How to use the discriminant to find out how many real number roots an equation has for $4 x^{2} - 4 x - 1 = 0$ $4x^2 -4x -1 = 0$ ?

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Answer 1 · 2015-06-03T15:56:21

$D = d^{2} = 16 + 16 = 32 \to d = \pm 4 \sqrt{2}$ $D = d^2 = 16 + 16 = 32 -> d = +-4sqrt2$

There are 2 real roots (2 x-intercepts).

$x = \frac{4}{8} \pm \frac{4 \sqrt{2}}{8} = \frac{1}{2} \pm \frac{\sqrt{2}}{2}$ $x = 4/8 +- (4sqrt2)/8 = 1/2 +- (sqrt2)/2$

How to use the discriminant to find out how many real number roots an equation has for 4x2−4x−1=04x^2 -4x -1 = 0?