Question

How do you write a quadratic equation in standard form with the given roots 8, –2?

Answer 1

Hint: We start solving the problem by recalling the fact that the equation of the quadratic equation with roots $\alpha $ and $\beta $ is $\left( x-\alpha \right)\left( x-\beta \right)=0$. We then substitute 8, –2 in place of $\alpha $ and $\beta $ to proceed through the problem. We then make the necessary calculations involving multiplication, addition and subtraction operations to get the required quadratic equation.

Complete step by step answer:
According to the problem, we are asked to find the quadratic equation in standard form with the given roots 8, –2.
We know that the equation of the quadratic equation with roots $\alpha $ and $\beta $ is $\left( x-\alpha \right)\left( x-\beta \right)=0$.
So, the required quadratic equation is $\left( x-8 \right)\left( x+2 \right)=0$.
$\Rightarrow {{x}^{2}}-8x+2x-16=0$.
$\Rightarrow {{x}^{2}}-6x-16=0$.
So, we have found the required quadratic equation as ${{x}^{2}}-6x-16=0$.
$\therefore $ The quadratic equation in standard form with the given roots 8, –2 is ${{x}^{2}}-6x-16=0$.

Note:
We can also solve the given problem as shown below:
Let us assume the required quadratic equation be ${{x}^{2}}-ax+b=0$ ---(1), where a is the sum of the roots and b is the product of the roots.
So, we have $a=8-2$.
$\Rightarrow a=6$ ---(2).
Now, we have $b=8\times \left( -2 \right)$.
$\Rightarrow b=-16$ ---(3).
Let us substitute equations (2) and (3) in equation (1).
So, we get the quadratic equation as ${{x}^{2}}-\left( 6 \right)x+\left( -16 \right)=0$.
$\Rightarrow {{x}^{2}}-6x-16=0$.
$\therefore $ The required quadratic equation with the given roots 8, –2 is ${{x}^{2}}-6x-16=0$.
Similarly, we can expect problems to find the quadratic equation with sum and product of the roots as 6 and -10.