How do you solve $4x² - 4x – 1 = 0$ ?

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Answer 1 · 2016-03-21T13:38:15

$x=(1+sqrt2)/(2)$

$color(blue)(4x^2-4x-1=0$

This is a Quadratic equation (in form $ax^2+bx+c=0$ )

$color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)$

Where

$color(red)(a=4,b=-4,c=-1$

$rarrx=(-(-4)+-sqrt(-4^2-4(4)(-1)))/(2(4))$

$rarrx=(4+-sqrt(-4^2-4(4)(-1)))/(8)$

$rarrx=(4+-sqrt(16-(-16)))/(8)$

$rarrx=(4+-sqrt(16+16))/(8)$

$rarrx=(4+-sqrt(32))/(8)$

$rarrx=(4+-sqrt(16*2))/(8)$

$rarrx=(4+-4sqrt2)/(8)$

$rarrx=(cancel(4)^1+-cancel(4)^1sqrt2)/(cancel8)^2$

$color(green)(rArrx=(1+-sqrt2)/(2)$

How do you solve 4x² - 4x – 1 = 04x² - 4x – 1 = 0?