## abbie1 Group Title . one year ago one year ago

1. satellite73

2. satellite73

start with $2x^2-8x+7=0$ then use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $$a=2,b=-8,c=7$$

3. saifoo.khan

@satellite73

4. satellite73

second step is $x=\frac{8\pm\sqrt{8^2-4\times 2\times 7}}{2\times 2}$ then a bunch of arithmetic

5. satellite73

oh ok we compute $\frac{8\pm\sqrt{64-56}}{4}$ $=\frac{8\pm\sqrt{8}}{4}$ $=\frac{8\pm2\sqrt{2}}{4}$ $=\frac{2(4+\sqrt{2})}{4}$ $=\frac{4+\sqrt{2}}{2}$

6. satellite73

rewrite $$\sqrt{8}$$ as $$2\sqrt{2}$$ then factor and cancel details are above

7. satellite73

it is because $$\sqrt{8}=\sqrt{4\times 2}=\sqrt{4}\sqrt{2}=2\sqrt{2}$$ in simplest radical form

8. satellite73

then be careful with factoring before you cancel the common factor of 2 top and bottom

9. candyme

Ok, you're going to multiply the terms in the radical$x=\frac{8 \pm \sqrt{64-56}}{4}$$x= \frac{8 \pm \sqrt{8}}{4}$$x=\frac{8 \pm \sqrt{2*4}}{4}$The 4 comes out as a 2 since the square root of 4 is 2$x= \frac{8 \pm 2\sqrt{2}}{4}$Divide by 2$x=\frac{4 \pm \sqrt{2}}{2}$

10. satellite73

it could be written as $2\pm\frac{\sqrt{2}}{2}$

11. candyme

Satellite is right, I was about to say that :)