How do you factor \[{{x}^{2}}-3x-88=0\] using completing the square method?
Answer
609k+ 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.
Recently Updated Pages
Difference Between Prokaryotic Cells and Eukaryotic Cells

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

