
How do you solve using the square method \[{{x}^{2}}+2x-63=0\] ?
Answer
558.9k+ views
Hint: To solve this question by the method of completing the square we will follow a few steps. First of all, write the given equation in decreasing order of the degree of \[x\]. then separate the constant or the number to the right side of the equation. Now, divide all the terms with the coefficient of \[{{x}^{2}}\]to make the coefficient of \[{{x}^{2}}\] one. After that we have to divide the coefficient of \[x\] by two and then square it, add this value to both sides of the equation. Now, take the square root on both sides. After that simplify to get the value of \[x\].
Complete step-by-step answer:
In this question the given equation is \[{{x}^{2}}+2x-63=0\]
Since, the coefficient \[{{x}^{2}}\] of is already one so, we don’t have to perform that step. Now moving the constant term to the other side and dividing the coefficient of \[x\] by two and adding value on both sides.
\[\Rightarrow {{\left( x+1 \right)}^{2}}={{\left( \dfrac{2}{2} \right)}^{2}}+63\]
\[\begin{align}
& \Rightarrow {{x}^{2}}+2x+{{\left( \dfrac{2}{2} \right)}^{2}}={{\left( \dfrac{2}{2} \right)}^{2}}+63 \\
& \\
\end{align}\]
After that we can see that on the left and side \[{{\left( x+1 \right)}^{2}}={{x}^{2}}+2x+{{\left( \dfrac{2}{2} \right)}^{2}}\]
Therefore, replacing it we get \[{{\left( x+1 \right)}^{2}}={{\left( \dfrac{2}{2} \right)}^{2}}+63\]. Now we will do squaring on both sides and we will get \[\left( x+1 \right)=\pm \sqrt{{{\left( \dfrac{2}{2} \right)}^{2}}+63}\] simplifying it further
\[\begin{align}
& \left( x+1 \right)=\pm \sqrt{{{\left( 1 \right)}^{2}}+63} \\
& \Rightarrow \left( x+1 \right)=\pm \sqrt{64} \\
& \Rightarrow \left( x+1 \right)=\pm \sqrt{{{8}^{2}}} \\
& \Rightarrow \left( x+1 \right)=\pm 8 \\
& \Rightarrow x=8-1\ or\ -8-1 \\
& \Rightarrow x=7\ or\ x=-9
\end{align}\]
Thus, we got two values for \[x\] after squaring both sides.
Hence after solving we got \[x=7\]or \[x=-9\]
Note: Since in this question we can see we have to follow a procedure to solve it by completing the square method therefore, keep in mind all the steps. While applying square root on both sides don’t forget to add both the signs as value from the root is always positive regardless of the number inside the square root. Also perform the calculations carefully in order to avoid silly mistakes.
Complete step-by-step answer:
In this question the given equation is \[{{x}^{2}}+2x-63=0\]
Since, the coefficient \[{{x}^{2}}\] of is already one so, we don’t have to perform that step. Now moving the constant term to the other side and dividing the coefficient of \[x\] by two and adding value on both sides.
\[\Rightarrow {{\left( x+1 \right)}^{2}}={{\left( \dfrac{2}{2} \right)}^{2}}+63\]
\[\begin{align}
& \Rightarrow {{x}^{2}}+2x+{{\left( \dfrac{2}{2} \right)}^{2}}={{\left( \dfrac{2}{2} \right)}^{2}}+63 \\
& \\
\end{align}\]
After that we can see that on the left and side \[{{\left( x+1 \right)}^{2}}={{x}^{2}}+2x+{{\left( \dfrac{2}{2} \right)}^{2}}\]
Therefore, replacing it we get \[{{\left( x+1 \right)}^{2}}={{\left( \dfrac{2}{2} \right)}^{2}}+63\]. Now we will do squaring on both sides and we will get \[\left( x+1 \right)=\pm \sqrt{{{\left( \dfrac{2}{2} \right)}^{2}}+63}\] simplifying it further
\[\begin{align}
& \left( x+1 \right)=\pm \sqrt{{{\left( 1 \right)}^{2}}+63} \\
& \Rightarrow \left( x+1 \right)=\pm \sqrt{64} \\
& \Rightarrow \left( x+1 \right)=\pm \sqrt{{{8}^{2}}} \\
& \Rightarrow \left( x+1 \right)=\pm 8 \\
& \Rightarrow x=8-1\ or\ -8-1 \\
& \Rightarrow x=7\ or\ x=-9
\end{align}\]
Thus, we got two values for \[x\] after squaring both sides.
Hence after solving we got \[x=7\]or \[x=-9\]
Note: Since in this question we can see we have to follow a procedure to solve it by completing the square method therefore, keep in mind all the steps. While applying square root on both sides don’t forget to add both the signs as value from the root is always positive regardless of the number inside the square root. Also perform the calculations carefully in order to avoid silly mistakes.
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