Question

Solve the following quadratic equation \[5{x^2} - 4x - 7 = 0\]

Answer 1

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.