Question

How do you solve \[{{\left( x-2 \right)}^{2}}-25=0?\]

Answer 1

Hint: We are given \[{{\left( x-2 \right)}^{2}}-25=0,\] and we are asked to find the solution of this, to do so we will first understand the type of equation we have, once we get that we reduce the given equation into the standard form to simplify we will assume x – 2 = t and after that we will find the greatest common factor from each term then in the remaining term be factor using the middle term and lastly we compare it with zero and solve further. We will use \[a\times c\] in such a way that its sum or difference from the ‘b’ of the equation \[a{{x}^{2}}+bx+c=0.\]

Complete answer:
We are given \[{{\left( x-2 \right)}^{2}}-25=0\] and we are asked to find the solution to it. We start out the solution by considering x – 2 as t. So, our equation \[{{\left( x-2 \right)}^{2}}-25=0\] becomes \[{{t}^{2}}-25=0.\] Now to find the solution of the equation, we should see that as the highest power is 2 so it is or 2-degree polynomial. So it is a quadratic equation. As our equation is quadratic equations and we know quadratic equations are given as \[a{{t}^{2}}+bt+c=0.\] And we reduce it to standard form. We can see that our equation \[{{t}^{2}}-25=0\] is already in standard form. Now, to find its solution we will first find the possible greatest common factor of all these.
In 1 and 25 we can see that 1 is the only possible term that can be separated and our equation stays like original as \[{{t}^{2}}-25=0.\] We can write this above equation as \[{{t}^{2}}+0t-25=0.\] Now, we will use the middle term to split. In the middle term split, apply on \[a{{x}^{2}}+bx+c,\] we produce ‘a’ by ‘c’ and then factor ‘ac’ in such a way that if the product is ‘ac’ while the sum or the difference is made up to ‘b’. Now, we have the middle term split on \[{{t}^{2}}+0t-25.\] So, we have a = 1, b = 0 and c = – 25. So,
\[a\times c=1\times \left( -25 \right)=-25\]
Now, we can see that \[5\times \left( -5 \right)=-25\] and also \[5+\left( -5 \right)=0.\] So, we will use this to split the middle term. So, we get,
\[{{t}^{2}}+0t-25={{t}^{2}}+\left( 5-5 \right)t-25\]
Opening the brackets, we get,
\[\Rightarrow {{t}^{2}}+5t-5t-25\]
We take the common in the first two terms and the last 2 terms. So, we get,
\[\Rightarrow t\left( t+5 \right)-5\left( t+5 \right)\]
As (t + 5) is the same, so we get,
\[\Rightarrow \left( t-5 \right)\left( t+5 \right)\]
So, we get,
\[\Rightarrow {{t}^{2}}-25=\left( t-5 \right)\left( t+5 \right)\]
So, as \[{{t}^{2}}-25\] is equal to zero, we compare the above equation with zero. So, we get,
\[\Rightarrow \left( t-5 \right)\left( t+5 \right)=0\]
Now, we use the zero product rule which says that \[a\times b=0\] means either a = 0 or b = 0. So, as
\[\Rightarrow \left( t-5 \right)\left( t+5 \right)=0\]
This means either t + 5 = 0 or t – 5 = 0.
On simplifying, we get,
\[t=-5;t=5\]
Now, as we considered x – 2 as t so replacing it, we get,
\[\Rightarrow x-2=-5;x-2=5\]
Adding 2 on each side, we get,
\[\Rightarrow x=-3;x=7\]
So, we get the solution as x = 7 and x = – 3.
Hence, for \[{{\left( x-2 \right)}^{2}}-25=0\] the solutions are 7 and – 3.

Note: While finding the middle term using the factor of \[a\times c\] we need to keep in mind that when the sign of ‘a’ and ‘c’ are the same then ‘b’ is obtained by addition only. If the sign of ‘a’ and ‘c’ is different then ‘b’ can be obtained using only subtraction. So, as we have a = 3 and c = – 6, different signs so ‘b’ is obtained as 9 – 2 = 7 using subtraction. The key point to remember is that the degree of the equation will also tell us about the number of the solution also. So, 2 degrees means the given equation can have only 2 solutions.