Question

Find the sum of the real values of $x$ satisfying the equation ${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}=1$
(a) $-5$
(b) 16
(c) 14
(d) $-4$

Answer 1

In this question, in order to find the sum of the real values of $x$ satisfying the equation ${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}=1$ we have to first find the values of $x$ satisfying the equation ${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}=1$. We will use the fact that for any real value $a$ we have ${{a}^{0}}=1$. Therefore we must have the equation in the form of ${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}={{2}^{0}}$. Now since the function ${{2}^{x}}$ is a one-to-one function therefore we can have $\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)=0$. We will then solve this equation to find the value of $x$ and then add all the values to get the desired answer.

Complete step-by-step answer :
We are given an equation in variable $x$, ${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}=1............(1)$
Since we know that for any real value $a$ we have ${{a}^{0}}=1$.
Therefore we have that the value of ${{2}^{0}}$ is equals to 1.
Now on substituting the value of 1 as ${{2}^{0}}$ in equation (1), we will get
${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}={{2}^{0}}$.
We also know that the function ${{2}^{x}}$ is a one-to-one function. That is
If ${{2}^{a}}={{2}^{b}}$, then we must have $a=b$.
Now on comparing the equation ${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}={{2}^{0}}$ with ${{2}^{a}}={{2}^{b}}$, we will have
$a=\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)$ and $b=0$.
Now using the above mention property of the function ${{2}^{x}}$, we will have
$\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)=0$
Now the above equation implies that either $x-1=0$ or ${{x}^{2}}+5x-50=0$.
Now if we consider the case when $x-1=0$, then we have
$x=1$
And if we consider the case when ${{x}^{2}}+5x-50=0$, we will have to factorise the equation ${{x}^{2}}+5x-50=0$ by splitting the middle term to get

& {{x}^{2}}+5x-50=0 \\\ & \Rightarrow {{x}^{2}}+10x-5x-50=0 \\\ & \Rightarrow x\left( x+10 \right)-5\left( x+10 \right)=0 \\\ & \Rightarrow \left( x+10 \right)\left( x-5 \right)=0 \\\ \end{aligned}$$ We now have $$\left( x+10 \right)\left( x-5 \right)=0$$. Therefore either we have $$x+10=0$$ or $$x-5=0$$. Therefore the possible values of $$x$$ are $$x=-10$$ or $$x=5$$. Hence in both the cases we have the following values of $$x$$ given by $$x=1,-10,5$$ which can satisfy the given equation $${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}=1$$. Now on adding all the possible values of $$x$$ which can satisfy the given equation $${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}=1$$, we get, $$\begin{aligned} & 1+5-10=6-10 \\\ & =-4 \end{aligned}$$ Therefore the sum of the real values of $$x$$ satisfying the equation $${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}=1$$ is equals to $$-4$$. **Hence option (d) is correct.** **Note** : In this problem, in order to find the sum of the real values of $$x$$ satisfying the equation $${{2}^{\left( x-1 \right)\left( {{x}^{2}}+5x-50 \right)}}=1$$ we are using the one-to-one property of the function $${{2}^{x}}$$ . Also we are using the fact that for any real value $$a$$ we have $${{a}^{0}}=1$$.