How do you find the solution set for 0 = x²+0.212x-0.0024 ?

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Answer 1 · 2015-05-16T09:14:26

${(x + 0.106)}^{2} = x^{2} + 0.212 x + 0.011236$ $(x+0.106)^2 = x^2+0.212x+0.011236$

So

$0 = x^{2} + 0.212 x - 0.0024$ $0 = x^2+0.212x-0.0024$

$= (x^{2} + 0.212 x + 0.011236) - 0.011236 - 0.0024$ $= (x^2+0.212x+0.011236)-0.011236-0.0024$

$= {(x + 0.106)}^{2} - (0.011236 + 0.0024)$ $= (x+0.106)^2-(0.011236+0.0024)$

$= {(x + 0.106)}^{2} - 0.013636$ $= (x+0.106)^2-0.013636$

Add $0.013636$ $0.013636$ to both sides to get

${(x + 0.106)}^{2} = 0.013636$ $(x+0.106)^2 = 0.013636$

So

$x + 0.106 = \pm \sqrt{0.013636}$ $x+0.106 = +-sqrt(0.013636)$

Hence

$x = - 0.106 \pm \sqrt{0.013636}$ $x = -0.106 +-sqrt(0.013636)$