How do you evaluate $\sqrt{b^{2} - 4 a c}$ $sqrt(b^2-4ac)$ for a=2, b=-5, c=2?

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Answer 1 · 2017-02-28T20:10:39

$\pm 3$ $+-3$

$\sqrt{b^{2} - 4 a c} \to \sqrt{{(- 5)}^{2} - 4 (2) (2)}$ $sqrt(b^2-4ac)" "->" "sqrt((-5)^2-4(2)(2))$

$\sqrt{+ 25 - 16} = \sqrt{9} = \pm 3$ $sqrt(+25-16)=sqrt(9)=+-3$

Answer 2 · 2017-02-28T20:14:35

See the entire solution process below:

To solve this you must substitute:

$2$ $color(red)(2)$ for $a$ $color(red)(a)$ in the expression.

$- 5$ $color(blue)(-5)$ for $b$ $color(blue)(b)$ in the expression.

$2$ $color(purple)(2)$ for $c$ $color(purple)(c)$ in the expression.

$\sqrt{b^{2} - 4 a c}$ $sqrt(color(blue)(b)^2 - 4color(red)(a)color(purple)(c))$ becomes:

$\sqrt{{- 5}^{2} - (4 \times 2 \times 2)}$ $sqrt(color(blue)(-5)^2 - (4xxcolor(red)(2)xxcolor(purple)(2)))$

$\sqrt{25 - 16}$ $sqrt(25 - 16)$

$\sqrt{9} = \pm 3$ $sqrt(9) = +-3$

How do you evaluate sqrt(b^2-4ac)√b2−4acsqrt(b^2-4ac) for a=2, b=-5, c=2?