How do you work out the x intercepts of the equation y=2(x-4)^2-3 by factorising? Thanks!

This is what I got up to: $y = 2 {\left(x - 4\right)}^{2} - 3$ $0 = 2 {\left(x - 4\right)}^{2} - {\sqrt{3}}^{2}$ $0 = 2 \left(x - 4 - \sqrt{3}\right) \left(x - 4 + \sqrt{3}\right)$ I know the third step is wrong because you have to include the 2 somewhere, but when I apply the null factor law the 2 sort of disappears and I get it wrong. Please help!!

Jun 12, 2018

The x intecepts are ${x}_{1} = 4 + \sqrt{\frac{3}{2}}$ and ${x}_{2} = 4 - \sqrt{\frac{3}{2}}$

Explanation:

As ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, you could write the function as:
$y = \left[\sqrt{2} \left(x - 4\right) - \sqrt{3}\right] \left[\sqrt{2} \left(x - 4\right) + \sqrt{3}\right]$

$y = 0$ whenever either of the parentheses is 0, i.e.
sqrt2(x-4)-sqrt3]=0, i.e ${x}_{1} = 4 + \sqrt{\frac{3}{2}}$
or sqrt2(x-4)+sqrt3]=0, i.e. ${x}_{2} = 4 + \sqrt{\frac{3}{2}}$

Jun 12, 2018

$x = 4 \pm \sqrt{\frac{3}{2}}$

Explanation:

The solution is a lot simpler than you think. Just add a 3 to both sides of the equation and then you can get rid of the 2. Like this :

$0 = 2 {\left(x - 4\right)}^{2} - 3$
$3 = 2 {\left(x - 4\right)}^{2}$
$\frac{3}{2} = {\left(x - 4\right)}^{2}$

Then :

${\left(x - 4\right)}^{2} - {\sqrt{\frac{3}{2}}}^{2} = 0$
$\left(x - 4 - \sqrt{\frac{3}{2}}\right) \cdot \left(x - 4 + \sqrt{\frac{3}{2}}\right) = 0$

so $x = 4 + \sqrt{\frac{3}{2}}$ or $x = 4 - \sqrt{\frac{3}{2}}$

Jun 12, 2018

color(red)(=> x=4+-sqrt(3/2)

color(magenta)(=>x=4+sqrt(3/2) or x=4-sqrt(3/2)

Explanation:

$y = 2 {\left(x - 4\right)}^{2} - 3$

Solving for the x set $\implies y = 0$

$\implies 0 = 2 {\left(x - 4\right)}^{2} - 3$

$\implies 2 {\left(x - 4\right)}^{2} = 3$

$\implies {\left(x - 4\right)}^{2} = \frac{3}{2}$

$\implies x - 4 = \sqrt{\frac{3}{2}}$

color(red)(=> x=4+-sqrt(3/2)

color(magenta)(=>x=4+sqrt(3/2) or x=4-sqrt(3/2)

~Hope this helps! :)

Jun 12, 2018

$x = 4 \pm \sqrt{\frac{3}{2}}$

Explanation:

$\text{to solve for x set } y = 0$

$2 {\left(x - 4\right)}^{2} - 3 = 0$

$\text{add 3 to both sides}$

$2 {\left(x - 4\right)}^{2} = 3$

$\text{divide both sides by 2}$

${\left(x - 4\right)}^{2} = \frac{3}{2}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x - 4\right)}^{2}} = \pm \sqrt{\frac{3}{2}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$x - 4 = \pm \sqrt{\frac{3}{2}}$

$\text{add 4 to both sides}$

$x = 4 \pm \sqrt{\frac{3}{2}} \leftarrow \textcolor{red}{\text{exact values}}$

$x = 4 + \sqrt{\frac{3}{2}} \approx 5.22 \text{ to 2 dec. places}$

$\text{or "x=4-sqrt(3/2)~~2.78" to 2 dec. places}$