How do you solve $4 x^{2} = 0$ $4x^2=0$ using the quadratic formula?

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Answer 1 · 2014-10-25T13:45:28

$x = \frac{- b \pm \sqrt[2]{b^{2} - 4 a c}}{2 a}$ $x = (-b +- root2 (b^2 - 4ac))/(2a)$

Your equation

$4 x^{2} = 0$ $4x^2 = 0$

can be re-written into

$4 x^{2} + 0 x + 0 = 0$ $4x^2 + 0x + 0 = 0$

Substituting the coefficients into the quadratic formula, we have

$x = \frac{- b \pm \sqrt[2]{b^{2} - 4 a c}}{2 a}$ $x = (-b +- root2 (b^2 - 4ac))/(2a)$

$\Rightarrow x = \frac{- 0 \pm \sqrt[2]{0^{2} - 4 (4) (0)}}{2 (4)}$ $=> x = (-0 +- root2 (0^2 - 4(4)(0)))/(2(4))$

$\Rightarrow x = \frac{- 0 \pm \sqrt[2]{0 - 0}}{8}$ $=> x = (-0 +- root2 (0 - 0))/(8)$

$\Rightarrow x = 0$ $=> x = 0$

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