Question

How do you solve the quadratic $5{x^2} - 3x = - 8$ using any method?

Answer 1

Hint:
Given a quadratic equation. We have to find the solution of the equation. First, we will rewrite the equation in the standard form of quadratic equation. We will apply the quadratic formula to the equation. Then, we will find the value of x.

Formula used:
The quadratic formula to find the value of x in the quadratic equation of the form $a{x^2} + bx + c = 0$ is given by:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
We are given the equation, $5{x^2} - 3x = - 8$.
First, we will rewrite the equation in the standard form of quadratic equation.
$ \Rightarrow 5{x^2} - 3x + 8 = 0$
Now, we will factorize the equation by applying the quadratic formula because splitting the middle term to factorize the equation does not work in this equation.
Compare the equation with standard form of equation, $a{x^2} + bx + c = 0$ we get:
Substitute $a = 5$, $b = - 3$ and $c = 8$ into the formula, we get:
$ \Rightarrow x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {{{\left( { - 3} \right)}^2} - 4\left( 5 \right)\left( 8 \right)} }}{{2\left( 5 \right)}}$
On simplifying the terms of the expression, we get:
\[ \Rightarrow x = \dfrac{{3 \pm \sqrt {9 - 160} }}{{10}}\]
$ \Rightarrow x = \dfrac{{3 \pm \sqrt { - 151} }}{{10}}$
Here, the value inside the radical symbol is negative which means the quadratic equation has no real solution.

Hence the quadratic equation $5{x^2} - 3x = - 8$ has no real solution.

Note:
The students please note that the roots of the quadratic equation will depend on the value of discriminant. The discriminant of the quadratic equation is given by $D = {b^2} - 4ac$. The value of discriminant can be classified into three categories.
1) If the value of discriminant is greater than zero or positive, then the equation has two real solutions.
2) If the value of discriminant is less than zero or negative, then the equation has no real solutions.
3) If the value of discriminant is equal to zero, then the equation has only one real solution.