Question

How do you solve $4{x^2} = 9x$ using quadratic formula?

Answer 1

In this given problem to solve the equation and find the value of $x$ using quadratic formula, first simplify the equation such that all terms should be on left hand side and then compare the quadratic coefficients with coefficients of the given equation and then simply put the values in the quadratic formula to get the solution.

Formula used:
Quadratic formula: $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Discriminant of the quadratic equation: $D = \sqrt {{b^2} - 4ac}$

Complete step by step solution:
In order to solve the given expression ${x^2} - 3 = 0$ with help of quadratic formula, we need to first simplify the equation as follows

$$\Rightarrow 4{x^2} = 9x \\\ \Rightarrow 4{x^2} - 9x + 0 = 0 \\\$$

Now comparing it with standard quadratic equation to get value of quadratic coefficients
$4{x^2} - 9x + 0 = 0\;{\text{and}}\;a{x^2} + bx + c$
Therefore respective values of quadratic coefficients are
$a = 4,\;b = - 9\;{\text{and}}\;c = 0$
Now we know that the solution for the quadratic equation $a{x^2} + bx + c = 0$ is given as
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Where $D$ is the discriminant of the quadratic expression, which can be calculated as
$D = \sqrt {{b^2} - 4ac}$
So first finding the value of discriminant,
$D = {( - 9)^2} - 4 \times 4 \times 0 \\\ D = 81 \\\$
Now substituting all the respective values in quadratic formula in order to get the solution for $x$
$x = \dfrac{{ - ( - 9) \pm \sqrt {81} }}{{2 \times 4}} \\\ = \dfrac{{9 \pm 9}}{8} \\\ = \dfrac{{9 + 9}}{8}\;{\text{or}}\;\dfrac{{9 - 9}}{8} \\\ = \dfrac{9}{4}\;{\text{or}}\;0 \\\$
Therefore $x = \dfrac{9}{4}\;{\text{and}}\;x = 0$ are the required solutions for the equation $4{x^2} = 9x$

Note: Quadratic formula is only applicable for quadratic equations that are equations with degree two. Sometimes students directly solve this type of question by cancelling out the degree of variable by division and getting the value, but in this process they left one solution for the equation that is $x = 0$, so never directly cancel the degree of variables.