Question

Write the equation of a quadratic in general form given two of its solutions.x= -2/3 ; x= 5

Answer 1

Given the roots/solutions of a quadratic equation as;

$x=\frac{-2}{3}\text{ and }x=5$

The general way of solving this kind of problem is by inserting/substituting it into the general form of factorized quadratic equation which is;

$\begin{gathered} (x-a)(x-b)=0 \\ \text{where a and b are the roots.} \end{gathered}$

substituting the roots given in the question we have;

$\begin{gathered} (x-a)(x-b)=0 \\ a=\frac{-2}{3}\text{ and b}=5 \\ \text{Then;} \\ (x-\frac{-2}{3})(x-5)=0 \\ (x+\frac{2}{3})(x-5)=0 \end{gathered}$

Then we can now expand to get the Quadratic equation.

$\begin{gathered} (x+\frac{2}{3})(x-5)=0 \\ x^2-5x+\frac{2}{3}x-\frac{10}{3}=0 \\ \text{multiplying through by 3, we have;} \\ 3x^2-15x+2x-10=0 \\ 3x^2-13x-10=0 \end{gathered}$

So, the quadratic equation with the roots/solutions x= -2/3 and x=5 is;

$3x^2-13x-10=0$