How do you solve $-2k ^ { 2} + 8= 0$ ?

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How do you solve $-2k ^ { 2} + 8= 0$ ?

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Answer 1 · 2018-04-24T13:10:08

$k=+2 or -2$

Explanation:

$-2k^2+8=0$

$-2k^2+8-8=0-8$

$-2k^2=-8$

$((-2k^2)/-2)=-8/-2$

$k^2=4$

$sqrt(k^2)= +-sqrt(4)$

$k=+-2$

Answer 2 · 2018-04-24T13:10:19

$k=+-2$

Explanation:

$"subtract 8 from both sides of the equation"$

$rArr-2k^2=-8$

$"divide both sides by "-2$

$rArrk^2=4$

$color(blue)"take the square root of both sides"$

$rArrk=+-sqrt4larrcolor(blue)"note plus or minus"$

$rArrk=+-2$

Answer 3 · 2018-04-24T13:12:19

$k=-2,2$

Explanation:

Solve:

$-2k^2+8=0$

Subtract $8$ from both sides.

$-2k^2=-8$

Divide both sides by $-2$ .

$k^2=(-8)/(-2)$

$k^2=8/2$

$k^2=4$

Take the square root of both sides.

$sqrt(k^2)=+-sqrt4$

Apply rule $sqrt(a^2)=a$ .

$k=+-2$

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How do you solve $-2k ^ { 2} + 8= 0$ ?

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How do you solve -2k ^ { 2} + 8= 0-2k ^ { 2} + 8= 0?

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How do you solve $-2k ^ { 2} + 8= 0$ ?