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I have this equation from this paper (Q.63)

Find the sum of all real values of $x$ satisfying the equation-$(x^2-5x+5)^{x^2+4x-60}=1$.

My attempt-

$(x^2-5x+5)^{x^2+4x-60}=(x^2-5x+5)^{(x-6)(x+10)}$

So, $x=6$ and $x=-10$ makes the power $0$ and hence the equation becomes one. So, the sum of the real values of $x$ satisfying the equation is $6+(-10)=-4$. But answer is $+3$.

Where have I gone wrong?

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  • $\begingroup$ I got the same paper for my mains exam :D $\endgroup$
    – Aditya Dev
    Commented May 21, 2016 at 7:56
  • $\begingroup$ @almagest you also missed $-1^\text{even}$. $x=2$ satisfies the condition. $\endgroup$
    – Aditya Dev
    Commented May 21, 2016 at 8:31
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    $\begingroup$ @AdityaDev Thanks. +1 for your soln. $\endgroup$
    – almagest
    Commented May 21, 2016 at 8:39
  • $\begingroup$ @AdityaDev How was the exam ? Did you get selected to advanced? $\endgroup$
    – N.S.JOHN
    Commented May 21, 2016 at 8:44
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    $\begingroup$ @AdityaDev what is the max score? $\endgroup$
    – N.S.JOHN
    Commented May 21, 2016 at 13:22

2 Answers 2

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You missed some cases.

The cases are: $\text{number}^0,1^\text{number},{-1}^{\text{even}}$

You have only considered the first case.

  1. For first case, $$x^2+4x-60=0$$ the roots are $6,-10$.

  2. For the second case, $$x^2-5x+5=1$$ the roots are $1,4$.

  3. For the third case, $$x^2-5x+5=-1$$ $x$ can be $2$ or $3$. If you substitute the value $2$ on the equation $x^2+4x-60$, you get $-48$ which satisfies the condition since $-1^{-48}=1$. If you substitute $x=3$, you get $-39$ which does not satisfy the given condition since $-1^{-39}\neq 1$. So only $x=2$ satisfies the condition.

The real values of $x$ are therefore: $1,2,4,6,-10$. The sum of real values of $x$ is therefore $3$.

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  • $\begingroup$ How was your advanced exam? $\endgroup$
    – Soham
    Commented Jun 3, 2016 at 15:30
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First of all, the equation in your link is : $(x^2-5x+5)^{x^2+4x-60}=1$.

Simply use $\ln$ in both sides :

$ \ln(x^2-5x+5)^{x^2+4x-60} = \ln1 \Leftrightarrow (x^2 + 4x - 60)\ln (x^2 - 5x + 5) = 0$.

That means that : $x^2 + 4x - 60 = 0$ or $(x^2 - 5x + 5) = 1$.

Of course, the functions inside of $\ln$ must be bigger than zero, do not forget to add that in your final answer after you've done these.

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    $\begingroup$ You missed $-1^\text{even}$ $\endgroup$
    – Aditya Dev
    Commented May 21, 2016 at 8:22

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