
How do you solve ${{x}^{2}}+x-72=0$?
Answer
542.4k+ views
Hint: The equation ${{x}^{2}}+x-72=0$ is a quadratic equation. So we can use the middle split method to solve it. According to the middle split method, we split the middle term of the quadratic polynomial as a sum of the two terms such that the product of the two is equal to the product of the first and the third terms. In the equation ${{x}^{2}}+x-72=0$ the first and the last terms are ${{x}^{2}}$ and $-72$ which makes the product equal to $-72{{x}^{2}}$. So the middle term $x$ is to be split as $x=9x-8x$. Then taking the factors common we will be able to factorize the quadratic polynomial and express it as a product of two linear factors. By equating each of the factors to zero, we will obtain two solutions of the equation.
Complete step by step answer:
The given equation is
${{x}^{2}}+x-72=0$
We use the middle term technique to factorize the quadratic polynomial in the above equation. We know from this technique that we have to split the middle term as a sum of two terms such that the product of the two terms is equal to the product of the first and the third terms.
From the above equation, we can observe that the first term is ${{x}^{2}}$ and the third term is $-72$. SO the required product is equal to $-72{{x}^{2}}$. So we split the middle term $x$ as $x=9x-8x$ in the above equation to get
$\Rightarrow {{x}^{2}}+9x-8x-72=0$
Taking $x$ common from the first two terms and $-8$ common from the last two terms, we get
$\Rightarrow x\left( x+9 \right)-8\left( x+9 \right)=0$
Now, taking $\left( x+9 \right)$ common, we get
$\Rightarrow \left( x+9 \right)\left( x-8 \right)=0$
So we have factorized the quadratic polynomial in the given equation. From the above equation we have the product of the two factors $\left( x+9 \right)$ and $\left( x-8 \right)$ equal to zero. So we can equate each to zero to get
\[\begin{align}
& \Rightarrow \left( x+9 \right)=0 \\
& \Rightarrow x=-9 \\
\end{align}\]
And
$\begin{align}
& \Rightarrow \left( x-8 \right)=0 \\
& \Rightarrow x=8 \\
\end{align}$
Hence, the solutions of the given equation are $x=-9$ and $x=8$.
Note: After splitting the middle term when you are taking a factor common from the either pair of the terms, the sign of the common term is to be the same as that of the first term of the pair encountered. For example, in the above case after splitting the middle term we got ${{x}^{2}}+9x-8x-72=0$. From the second pair of terms we took $-8$ common and not \[8\] due to the negative sign of the first term of the second pair, \[-8x\].
Complete step by step answer:
The given equation is
${{x}^{2}}+x-72=0$
We use the middle term technique to factorize the quadratic polynomial in the above equation. We know from this technique that we have to split the middle term as a sum of two terms such that the product of the two terms is equal to the product of the first and the third terms.
From the above equation, we can observe that the first term is ${{x}^{2}}$ and the third term is $-72$. SO the required product is equal to $-72{{x}^{2}}$. So we split the middle term $x$ as $x=9x-8x$ in the above equation to get
$\Rightarrow {{x}^{2}}+9x-8x-72=0$
Taking $x$ common from the first two terms and $-8$ common from the last two terms, we get
$\Rightarrow x\left( x+9 \right)-8\left( x+9 \right)=0$
Now, taking $\left( x+9 \right)$ common, we get
$\Rightarrow \left( x+9 \right)\left( x-8 \right)=0$
So we have factorized the quadratic polynomial in the given equation. From the above equation we have the product of the two factors $\left( x+9 \right)$ and $\left( x-8 \right)$ equal to zero. So we can equate each to zero to get
\[\begin{align}
& \Rightarrow \left( x+9 \right)=0 \\
& \Rightarrow x=-9 \\
\end{align}\]
And
$\begin{align}
& \Rightarrow \left( x-8 \right)=0 \\
& \Rightarrow x=8 \\
\end{align}$
Hence, the solutions of the given equation are $x=-9$ and $x=8$.
Note: After splitting the middle term when you are taking a factor common from the either pair of the terms, the sign of the common term is to be the same as that of the first term of the pair encountered. For example, in the above case after splitting the middle term we got ${{x}^{2}}+9x-8x-72=0$. From the second pair of terms we took $-8$ common and not \[8\] due to the negative sign of the first term of the second pair, \[-8x\].
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