Question

The sum of all real values of x satisfying the equation (x^2-5x+5)^x^2+4x-60=1 is

Answer 1

Answer:

sum = 3

Step-by-step explanation:

given in the question the equation:

(x² - 5x + 5)^(x² + 4x - 60) = 1

we know that any integer or expression when have power 0 is equal to 1

so,

(x² + 4x - 60) = 0

by using quadratic factorisation

(x-6)(x+10) = 0

x = 6

x = -10

= {−10,1,2,4,6}

Sum = -10+1+2+4+6

        = 3

so sum of all real values of x satisfying the equation = 3

Answer 2

$\large\underline\pink{Given:-}$

• The sum of all real values of x satisfying.the equation (x^2-5x+5)^x^2+4x-60=1 is

$\large\underline\pink{To find:-}$

• Fine the equation (x^2-5x+5)^x^2+4x-60=1 is

....?

$\large\underline\pink{Solutions:-}$

Given;

• (x^2-5x+5)^x^2+4x-60=0 clearly, this is possible when

1. x² + 4x - 60 and x² - 5x + 5 doesn't equal to 0
2. x² - 5x + 5 = 1
3. x² - 5x + 5 = -1 and x² + 4x - 60 = Ever integer.

Case 1.

when, x² + 4x - 60 = 0

⟹ x² + 10x - 6x - 60 = 0

⟹ x(x + 10) - 6(x + 10) = 0

⟹ (x + 10) (x - 6) = 0

⟹ x = -10 or x = 6

Note that, for these two value of x,x² - 5x + 5 doesn't equal to 0

Case 2.

when, x² - 5x + 5 = 0

⟹ x² - 5x + 4 = 0

⟹ x² - 4x - x + 4 = 0

⟹ x(x - 4) - 1(x - 4) = 0

⟹ (x - 4) (x - 1) = 0

⟹ x = 4 or x = 1

Case 3.

when, x² - 5x + 5 = 0

⟹ x² - 5x + 6 = 0

⟹ x² - 2x - 3x + 6 = 0

⟹ x(x - 2) - 3(x - 2) = 0

⟹ (x - 2) (x - 3) = 0

⟹ x = 2 or x = 3

Now,

when x = 2, x² + 4x - 60 = 4 + 8 - 60 = -48, which is an even integer.

when x = 3, x² + 4x - 60 = 9 + 12 - 60 = -39, which in not an even integer.

Thus, in this case, we get x = 2

hence, the sum of all real value of x = -10 + 6 + 4 + 1 + 2 = 3