Question

What are the solutions to the equation: ${{x}^{2}}-8x=24$?

Answer 1

To solve the given equation, we are going to use the Sri Dharacharya rule. From Sri Dharacharya rule for the quadratic equation $a{{x}^{2}}+bx+c=0$ we can find the roots of the quadratic equation as follows: $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, first of all, we are going to convert the above quadratic equation in the standard form and then proceed. Now, we will get two values of x (positive and negative). And this is how we will get the solutions of the given equation.

Complete step-by-step solution:
The quadratic equation given in the above problem of which we have to find the solutions is equal to:
${{x}^{2}}-8x=24$
Rearranging the above equation we get,
${{x}^{2}}-8x-24=0$
We know that we can find the solutions using Shree Dharacharya rule for the quadratic equation written in standard form $a{{x}^{2}}+bx+c=0$ as follows:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Applying Shree Dharacharya rule in the above quadratic equation we get the following values of $a,b\And c$:
$a=1,b=-8,c=-24$
$\begin{aligned} & x=\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 1 \right)\left( -24 \right)}}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{8\pm \sqrt{64-4\left( 1 \right)\left( -24 \right)}}{2} \\\ & \Rightarrow x=\dfrac{8\pm \sqrt{64+96}}{2} \\\ & \Rightarrow x=\dfrac{8\pm \sqrt{160}}{2} \\\ \end{aligned}$
Taking 16 out from the square root expression and we get,
$\Rightarrow x=\dfrac{8\pm 4\sqrt{10}}{2}$
Now, in the numerator, 8 and 4 are divisible by 2 and we get,
$x=4\pm 2\sqrt{10}$
Taking the value of x with positive sign we get,
$x=4+2\sqrt{10}$
Taking the value of x with negative sign we get,
$x=4-2\sqrt{10}$
**Hence, the solutions for the given quadratic equation are:
$x=4+2\sqrt{10}$ and $x=4-2\sqrt{10}$ **

Note: The mistake that could be possible in the above problem is in misplacing the signs in the Shree Dharacharya rule. The solutions of x for the standard quadratic equation $a{{x}^{2}}+bx+c=0$ are as follows:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
The trick to remember the above formula is that there is always a negative sign like before “b” written out from the square root and before “4ac”. Generally, people miss the negative sign before “b” so you can remember that the “b” written outside the square root must have a negative sign.