Question

How‌ ‌do‌ ‌you‌ ‌solve‌ the quadratic equation‌ ${{\left(‌ ‌x-5‌ ‌\right)}^{2}}=2$?‌

Answer 1

Take all the terms to the L.H.S. so that in the R.H.S. we get 0. Now, write 2 in exponential form such that it has an exponent equal to 2. To do this, take square root of 2 and to balance take the exponent equal to 2. Now, use the algebraic identity: - ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factorize the quadratic equation. Finally, substitute each term equal to 0 and find the two values of x to get the answer.

Complete step-by-step solution:
Here, we have been provided with the equation: - ${{\left( x-5 \right)}^{2}}=2$ and we are asked to solve it. That means we have to find the value of x.
Now, we can say that the given equation is a quadratic equation because if we will expand the expression ${{\left( x-5 \right)}^{2}}$ we will get the exponent equal to 2. Let us solve this equation without expanding the whole square expression in the L.H.S. We will use the factorization method to solve the question.
$\because {{\left( x-5 \right)}^{2}}=2$
Taking all the term to the L.H.S., we get,
$\Rightarrow {{\left( x-5 \right)}^{2}}-2=0$
Now, here in the above expression, to factorize the L.H.S. using the algebraic identity: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we need to make the exponent of 2 equal to 2. So, to make this happen let us make the square root of 2 and to balance we will square the term, so we get,
$\Rightarrow {{\left( x-5 \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}=0$
Using the algebraic identity: - ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we get,

& \Rightarrow \left( x-5+\sqrt{2} \right)\left( x-5-\sqrt{2} \right)=0 \\\ & \Rightarrow \left[ x-\left( 5-\sqrt{2} \right) \right]\left[ x-\left( 5+\sqrt{2} \right) \right]=0 \\\ \end{aligned}$$ Substituting each term equal to 0, we get, $$\Rightarrow x-\left( 5-\sqrt{2} \right)=0$$ or $$x-\left( 5+\sqrt{2} \right)=0$$ $$\Rightarrow x=5-\sqrt{2}$$ or $$x=5+\sqrt{2}$$ **Hence, the roots of the given quadratic equation are $$\left( 5-\sqrt{2} \right)$$ and $$\left( 5+\sqrt{2} \right)$$.** **Note:** One may note that we can also solve the quadratic equation given by expanding the L.H.S. using the formula $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$. Using this formula, we will form a quadratic equation in x as $$a{{x}^{2}}+bx+c=0$$. In the next step we will use the discriminant formula given as $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ to get the two values of x. Note that it would have been difficult to use the middle term split method because it is difficult to think of the roots like: - $$\left( 5-\sqrt{2} \right)$$ and $$\left( 5+\sqrt{2} \right)$$.