How do you solve the quadratic $4x^2-12x=16$ using any method?

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Answer 1 · 2016-09-18T09:53:39

$x=4$
$x=-1$

Given -

$4x^2-12x=16$

$4x^2-12x-16=0$

$4x^2+4x-16x-16=0$

$4x(x+1)-16(x+1)=0$

$(4x-16)(x+1)=0$

$4x-16=0$

$x=16/4=4$

$x+1=0$

$x=-1$

Answer 2 · 2016-09-18T10:03:42

$x=4$ or $x = -1$

$4x^2-12x=16$

$4x^2 -12x-16 =0" "larr"div4$

$x^2-3x-4=0" "larr$ factorise

Find factors of 4 which subtract to give 3.

$(x-4)(x+1) =0$

If $x-4 = 0, " then " x = 4$

If $x+1 = 0, " then " x = -1$

How do you solve the quadratic 4x^2-12x=164x^2-12x=16 using any method?