Theory of Equations 1 Question 15
16. The sum of all real values of $x$ satisfying the equation $\left(x^{2}-5 x+5\right)^{x^{2}+4 x-60}=1$ is
(2016 Main)
(a) 3
(b) -4
(c) 6
(d) 5
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Solution:
- Given, $\left(x^{2}-5 x+5\right)^{x^{2}+4 x-60}=1$
Clearly, this is possible when I. $x^{2}+4 x-60=0$ and $x^{2}-5 x+5 \neq 0$ or II. $x^{2}-5 x+5=1$ or
III. $x^{2}-5 x+5=-1$ and $x^{2}+4 x-60=$ Even integer.
$$ \begin{array}{lrl} \text { Case I } & \text { When } x^{2}+4 x-60 & =0 \\ \Rightarrow & x^{2}+10 x-6 x-60 & =0 \\ \Rightarrow & x(x+10)-6(x+10) & =0 \\ \Rightarrow & (x+10)(x-6) & =0 \\ \Rightarrow & x=-10 \text { or } x & =6 \end{array} $$
Note that, for these two values of $x, x^{2}-5 x+5 \neq 0$
$$ \begin{array}{lr} \text { Case II When } & x^{2}-5 x+5=1 \\ \Rightarrow & x^{2}-5 x+4=0 \\ \Rightarrow & x^{2}-4 x-x+4=0 \\ \Rightarrow & x(x-4)-1(x-4)=0 \\ \Rightarrow & (x-4)(x-1)=0 \Rightarrow x=4 \text { or } x=1 \\ \text { Case III When } & x^{2}-5 x+5=-1 \\ \Rightarrow & x^{2}-5 x+6=0 \\ \Rightarrow & x^{2}-2 x-3 x+6=0 \\ \Rightarrow & x(x-2)-3(x-2)=0 \\ \Rightarrow & (x-2)(x-3)=0 \\ \Rightarrow & x=2 \text { or } x=3 \end{array} $$
Now, when $x=2, x^{2}+4 x-60=4+8-60=-48$, which is an even integer.
When $x=3, x^{2}+4 x-60=9+12-60=-39$, which is not an even integer.
Thus, in this case, we get $x=2$.
Hence, the sum of all real values of
$$ x=-10+6+4+1+2=3 $$
