```Question 94447
Solve the absolute value equation or indication the equation has no solution. The answers are in solution sets with square roots in them:

|2x^2 - x - 1| = 3; vertical lines are uppercase back-slash on my keyboard
:
We have two quadratic equation to solve
2x^2 - x - 1 = +3
2x^2 - x - 1 - 3 = 0
2x^2 - x - 4 = 0
and
2x^2 - x - 1 = -3
2x^2 - x - 1 + 3 = 0
2x^2 - x + 2 = 0
:
The first one can be solved using the quadratic formula:
a = 2; b = -1; c = -4
:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
{{{x = (-(-1) +- sqrt(-1^2 - 4 * 2 * -4 ))/(2*2) }}}
:
{{{x = (+1 +- sqrt(1 - (-32) ))/(4) }}}
:
{{{x = (+1 +- sqrt(1 + 32 ))/(4) }}}
:
{{{x = (+1 +- sqrt(33))/(4) }}}
:
Two solutions:
{{{x = (1 + sqrt(33))/(4) }}}
and
{{{x = (1 - sqrt(33))/(4) }}}
:
:
The 2nd equation: 2x^2 - x + 2; has no real roots. We can see that by looking
at the discriminant: b^2 - 4*a*c. In this equation: a=2, b=-1, c=+2 so we have:
-1^2 - 4*2*2 = 1 - 16 = -15
Remember the discriminant rules. Less than 0, no real roots.
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