How do you solve $4+4x=2x^2$ using the quadratic formula?

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Answer 1 · 2017-05-18T15:42:58

$x=1+-sqrt(3)" "$ Exact value

$x~~-0.732050....-> -0.73$ to 2 decimal places

$x~~color(white)(-)2.732050...... ->+2.73$ to 2 decimal places

Subtract 4 and $4x$ from both sides giving:

$y=0=2x^2-4x-4$

Consider the standard form of $y=ax^2+bx+c$

That at $y=0$ we have $x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case: $a=2"; "b=-4"; "c=-4$

so by substitution we have:

$x=(+4+-sqrt((-4)^2-4(2)(-4)))/(2(2))$

$x=1+-sqrt(48)/4$

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Note that $sqrt(48)/4$ is the same as $sqrt(48/4^2)=sqrt(3)$

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$x=1+-sqrt(3)" "$ Exact value

$x~~-0.732050....-> -0.73$ to 2 decimal places

$x~~color(white)(-)2.732050...... ->+2.73$ to 2 decimal places

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