How do you solve $8 x^{2} - 5 = - 4 x$ $8x^2-5=-4x$ ?

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Answer 1 · 2018-06-15T05:26:56

$\Rightarrow x \approx 0.57915$ $=> x approx 0.57915$

or

$x \approx - 1.07915$ $x approx -1.07915$

$8 x^{2} - 5 = - 4 x$ $8x^2-5=-4x$

$\Rightarrow 8 x^{2} - 5 + 4 x = 0$ $=> 8x^2-5+4x=0$

$\Rightarrow 8 x^{2} + 4 x - 5 = 0$ $=> 8x^2+4x -5=0$ --------(1)

using quadratic equation formula:

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

Substitute the corresponding values of a,b, and c from (1)

$\Rightarrow x = \frac{- 4 \pm \sqrt{4^{2} - 4 \times 8 \times (- 5)}}{2 \times 8}$ $=> x= (-4+- sqrt(4^2-4xx8xx(-5)))/(2xx8)$

$\Rightarrow x = \frac{- 4 \pm \sqrt{16 + 160}}{16}$ $=> x= (-4+- sqrt(16+ 160))/(16)$

$\Rightarrow x = \frac{- 4 \pm \sqrt{176}}{16}$ $=> x= (-4+- sqrt(176))/(16)$

$\Rightarrow x \approx - \frac{4}{16} \pm \frac{13.266499}{16}$ $=> xapprox -4/16 +- 13.266499/16$ ---truncating value of $\sqrt{176}$ $sqrt(176)$

$\Rightarrow x \approx - \frac{1}{4} \pm \frac{13.2665}{16}$ $=>xapprox -1/4 +- 13.2665/16$

$\Rightarrow x \approx - 0.25 \pm 0.82915$ $=> x approx - 0.25 +- 0.82915$

$\Rightarrow x \approx - 0.25 + 0.82915 or x \approx - 0.25 - 0.82915$ $=> x approx - 0.25 + 0.82915 or x approx - 0.25 - 0.82915$

$\Rightarrow x \approx 0.57915$ $=> x approx 0.57915$

or

$x \approx - 1.07915$ $x approx -1.07915$

How do you solve 8x^2-5=-4x8x2−5=−4x8x^2-5=-4x?