QUESTION

# Solve the following quadratic equation $5{x^2} - 4x - 7 = 0$

Hint: First of all, make the coefficient of ${x^2}$ term equal to one by doing simple math applications like multiplication and division. Then add the square of the coefficient of $2x$ on both sides to have the complete square. So, use this concept to reach the solution of the problem.

Complete step-by-step solution -
Given quadratic equation is $5{x^2} - 4x - 7 = 0$
Dividing both sides with 5, we get
$\Rightarrow \dfrac{{5{x^2} - 4x - 7}}{5} = \dfrac{0}{5} \\ \Rightarrow \dfrac{{5{x^2}}}{5} - \dfrac{{4x}}{5} = \dfrac{7}{5} \\ \Rightarrow {x^2} - 2x\left( {\dfrac{2}{5}} \right) = \dfrac{7}{5} \\$
Adding ${\left( {\dfrac{2}{5}} \right)^2} = \dfrac{4}{{25}}$ on both sides we get
$\Rightarrow {x^2} - 2\left( {\dfrac{{2x}}{5}} \right) + \dfrac{4}{{25}} = \dfrac{7}{5} + \dfrac{4}{{25}} \\ \Rightarrow {x^2} - 2\left( {\dfrac{{2x}}{5}} \right) + {\left( {\dfrac{2}{5}} \right)^2} = \dfrac{{7\left( 5 \right) + 4}}{{25}} \\ \Rightarrow {x^2} - 2\left( {\dfrac{{2x}}{5}} \right) + {\left( {\dfrac{2}{5}} \right)^2} = \dfrac{{39}}{{25}} \\$
We know that ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
$\Rightarrow {\left( {x - \dfrac{2}{5}} \right)^2} = \dfrac{{39}}{{25}}$
Rooting on both sides, we get
$\Rightarrow x - \dfrac{2}{5} = \pm \sqrt {\dfrac{{39}}{{25}}} \\ \Rightarrow x = \dfrac{2}{5} \pm \sqrt {\dfrac{{39}}{{25}}} \\ \Rightarrow x = \dfrac{2}{5} \pm \dfrac{{\sqrt {39} }}{5} \\ \therefore x = \dfrac{{2 \pm \sqrt {39} }}{5} \\$
Thus, the solutions of the quadratic equation $5{x^2} - 4x - 7 = 0$ are $\dfrac{{2 \pm \sqrt {39} }}{5}$.

Note: We can also solve this problem by another method i.e., by using the formula that the two roots of the quadratic equation $a{x^2} + bx + c = 0$ are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. The quadratic equation has only two roots.