Question

How do you find all real solutions to the following equation: ${{\left( {{x}^{2}}-7x+11 \right)}^{{{x}^{2}}-2x-35}}=1?$

Answer 1

Hint: What we have to remember is that if the power of any number or variable is zero, then the whole value becomes one. That is, in mathematical terms we say, \[{{x}^{0}}=1,\] for all non-zero values of $x.$ We make use of quadratic formulas to find the solution.

Complete step-by-step solution:
Consider the given algebraic equation ${{\left( {{x}^{2}}-7x+11 \right)}^{{{x}^{2}}-2x-35}}=1.$
From what we have learnt, we know that \[{{x}^{0}}=1,\] for all non-zero values of $x.$
So, here, what we have to do is to find the values of $x$ so that the value of the expression ${{x}^{2}}-2x-35$ becomes zero.
That is, ${{x}^{2}}-2x-35=0$ which is a quadratic equation.
Therefore, we are going to use the quadratic formula to find the solutions of the expression.
 For an equation of the form $a{{x}^{2}}+bx+c=0,$ the quadratic formula is,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here we can see, $a=1,b=-2,c=-35$
Therefore, we write,
$\Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4.1.\left( -35 \right)}}{2.1}$
From this we will get the following equation,
\[\Rightarrow x=\dfrac{-2\pm \sqrt{4+4.35}}{2}\] we know that $x-\left( -y \right)=x+y$
Now, we will get,
$\Rightarrow x=\dfrac{-2\pm \sqrt{4+140}}{2}$
Thus, we get,
$\Rightarrow x=\dfrac{-2\pm \sqrt{144}}{2}$
We know the square root of $144$ is $12.$
Therefore, we are led to,
$\Rightarrow x=\dfrac{-2\pm 12}{2}=-1\pm 6,$ by cutting out the common factor, that is, $2.$
So, we conclude that the values of $x$ are $x=-1+6=5$ and $x=-1-6=-7.$
We say that the values of $x=\left\{ -7,5 \right\}$ that make ${{x}^{2}}-2x-35=0.$
Now we get, ${{\left( {{x}^{2}}-7x+11 \right)}^{{{x}^{2}}-2x-35}}={{\left( {{x}^{2}}-7x+11 \right)}^{0}}=1.$
As we have said earlier, ${{x}^{0}}=1$ is true only for non-zero values of $x.$
 Therefore, it is confirmed that ${{x}^{2}}-7x+11\ne 0.$
Now, we have to find for which values of $x,$ ${{x}^{2}}-7x+11=0.$
Then we exclude those values from the solution set.
Take ${{x}^{2}}-7x+11=0.$
Use quadratic formula,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now we get,
 $\Rightarrow x=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4.1.11}}{2.1}$
From this, we obtain,
$\Rightarrow x=\dfrac{7\pm \sqrt{49-44}}{2}$ since $-\left( -x \right)=x$
Now we get,
$\Rightarrow x=\dfrac{7\pm \sqrt{5}}{2}$
We understand that whenever $x\ne \dfrac{7\pm \sqrt{5}}{2},$ ${{x}^{2}}-7x+11\ne 0.$
And whenever $x=\left\{ -7,5 \right\},$ ${{x}^{2}}-2x-35=0$
Therefore, $x=\left\{ -7,5 \right\}$ is the solution of the given equation.

Note: We should remember that ${{x}^{0}}=1$ is not true when $x=0.$ We use a quadratic formula to find the solutions of a quadratic equation. The equation has real roots only if $\sqrt{{{b}^{2}}-4ac}>0.$