Question

2. The sum of all possible values of a satisfyin
(x²-5x+6)x²-1 =1​

Answer 1

Answer:

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Answer 2

ANSWER

Given:

$\:\:\:\bullet\:\:\:(x^2-5x+6)^{x^2-1}=1$

To Find:

• Sum of all possible values of x

Solution:

We are given that,

$\implies(x^2-5x+6)^{x^2-1}=1$

So, first of all we will find the possible values of x, and then add them all. For finding the possible values we need to understand the following points to get the answer equal to 1.

• If a number is raised to the power 0, it equals 1. That is, a^0 = 1, where a ≠ 0
• If the number 1 is raised to any power, it equals 1. That is 1^a = 1
• If the number -1 is raised to any even power, it equals 1. That is (-1)^a = 1, where a is an even no.

So, now we can see that there are 3 cases for which, value of the given expression equals 1. We will take each case and discuss it one by one.

CASE 1: $\bf x^2-5x+6\neq0\:and\:x^2-1=0$

So,

$\implies x^2-5x+6\neq0\:and\:x^2-1=0$

$\implies x^2-2x-3x+6\neq0\:and\:x^2=1$

$\implies x(x-2)+3(x-2)\neq0\:and\:x^2=1$

$\implies (x-2)(x+3)\neq0\:and\:x=\sqrt1$

Hence,

$\implies x\neq2,-3\:and\:x=\pm1$

Therefore,

$\bf\implies x=\pm1$

CASE 2: $\bf x^2-5x+6=1$

So,

$\implies x^2-5x+6=1$

$\implies x^2-5x+6-1=0$

$\implies x^2-5x+5=0$

Using Quadratic formula:

$\implies x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

So,

$\implies x=\dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(5)}}{2(1)}$

Hence,

$\implies x=\dfrac{5\pm\sqrt{25-20}}{2}$

Therefore,

$\bf\implies x=\dfrac{5\pm\sqrt{5}}{2}$

CASE 3: $\bf x^2-5x+6=-1\:and\:x^2-1=even$

So,

$\implies x^2-5x+6=-1\:and\:x^2-1=even$

$\implies x^2-5x+6+1=0\:and\:x^2=even+1$

$\implies x^2-5x+7=0\:and\:x^2=odd$

Using Quadratic formula:

$\implies x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

So,

$\implies x=\dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(7)}}{2(1)} \:and\:x^2=odd$

$\implies x=\dfrac{5\pm\sqrt{25-28}}{2} \:and\:x^2=odd$

Therefore,

$\bf\implies x=\dfrac{5\pm\sqrt{-3}}{2} \:and\:x^2=odd$

As, the roots of the equation are unreal(or imaginary), we will ignore this case.

So, the possible values of x are:

$\:\:\bullet\:\:x=1$

$\:\:\bullet\:\:x=-1$

$\:\:\bullet\:\:x=\dfrac{5+\sqrt5}{2}$

$\:\:\bullet\:\:x=\dfrac{5-\sqrt5}{2}$

Now, we need to find that,

⇒ Sum of all possible values of x

So,

$\implies\text{Sum of all possible values of x =} (1)+(-1)+\dfrac{5+\sqrt5}{2}+\dfrac{5-\sqrt5}{2}$

Hence,

$\implies\text{Sum of all possible values of x =} 1-1+\dfrac{5+\sqrt5+5-\sqrt5}{2}$

$\implies\text{Sum of all possible values of x =} \dfrac{10}{2}$

Therefore,

⇒ Sum of all possible values of x = 5