How do you evaluate $\sqrt{b^{2} - 4 a c}$ $sqrt(b^2-4ac)$ for a=2, b=-4, c=-6?

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How do you evaluate $\sqrt{b^{2} - 4 a c}$ $sqrt(b^2-4ac)$ for a=2, b=-4, c=-6?

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Answer 1 · 2016-12-19T23:45:16

The discriminant $D = {(- 4)}^{2} - 4 \cdot 2 \cdot (- 6)$ $D=(-4)^2-4*2*(-6)$

Explanation:

$D = 16 - (- 48) = 16 + 48 = 64$ $D=16-(-48)=16+48=64$ This is $> 0$ $>0$
so there will be two solutions to the equation $2 x^{2} - 4 x - 6 = 0$ $2x^2-4x-6=0$

The square root $\sqrt{D} = \sqrt{64} = 8$ $sqrtD=sqrt64=8$

The two solutions being:

$x_{1, 2} = \frac{- b \pm \sqrt{D}}{2 a} = \frac{+ 4 \pm 8}{4} = 1 \pm 2$ $x_(1,2)=(-b+-sqrtD)/(2a)=(+4+-8)/4=1+-2$

$\to x = + 3 or x = - 1$ $->x=+3orx=-1$

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How do you evaluate $\sqrt{b^{2} - 4 a c}$ $sqrt(b^2-4ac)$ for a=2, b=-4, c=-6?

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How do you evaluate $\sqrt{b^{2} - 4 a c}$ $sqrt(b^2-4ac)$ for a=2, b=-4, c=-6?