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

Question

5244 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-12T19:06:29

$x = \frac{3}{2} \pm \sqrt{3}$ $x=3/2+-sqrt(3)$

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

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

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

It uses different numbers.

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

$y = 0 = 4 {(x - \frac{12}{4 \times 2})}^{2} + k - 3$ $y=0=4(x-12/(4xx2))^2+k-3$

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

Set $4 {(- \frac{3}{2})}^{2} + k = 0$ $" "4(-3/2)^2+k=0$

$4 \times \frac{9}{4} + k = 0 \Rightarrow k = - 9$ $4xx9/4+k=0" "=>" "k =-9$
.......................................................................

$y = 0 = 4 {(x - \frac{3}{2})}^{2} - 12$ $y=0=4(x-3/2)^2-12$

$+ \frac{12}{4} = {(x - \frac{3}{2})}^{2}$ $+12/4=(x-3/2)^2$

$x - \frac{3}{2} = \pm \sqrt{3}$ $x-3/2=+-sqrt(3)$

Exact answer is:
$x = \frac{3}{2} \pm \sqrt{3}$ $x=3/2+-sqrt(3)$

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

Tony B

How do you solve 4x^2 - 12x - 3 = 04x2−12x−3=04x^2 - 12x - 3 = 0 by completing the square?