How do you solve $4x^2 - 12x - 3 = 0$ by completing the square?

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Answer 1 · 2017-06-12T19:06:29

$x=3/2+-sqrt(3)$

$x~~-0.23 and x~~+3.23$

Given: $" "0=4x^2-12x-3$

For a complete explanation of the method see:
https://socratic.org/s/aFvmbGAs

It uses different numbers.

Set $y=0=4x^2-12x-3$

$y=0=4(x-12/(4xx2))^2+k-3$

$y=0=4(x-3/2)^2+k-3$
.........................................................................

Set $" "4(-3/2)^2+k=0$

$4xx9/4+k=0" "=>" "k =-9$
.......................................................................

$y=0=4(x-3/2)^2-12$

$+12/4=(x-3/2)^2$

$x-3/2=+-sqrt(3)$

Exact answer is:
$x=3/2+-sqrt(3)$

Approximate answer is:
$x=-0.232050.. and x=+3.232-50....$

Tony B

How do you solve 4x^2 - 12x - 3 = 04x^2 - 12x - 3 = 0 by completing the square?