Question

`\sqrt{{x}^(2) + 3x - 28 } + \sqrt{ {x}^(2) - 2x - 8 } = \sqrt{ {6x}^(2) - 11x - 52}`

find x in the equation above​

Answer 1

The calculated values of x in the equation are x = 4, x = 2 and x = -5/3

How to determine the value of x in the equation

From the question, we have the following parameters that can be used in our computation:

`\sqrt{{x}^(2) + 3x - 28 } + \sqrt{ {x}^(2) - 2x - 8 } = \sqrt{ {6x}^(2) - 11x - 52}`

Take the square of both sides of the equation

This gives

`{x}^(2) + 3x - 28 + {x}^(2) - 2x - 8 + 2\sqrt{({x}^(2) + 3x - 28) * ({x}^(2) - 2x - 8})} = {6x}^(2) - 11x - 52`

This gives

`2x^2 + x - 36 + 2√(x^4 + x^3 - 42x^2 + 32x + 224) = {6x}^(2) - 11x - 52`

Evaluating the like terms, we have

`2√(x^4 + x^3 - 42x^2 + 32x + 224) = 4x^2 - 12x -16`

So, we have

`√(x^4 + x^3 - 42x^2 + 32x + 224) = 2x^2 - 6x -8`

Square both sides

`x^4 + x^3 - 42x^2 + 32x + 224 = 4x^4 - 24x^3 + 4x^2 + 96x + 64`

Evaluating the like terms, we have

`3x^4 - 25x^3 + 46x^2 + 64x - 160 = 0`

Solving graphically, we have

x = 4, x = 2 and x = -5/3

Hence, the solutions are x = 4, x = 2 and x = -5/3