Question

Solve :- 1. | -x² + x - 1 | = |x + 5|

2. | x² + 3x + 5| + x² + 3x + 5 = 0.​

Answer 1

Solution :-1

• | m | = - m if m < 0 and m if m > 0.

• - x² + x - 1 is always negative ( D < 0 and a > 0. )

• Hence, | -- x² + x - 1 | = x² - x + 1

Case 1:- x > -5 => | x + 5| = x + 5

The quadratic equation becomes

=> x² - x + 1 = x + 5

=> x² - 2x - 4 = 0.

By using quadratic formula,

x = $\frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a}$ or x = $\frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

=> x = $\frac{ 2 + \sqrt{ 20 } }{2}$ or x = $\frac{ 2 - \sqrt{ 20 } }{2}$

=> x = 1 + $\sqrt{5}$ or x = 1 - $\sqrt{5}$

Both solution are acceptable since x > -5 in both cases.

Case 2:- When x < - 5 , |x + 5 |= -x - 5

The equation becomes

=> x² - x + 1 = -x - 5

=> x² + 6 = 0 ( not possible )

Hence, solution to the equation is x = 1 + √5 or 1 - √5

Solution :- 2

x² + 3x + 5 > 0 (always, since D > 0 and a > 0 )

| x² + 3x + 5| = x² + 3x + 5

The equation becomes

=> x² + 3x + 5 + x² + 3x + 5 = 0

=> 2(x² + 3x + 5) = 0

=> x² + 3x + 5 = 0 ( not possible , since x² + 3x + 5 > 0)

Hence, no solution in this case.