Solveeit Logo

Question

Question: How do you solve \({{x}^{2}}-6=x\) using the quadratic formula?...

How do you solve x26=x{{x}^{2}}-6=x using the quadratic formula?

Explanation

Solution

By quadratic formula, we mean the Sridhar Acharya formula, which is more commonly known. We compare the given equation with the general form of a quadratic equation to find out the coefficients and put them in the formula x=b±b24ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} . After that, we get two solutions to the given equation.

Complete step-by-step solution:
The given equation is
x26=x{{x}^{2}}-6=x
The equation can be rearranged by moving the xx term from the right hand side to the left hand side. The equation thus becomes,
x2x6=0....equation1\Rightarrow {{x}^{2}}-x-6=0....equation1
Now, let us consider the general form of a quadratic equation which is
ax2+bx+c=0a{{x}^{2}}+bx+c=0
By quadratic formula, we mean the Sridhar Acharya formula which states that the two roots of a quadratic equation can be expressed as
x=b±b24ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Where, a,ba,b are the respective coefficients of the terms x2{{x}^{2}} , xx and cc is the constant term. Comparing the given equation x2x6=0{{x}^{2}}-x-6=0 with the general quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 , we get
a=1 b=1 c=6 \begin{aligned} & a=1 \\\ & b=-1 \\\ & c=-6 \\\ \end{aligned}
Having found out the coefficients, we put them in the Sridhar Acharya formula to get the two roots of the equation or the two values of xx that satisfy the given equation. Thus,
x=(1)±(1)24(1)(6)2(1)\Rightarrow x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -6 \right)}}{2\left( 1 \right)}
Simplifying the terms, we get,
x=1±1+4(6)2\Rightarrow x=\dfrac{1\pm \sqrt{1+4\left( 6 \right)}}{2}
Further simplification gives
x=1±1+242\Rightarrow x=\dfrac{1\pm \sqrt{1+24}}{2}
Adding the two numbers 2424 and 11 and implementing the square root, we get,
x=1±52\Rightarrow x=\dfrac{1\pm 5}{2}
Now, the two values of xx after separating the ++ and the - from the ±\pm , are
x=(1+52)\Rightarrow x=\left( \dfrac{1+5}{2} \right) and x=(152)x=\left( \dfrac{1-5}{2} \right)
On simplification, we get,
x=3\Rightarrow x=3 and x=2x=-2
Therefore, we can conclude that the two roots or solutions of the given equation x26=x{{x}^{2}}-6=x are x=3x=3 and x=2x=-2 .

Note: We should remember the quadratic formula accurately and should not interpret it wrongly. As the coefficients play a vital role here, we must compare our equation with the general quadratic equation correctly. Also, there are various other ways to find the solution like, by plotting the corresponding graph and see where it intersects the xaxisx-axis , by completing the square, or by the middle term method.