Question

Solve the equation: ${{\left( x+5 \right)}^{2}}-4\left( x+5 \right)=0$

Answer 1

Hint: We have to convert the given equation into a quadratic equation by expanding and then solve for x using the general formula for solving a quadratic equation. Also, we can define a new variable to be equal to x+5 as it is this combination of x which appears in the entire equation. Then, the equation would get transformed to a simpler quadratic equation which can be solved using the general methods for solving quadratic equations.

Complete step-by-step answer:

The given equation is : ${{\left( x+5 \right)}^{2}}-4\left( x+5 \right)=0\text{ }...................\text{(1}\text{.1)}$
Using the expansion formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we can expand equation (1.1) as
$\begin{align}
  & {{x}^{2}}+{{5}^{2}}+2\times 5x-4x-20=0 \\
 & \Rightarrow {{x}^{2}}+10x-4x+25-20=0 \\
 & \Rightarrow {{x}^{2}}+6x+5=0\text{ }...................\text{(1}\text{.2)} \\
\end{align}$
Now, the general formula to find the solution of a quadratic equation $a{{x}^{2}}+bx+c=0$ is given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\text{ }.............\text{(1}\text{.3)}$
In this case from equation (1.2), a=1, b=6 and c=5. Thus the solution is given by
\[x=\dfrac{-6\pm \sqrt{{{6}^{2}}-4\times 5}}{2}=\dfrac{-6\pm \sqrt{36-20}}{2}=\dfrac{-6\pm \sqrt{16}}{2}=\dfrac{-6\pm 4}{2}\]
Thus, the value of x will be $x=\dfrac{-6+4}{2}=-1$ or $x=\dfrac{-6-4}{2}=\dfrac{-10}{2}=-5$
Thus the solution of the given equation is x=-1 or x=-5.

Note: We could also have solved this equation by the following method
We can define another variable z as \[z=\left( x+5 \right)\ldots \ldots \ldots \ldots \ldots \ldots .\left( 1.4 \right)\]
Then the equation in the question becomes
${{z}^{2}}-4z=0$
Now, using the formula in equation (1.3), we get the solution of the above equation as
\[z=\dfrac{4\pm \sqrt{{{4}^{2}}-4\times 1\times 0}}{2}=\dfrac{4\pm 4}{2}\]
As here a=1, b=-4 and c=0. Therefore, we get the values of z as
$z=\dfrac{4-4}{2}=0$ or $z=\dfrac{4+4}{2}=4$
Now, using the relation of x and z using the equation (1.4), we obtain
$z=0\Rightarrow x+5=0\Rightarrow x=-5$ or $z=4\Rightarrow x+5=4\Rightarrow x=-1$
Which is the same as obtained by the previous method.
We should be careful to use the correct sign of the terms in equation (1.3) as each sign would correspond to a different solution of the same equation.