How do you solve the quadratic $3 r^{2} - 5 = - 10$ $3r^2-5=-10$ using any method?

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Answer 1 · 2016-12-02T04:02:08

$r = \pm \frac{\sqrt{15}}{3} i$ $r=+-sqrt15/3i$

$3 r^{2} - 5 = - 10$ $3r^2-5=-10$
$a^{2} a + 5 a a^{2} a + 5 a a a$ $color(white)(a^2a)+5color(white)(aa^2a)+5color(white)(aaa)$ Add 5 to both sides

$3 r^{2} = - 5$ $3r^2=-5$

$\frac{3 r^{2}}{3} = \frac{- 5}{3} a a a$ $(3r^2)/3=(-5)/3color(white)(aaa)$ Divide both sides by 3

$r^{2} = - \frac{5}{3}$ $r^2=-5/3$

$\sqrt{r^{2}} = \sqrt{- \frac{5}{3}} a a a$ $sqrt(r^2)=sqrt(-5/3)color(white)(aaa)$ Square root both sides

$r = \pm i \sqrt{\frac{5}{3}} a a$ $r=+-isqrt(5/3)color(white)(aa)$ The negative inside the square root comes out as $i$ $i$ .

$r = \pm i \sqrt{\frac{5}{3}} \cdot \sqrt{\frac{3}{3}} a a a$ $r=+-isqrt(5/3)*sqrt(3/3)color(white)(aaa)$ Rationalize the denominator

$r = \pm i \frac{\sqrt{15}}{3} = \pm \frac{\sqrt{15}}{3} i$ $r=+-isqrt(15)/3=+-sqrt15/3i$

How do you solve the quadratic 3r^2-5=-103r2−5=−103r^2-5=-10 using any method?