text{are}$ given above is:

  ${\text{x}}^{\text{2}}\text{-sx+p=0}$ ………. $\left(\text{i}\right)$

To form a quadratic equation, let $\text{α}$ and $\text{β}$ be the two roots.

  $\text{s=sum of root=α+β}$ and $\text{p=product of the roots=α×β}$ .

Calculation:

Two roots according to the given as:

  $-\frac{2}{5},6$ .

The equation for the quadratic equation whose roots $\text{are}$ given above is:

  ${\text{x}}^{\text{2}}\text{-sx+p=0}$ ………. $\left(\text{i}\right)$

To form a quadratic equation, let $\text{α}$ and $\text{β}$ be the two roots.

  $\text{s=sum of root=α+β}$ and $\text{p=product of the roots=α×β}$ .

  $\text{s=sum of root=α+β=-}\frac{2}{5}+6=\frac{-2+30}{5}=\frac{28}{5}$ …… $\left(\text{ii}\right)$

  $\text{p=product of the roots=α×β=-}\frac{2}{5}\times 6=-\frac{12}{5}$ ……… $\left(\text{iii}\right)$

Putting the value of equation $\left(\text{ii}\right)$ and $\left(\text{iii}\right)$ in equation $\left(\text{i}\right)$ :

  ${\text{x}}^{\text{2}}\text{+}\frac{28}{5}\text{x-}\frac{12}{5}\text{=0}$ can also written as:

  $\frac{{\text{5x}}^{\text{2}}\text{+28x-12}}{5}\text{=0}$ .

  ${\text{5x}}^{\text{2}}\text{+28x-12=0}$ .

Hence the quadratic equation is ${\text{5x}}^{\text{2}}\text{+28x-12=0}$ .