How do you factor \[{{x}^{2}}-3x-88=0\] using completing the square method?
Answer
593.7k+ views
Hint: To find the factors of the quadratic equation we have to create a trinomial square on the left side of the equation, we have to find a value that is equal to the square of half of b from the equation \[a{{x}^{2}}+bx+c=0\] and from this we may get real roots or imaginary roots based on the given quadratic equation. The equation is further simplified to obtain the required roots of the equation.
Complete step by step solution:
The given equation to find the factor is as follows,
\[{{x}^{2}}-3x-88=0\]
\[\Rightarrow {{x}^{2}}-3x=88\]
Using completing the square method, we will follow certain steps as below:
To create a trinomial square on the left side of the equation, we have to find a value that is equal to the square of half of b in equation \[a{{x}^{2}}+bx+c=0\],
\[{{\left( \dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{-3}{2} \right)}^{2}}\]
Now let us add the term to each side of the equation,
\[{{x}^{2}}-3x+{{\left( \dfrac{-3}{2} \right)}^{2}}=88+{{\left( \dfrac{-3}{2} \right)}^{2}}\]
Now simplify the above equation,
\[\Rightarrow {{x}^{2}}-3x+\left( \dfrac{9}{4} \right)=\left( \dfrac{361}{4} \right)\]
We can factor the perfect trinomial square into \[{{\left( x-\dfrac{3}{2} \right)}^{2}}\].
\[\Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{361}{4}\]
We can write the above equation as below,
\[\Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}={{\left( \dfrac{19}{2} \right)}^{2}}\]
\[\Rightarrow \left( x-\dfrac{3}{2} \right)=\pm \left( \dfrac{19}{2} \right)\]
Therefore, the value of x can be as follows,
\[\Rightarrow \]\[x=+\dfrac{19}{2}+\dfrac{3}{2}\] and \[x=-\dfrac{19}{2}+\dfrac{3}{2}\]
\[\Rightarrow \]\[x=\dfrac{22}{2}\] and \[x=\dfrac{-16}{2}\]
\[\Rightarrow \]\[x=11\] and -8
So the roots of the equation are 11 and -8 for the given equation.
Note: While solving this problem students should be aware of creating a trinomial square on the left side of the equation, we have to find a value that is equal to the square of half of b. Then it is important for students to know the method of simplification to solve in further steps to get the real roots for the given quadratic equation.
Complete step by step solution:
The given equation to find the factor is as follows,
\[{{x}^{2}}-3x-88=0\]
\[\Rightarrow {{x}^{2}}-3x=88\]
Using completing the square method, we will follow certain steps as below:
To create a trinomial square on the left side of the equation, we have to find a value that is equal to the square of half of b in equation \[a{{x}^{2}}+bx+c=0\],
\[{{\left( \dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{-3}{2} \right)}^{2}}\]
Now let us add the term to each side of the equation,
\[{{x}^{2}}-3x+{{\left( \dfrac{-3}{2} \right)}^{2}}=88+{{\left( \dfrac{-3}{2} \right)}^{2}}\]
Now simplify the above equation,
\[\Rightarrow {{x}^{2}}-3x+\left( \dfrac{9}{4} \right)=\left( \dfrac{361}{4} \right)\]
We can factor the perfect trinomial square into \[{{\left( x-\dfrac{3}{2} \right)}^{2}}\].
\[\Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{361}{4}\]
We can write the above equation as below,
\[\Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}={{\left( \dfrac{19}{2} \right)}^{2}}\]
\[\Rightarrow \left( x-\dfrac{3}{2} \right)=\pm \left( \dfrac{19}{2} \right)\]
Therefore, the value of x can be as follows,
\[\Rightarrow \]\[x=+\dfrac{19}{2}+\dfrac{3}{2}\] and \[x=-\dfrac{19}{2}+\dfrac{3}{2}\]
\[\Rightarrow \]\[x=\dfrac{22}{2}\] and \[x=\dfrac{-16}{2}\]
\[\Rightarrow \]\[x=11\] and -8
So the roots of the equation are 11 and -8 for the given equation.
Note: While solving this problem students should be aware of creating a trinomial square on the left side of the equation, we have to find a value that is equal to the square of half of b. Then it is important for students to know the method of simplification to solve in further steps to get the real roots for the given quadratic equation.
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