
How do you solve \[{{x}^{2}}=12x-7\] by completing the square?
Answer
556.5k+ views
Hint: To solve a quadratic equation in \[x\] by completing the square method we need to follow the below steps,
Step 1: Take the terms having \[{{x}^{2}}\], and \[x\]to one side of the equation leaving a constant term to the other side.
Step 2: Add the term \[{{\left( \dfrac{b}{2} \right)}^{2}}\] to both sides of the equation, \[b\]is the coefficient of \[x\].
Step 3: The left-hand side will become a perfect square, solving the square will give the roots of the equation.
Complete step by step answer:
We are given the quadratic equation \[{{x}^{2}}=12x-7\], we need to solve the equation by completing the square method. We know the steps required to solve a quadratic by this method.
First, we have to take the terms having \[{{x}^{2}}\], and \[x\]to one side of the equation leaving a constant term to the other side.
\[{{x}^{2}}=12x-7\]
Subtracting \[12x\] from both sides of the equation, we get
\[\Rightarrow {{x}^{2}}-12x=12x-7-12x\]
\[\Rightarrow {{x}^{2}}-12x=-7\]
In the next step, we have to add the term \[{{\left( \dfrac{b}{2} \right)}^{2}}\] to both sides of the equation, \[b\]is the coefficient of \[x\]. Here, \[b=-12\]. \[{{\left( \dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36\], adding 36 to both sides of the equation, we get
\[\begin{align}
& \Rightarrow {{x}^{2}}-12x+36=-7+36 \\
& \Rightarrow {{x}^{2}}-12x+36=29 \\
\end{align}\]
The left-hand side of the above equation is the square of the term \[\left( x-6 \right)\]. Hence, we can express it as
\[\Rightarrow {{\left( x-6 \right)}^{2}}=29\]
Taking the square root of both sides of the above equation, we get
\[\Rightarrow x-6=\pm \sqrt{29}\]
\[\Rightarrow x-6=\sqrt{29}\] or \[x-6=-\sqrt{29}\]
\[\therefore x=\sqrt{29}+6\] or \[\Rightarrow x=6-\sqrt{29}\].
Hence, the roots of the quadratic equation are \[x=\sqrt{29}+6\] or \[x=6-\sqrt{29}\].
Note:
There are many methods to solve a quadratic equation, as the factorization method, completing the square method, and the formula method. We can use any of them to solve a quadratic equation. The formula method should be preferred because it gives two roots of the equation, whether they are real or imaginary.
Step 1: Take the terms having \[{{x}^{2}}\], and \[x\]to one side of the equation leaving a constant term to the other side.
Step 2: Add the term \[{{\left( \dfrac{b}{2} \right)}^{2}}\] to both sides of the equation, \[b\]is the coefficient of \[x\].
Step 3: The left-hand side will become a perfect square, solving the square will give the roots of the equation.
Complete step by step answer:
We are given the quadratic equation \[{{x}^{2}}=12x-7\], we need to solve the equation by completing the square method. We know the steps required to solve a quadratic by this method.
First, we have to take the terms having \[{{x}^{2}}\], and \[x\]to one side of the equation leaving a constant term to the other side.
\[{{x}^{2}}=12x-7\]
Subtracting \[12x\] from both sides of the equation, we get
\[\Rightarrow {{x}^{2}}-12x=12x-7-12x\]
\[\Rightarrow {{x}^{2}}-12x=-7\]
In the next step, we have to add the term \[{{\left( \dfrac{b}{2} \right)}^{2}}\] to both sides of the equation, \[b\]is the coefficient of \[x\]. Here, \[b=-12\]. \[{{\left( \dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36\], adding 36 to both sides of the equation, we get
\[\begin{align}
& \Rightarrow {{x}^{2}}-12x+36=-7+36 \\
& \Rightarrow {{x}^{2}}-12x+36=29 \\
\end{align}\]
The left-hand side of the above equation is the square of the term \[\left( x-6 \right)\]. Hence, we can express it as
\[\Rightarrow {{\left( x-6 \right)}^{2}}=29\]
Taking the square root of both sides of the above equation, we get
\[\Rightarrow x-6=\pm \sqrt{29}\]
\[\Rightarrow x-6=\sqrt{29}\] or \[x-6=-\sqrt{29}\]
\[\therefore x=\sqrt{29}+6\] or \[\Rightarrow x=6-\sqrt{29}\].
Hence, the roots of the quadratic equation are \[x=\sqrt{29}+6\] or \[x=6-\sqrt{29}\].
Note:
There are many methods to solve a quadratic equation, as the factorization method, completing the square method, and the formula method. We can use any of them to solve a quadratic equation. The formula method should be preferred because it gives two roots of the equation, whether they are real or imaginary.
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