
How do you solve \[{{x}^{2}}+5x-84=0\]?
Answer
540.6k+ views
Hint: Use the middle term split method to factorise \[{{x}^{2}}+5x-84\] and write it as the product of two terms given as \[\left( x-a \right)\left( x-b \right)\], where ‘a’ and ‘b’ are called zeroes of the equation. Now, substitute each term equal to 0 and find the two values of x and get the answer.
Complete step-by-step solution:
Here, we have been provided with the quadratic equation \[{{x}^{2}}+5x-84=0\] and we are asked to solve it. That means we have to find the values of x.
Now, let us apply the middle term split method to factorize the given equation first. Let us assume \[{{x}^{2}}+5x-84=f\left( x \right)\], so we have,
\[\Rightarrow f\left( x \right)={{x}^{2}}+5x-84\]
According to the middle term split method we have to break 5x into two terms such that their sum is 5x and the product is \[-84{{x}^{2}}\]. So, breaking 5x into 12x and (-7x), we have,
(i) \[12x+\left( -7x \right)=5x\]
(ii) \[12x\times \left( -7x \right)=-84{{x}^{2}}\]
Therefore, both the conditions of the middle term split method are satisfied, so we can write the given expression as,
\[\begin{align}
& \Rightarrow f\left( x \right)={{x}^{2}}+12x-7x-84 \\
& \Rightarrow f\left( x \right)=x\left( x+12 \right)-7\left( x+12 \right) \\
& \Rightarrow f\left( x \right)=\left( x+12 \right)\left( x-7 \right) \\
& \Rightarrow f\left( x \right)=0 \\
& \Rightarrow \left( x+12 \right)\left( x-7 \right)=0 \\
\end{align}\]
Substituting each term equal to 0, we get,
\[\Rightarrow \left( x+12 \right)=0\] or \[\left( x-7 \right)=0\]
\[\Rightarrow x=-12\] or \[x=7\]
Hence, the solutions of the given equation are: - x = 7 or x = -12.
Note: One may note that here we have applied the middle term split method to get the values of x. You can also apply the discriminant method to get the answer. In that condition assume the coefficient of \[{{x}^{2}}\] as ‘a’, the coefficient of x as b and the constant term as c, and apply the formula: - \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to solve for the values of x. There can be a third method also, known as completing the square method, to solve the question. Note that the discriminant formula is obtained from completing the square method.
Complete step-by-step solution:
Here, we have been provided with the quadratic equation \[{{x}^{2}}+5x-84=0\] and we are asked to solve it. That means we have to find the values of x.
Now, let us apply the middle term split method to factorize the given equation first. Let us assume \[{{x}^{2}}+5x-84=f\left( x \right)\], so we have,
\[\Rightarrow f\left( x \right)={{x}^{2}}+5x-84\]
According to the middle term split method we have to break 5x into two terms such that their sum is 5x and the product is \[-84{{x}^{2}}\]. So, breaking 5x into 12x and (-7x), we have,
(i) \[12x+\left( -7x \right)=5x\]
(ii) \[12x\times \left( -7x \right)=-84{{x}^{2}}\]
Therefore, both the conditions of the middle term split method are satisfied, so we can write the given expression as,
\[\begin{align}
& \Rightarrow f\left( x \right)={{x}^{2}}+12x-7x-84 \\
& \Rightarrow f\left( x \right)=x\left( x+12 \right)-7\left( x+12 \right) \\
& \Rightarrow f\left( x \right)=\left( x+12 \right)\left( x-7 \right) \\
& \Rightarrow f\left( x \right)=0 \\
& \Rightarrow \left( x+12 \right)\left( x-7 \right)=0 \\
\end{align}\]
Substituting each term equal to 0, we get,
\[\Rightarrow \left( x+12 \right)=0\] or \[\left( x-7 \right)=0\]
\[\Rightarrow x=-12\] or \[x=7\]
Hence, the solutions of the given equation are: - x = 7 or x = -12.
Note: One may note that here we have applied the middle term split method to get the values of x. You can also apply the discriminant method to get the answer. In that condition assume the coefficient of \[{{x}^{2}}\] as ‘a’, the coefficient of x as b and the constant term as c, and apply the formula: - \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to solve for the values of x. There can be a third method also, known as completing the square method, to solve the question. Note that the discriminant formula is obtained from completing the square method.
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