How do you use the quadratic formula to solve $x^{2} + 5 = 4 x$ $x^2+5=4x$ ?

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How do you use the quadratic formula to solve $x^{2} + 5 = 4 x$ $x^2+5=4x$ ?

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Answer 1 · 2015-05-03T19:43:44

The quadratic formula:
$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$
can be used to find root solutions to quadratics in the form
$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

So first we need to convert the given equation into this form:
$x^{2} + 5 = 4 x$ $x^2+5 = 4x$

$\to x^{2} - 4 x + 5 = 0$ $rarr x^2-4x+5 = 0$

Solutions are
$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (5)}}{2 (1)}$ $x= (-(-4) +- sqrt((-4)^2 -4(1)(5)))/(2(1))$

$x = 2 \pm \frac{\sqrt{(- 4)}}{2}$ $x= 2+-sqrt((-4))/2$

Since the discriminant is negative
there are no Real value solutions to this equation
but within Complex numbers, the solution is
$x = 2 \pm i$ $x=2+-i$

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How do you use the quadratic formula to solve $x^{2} + 5 = 4 x$ $x^2+5=4x$ ?

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