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Question: How do you solve \({{\left( x-6 \right)}^{2}}=25\) ?...

How do you solve (x−6)2=25{{\left( x-6 \right)}^{2}}=25 ?

Explanation

Solution

To solve the above equation, we are going to first subtract 25 on both the sides. Then we will require the following algebraic identity: a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right). After applying this algebraic identity we can easily find the values of x.

Complete step-by-step solution:
The quadratic equation given in the above problem which we have to solve is as follows:
(x−6)2=25{{\left( x-6 \right)}^{2}}=25
Subtracting 25 on both the sides of the above equation we get,
(x−6)2−25=25−25 ⇒(x−6)2−25=0 \begin{aligned} & {{\left( x-6 \right)}^{2}}-25=25-25 \\\ & \Rightarrow {{\left( x-6 \right)}^{2}}-25=0 \\\ \end{aligned}
In the above equation, we can write 25 as 52{{5}^{2}} so we are writing 25 in this manner we get,
(x−6)2−52=0{{\left( x-6 \right)}^{2}}-{{5}^{2}}=0
As you can see that the expression written in the L.H.S of the above equation is of the form a2−b2{{a}^{2}}-{{b}^{2}} so we can use the identity of a2−b2{{a}^{2}}-{{b}^{2}} which is equal to:
a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)
⇒(x−6)2−52=0 ⇒(x−6−5)(x−6+5)=0 ⇒(x−11)(x−1)=0 \begin{aligned} & \Rightarrow {{\left( x-6 \right)}^{2}}-{{5}^{2}}=0 \\\ & \Rightarrow \left( x-6-5 \right)\left( x-6+5 \right)=0 \\\ & \Rightarrow \left( x-11 \right)\left( x-1 \right)=0 \\\ \end{aligned}
To find the value of x in the above equation, we are going to equate each bracket to 0 and we get,
x−11=0 ⇒x=11; \begin{aligned} & x-11=0 \\\ & \Rightarrow x=11; \\\ \end{aligned}
x−1=0 ⇒x=1 \begin{aligned} & x-1=0 \\\ & \Rightarrow x=1 \\\ \end{aligned}
From the above solutions, we have got the solutions of x as 1 and 11.
Hence, we have solved the equation given in the above problem and its solutions are 1 and 11.

Note: To check the solutions of the quadratic equation, we are going to substitute the values of x which we have obtained above in the given equation and then check whether these values of x are either satisfying the given equation or not.
The values of x which we have obtained above are:
1 and 11
Substituting the value of x as 1 in the above equation we get,
(x−6)2=25 ⇒(1−6)2=25 ⇒(−5)2=25 ⇒25=25 \begin{aligned} & {{\left( x-6 \right)}^{2}}=25 \\\ & \Rightarrow {{\left( 1-6 \right)}^{2}}=25 \\\ & \Rightarrow {{\left( -5 \right)}^{2}}=25 \\\ & \Rightarrow 25=25 \\\ \end{aligned}
From the above, it is proven that the value of x equal to 1 is satisfying the given equation.
Similarly, you can check the other value of x (which is equal to 11) whether this value is satisfying the given equation or not.