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Question

Question: How do you solve by completing the square \[{{x}^{2}}-2x-24=0\]?...

How do you solve by completing the square x22x24=0{{x}^{2}}-2x-24=0?

Explanation

Solution

In the given question, we have been asked to find the value of ‘x’ by solving the given equation i.e. x22x24=0{{x}^{2}}-2x-24=0 using the completing the square method. Completing the square method is used to solve the quadratic equation by converting the form of the equation so that it will become a perfect trinomial. And then after applying the square formula, we will get our required solution.

Formula used:
a22ab+b2=(ab)2{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}

Complete step by step solution:
We have given that,
x22x24=0\Rightarrow {{x}^{2}}-2x-24=0
Adding 24 to both the side of the equation, we get
x22x24+24=0+24\Rightarrow {{x}^{2}}-2x-24+24=0+24
Simplifying the above, we get
x22x=24\Rightarrow {{x}^{2}}-2x=24
Now, for completing the square adding (1)2{{\left( 1 \right)}^{2}} to both the sides of the equation, we get
x22x+12=24+1\Rightarrow {{x}^{2}}-2x+{{1}^{2}}=24+1
x22x+12=25\Rightarrow {{x}^{2}}-2x+{{1}^{2}}=25
As we know that, a22ab+b2=(ab)2{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}
Therefore,
(x1)2=25\Rightarrow {{\left( x-1 \right)}^{2}}=25
Transposing the power 2 on the right side of the equation, we get
(x1)=25\Rightarrow \left( x-1 \right)=\sqrt{25}
As we know that 25=±5\sqrt{25}=\pm 5
x1=±5\Rightarrow x-1=\pm 5
Adding 1 to both side of the equation, we get
x=±5+1\Rightarrow x=\pm 5+1
Now, we have
x=5+1\Rightarrow x=-5+1 or x=5+1x=5+1
x=4 or 6\Rightarrow x=-4\ or\ 6
Therefore, the possible values of x are 6 or 4x\ are\ 6\ or\ -4.

Note: While solving these types of questions, students should carefully observe when considering the terms, which one is the ‘a’ and the ‘b’. To check whether the obtained possible values are correct or not, we can verify the result by solving the quadratic equation with the roots of the quadratic equation formula. Standard form of quadratic equation; ax2+bx+c=0a{{x}^{2}}+bx+c=0, then the roots of the quadratic equation is given by, x=b±b24ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.