Question

How do you factor and solve ${{x}^{2}}-8x=-3$?

Answer 1

To find the factors of equation ${{x}^{2}}-8x=-3$, first of all we should be aware of basic quadratic equation and formula for finding roots for a quadratic equation. Then we have to compare the given equation with the basic quadratic equation and substitute the values in the formula. So that we can get the required roots.

Complete step by step answer:
From the given question we are given to factor and solve the equation ${{x}^{2}}-8x=-3$. For that
Let us consider
${{x}^{2}}-8x=-3.........\left( 1 \right)$
By observing equation (1) we can clearly say that the given equation is a quadratic equation.
As we know that

& For\text{ a}{{\text{x}}^{2}}+bx+c=0; \\\ & \text{The roots of the equation are }\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\text{ } \\\ \end{aligned}$$ Let us consider $$\text{a}{{\text{x}}^{2}}+bx+c=0........\left( f1 \right)$$ $$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}...........(f2)$$ By comparing the equation (1) with formula (f1). We can observe that RHS (right hand side) should be equal to 0. So, transferring -3 from RHS to LHS (left hand side), we get $${{x}^{2}}-8x+3=0$$ Let us consider $${{x}^{2}}-8x+3=0..........\left( 2 \right)$$ By comparing the equation (2) with (f1) we can observe a, b, c values. Therefore $$a=1\text{ b=-8 c=3}$$ Let us substitute above a, b, c values in formula (f2), we get $$\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}$$ By simplifying it , we get $$\Rightarrow \dfrac{8\pm \sqrt{64-12}}{2}$$ Subtracting 64-12 and then, we get $$\Rightarrow \dfrac{8\pm \sqrt{52}}{2}$$ Rewriting 52 as 4.13, we get $$\Rightarrow \dfrac{8\pm \sqrt{4.13}}{2}$$ Getting out 4 from the square root, we have $$\Rightarrow \dfrac{8}{2}\pm \dfrac{2\sqrt{13}}{2}$$ Therefore factors or roots of the equation (1) is $$\Rightarrow 4\pm \sqrt{13}$$ Which can be written as $$4+\sqrt{13}\text{, 4-}\sqrt{13}$$ Let us consider $$\begin{aligned} & 4+\sqrt{13}.........\left( 3 \right) \\\ & \text{4-}\sqrt{13}............\left( 4 \right) \\\ \end{aligned}$$ So, therefore equation (3) and (4) are the factors of equation (1). **Note:** Students should be aware of quadratic equations concepts. Questionnaires may ask this type of problems in many types i.e. sum of the roots or multiplication of roots or in many ways. So students should aware of all these types of concepts.