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

Hint: 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
\[\begin{align}
  & {{x}^{2}}+5x-50=0 \\
 & \Rightarrow {{x}^{2}}+10x-5x-50=0 \\
 & \Rightarrow x\left( x+10 \right)-5\left( x+10 \right)=0 \\
 & \Rightarrow \left( x+10 \right)\left( x-5 \right)=0 \\
\end{align}\]
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,
\[\begin{align}
  & 1+5-10=6-10 \\
 & =-4
\end{align}\]
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 \[{{a}^{0}}=1\].