Question

How do you solve ${{x}^{2}}-1=3x$?

Answer 1

First we will convert the given equation into general form which is given as $a{{x}^{2}}+bx+c=0$. Then the obtained equation is a quadratic equation so we will solve the given equation by using the quadratic formula method. The quadratic formula is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step by step solution:
We have been given an equation ${{x}^{2}}-1=3x$.
We know that the general quadratic equation is of the form $a{{x}^{2}}+bx+c=0$. Then by converting the given equation into general form we will get
$\Rightarrow {{x}^{2}}-3x-1=0$
Now, on comparing the obtained equation with the general equation we will get the values as
$a=1,b=-3,c=-1$
Now, the quadratic formula to solve the quadratic equation is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, substituting the values we will get
$\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times -1}}{2\times 1}$
Now, simplifying the above obtained equation we will get
$\begin{aligned} & \Rightarrow x=\dfrac{3\pm \sqrt{9+4}}{2} \\\ & \Rightarrow x=\dfrac{3\pm \sqrt{13}}{2} \\\ \end{aligned}$
Now, we have to consider both signs one by one then we will get
$\Rightarrow x=\dfrac{3+\sqrt{13}}{2},x=\dfrac{3-\sqrt{13}}{2}$
Hence on solving the given equation we get the two values of x as $x=\dfrac{3+\sqrt{13}}{2},\dfrac{3-\sqrt{13}}{2}$.

Note: The given quadratic equation is of order two so it has two solutions. The number of solutions of an equation depends on its order or degree. The order of the equation is defined by the highest power of the variable the equation has. Also we have to consider both signs one by one to obtain the solution. We have alternate methods also to solve the equation such as completing the square method, splitting the middle term method and factorization method.