
How do you solve using the completing square method \[{{z}^{2}}+10z=-24\]?
Answer
552k+ views
Hint: In this problem we have to solve the given quadratic equation and find the value of x. we can first find the complete square equation to find the value of x. we can then change it to a whole square format and we can take square root on both the sides to get the value of x.
Complete step by step solution:
We know that the given quadratic equation to be solved is,
\[{{z}^{2}}+10z=-24\] ……. (1)
We can now take the first two terms \[{{z}^{2}}+10z\]
We know that,
\[{{\left( z+a \right)}^{2}}={{z}^{2}}+2az+{{a}^{2}}\]
We can see that,
\[\begin{align}
& \Rightarrow 2a=10 \\
& \Rightarrow a=5 \\
& \Rightarrow {{a}^{2}}=25 \\
\end{align}\]
We can add the above value in both the left-hand side and the right-hand side of equation (1), to get a perfect square equation, we get
Now we can add 25 on both sides in the above equation (1), we get
\[\Rightarrow {{z}^{2}}+10z+25=-24+25\]….. (2)
We can write the left-hand side in whole square form as it is a perfect square equation, we get
\[\Rightarrow {{\left( z+5 \right)}^{2}}=1\]
Now we can take square root on both sides, we get
\[\Rightarrow z+5=\pm \sqrt{1}=\pm 1\]
Now we can separate the terms, we get
\[\begin{align}
& \Rightarrow z=-5+1=-4 \\
& \Rightarrow z=-5-1=-6 \\
\end{align}\]
Therefore, the value of \[z=-4,-6\].
Note: We can now verify to check whether the values we got are correct or not. We can substitute the values of x in the equation to check.
We can now take the equation (1) and substitute the value of \[z=-4,-6\], we get
When z = -4,
\[\begin{align}
& \Rightarrow {{\left( -4 \right)}^{2}}+10\left( -4 \right)=-24 \\
& \Rightarrow 16-40=-24 \\
\end{align}\]
When z = -3,
\[\begin{align}
& \Rightarrow {{\left( -6 \right)}^{2}}+10\left( -6 \right)=-24 \\
& \Rightarrow 36-60=-24 \\
\end{align}\]
Therefore, we can verify that the value for is \[z=-4,-6\].
Complete step by step solution:
We know that the given quadratic equation to be solved is,
\[{{z}^{2}}+10z=-24\] ……. (1)
We can now take the first two terms \[{{z}^{2}}+10z\]
We know that,
\[{{\left( z+a \right)}^{2}}={{z}^{2}}+2az+{{a}^{2}}\]
We can see that,
\[\begin{align}
& \Rightarrow 2a=10 \\
& \Rightarrow a=5 \\
& \Rightarrow {{a}^{2}}=25 \\
\end{align}\]
We can add the above value in both the left-hand side and the right-hand side of equation (1), to get a perfect square equation, we get
Now we can add 25 on both sides in the above equation (1), we get
\[\Rightarrow {{z}^{2}}+10z+25=-24+25\]….. (2)
We can write the left-hand side in whole square form as it is a perfect square equation, we get
\[\Rightarrow {{\left( z+5 \right)}^{2}}=1\]
Now we can take square root on both sides, we get
\[\Rightarrow z+5=\pm \sqrt{1}=\pm 1\]
Now we can separate the terms, we get
\[\begin{align}
& \Rightarrow z=-5+1=-4 \\
& \Rightarrow z=-5-1=-6 \\
\end{align}\]
Therefore, the value of \[z=-4,-6\].
Note: We can now verify to check whether the values we got are correct or not. We can substitute the values of x in the equation to check.
We can now take the equation (1) and substitute the value of \[z=-4,-6\], we get
When z = -4,
\[\begin{align}
& \Rightarrow {{\left( -4 \right)}^{2}}+10\left( -4 \right)=-24 \\
& \Rightarrow 16-40=-24 \\
\end{align}\]
When z = -3,
\[\begin{align}
& \Rightarrow {{\left( -6 \right)}^{2}}+10\left( -6 \right)=-24 \\
& \Rightarrow 36-60=-24 \\
\end{align}\]
Therefore, we can verify that the value for is \[z=-4,-6\].
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