## anonymous one year ago HELP MEDAL GIVEN!!! Question Down Below I v

1. anonymous

The Quadratic Equation : $2x^2 + 4x + 3 = 0$ I have done part a) and b) so part c question is : find the value of : $\alpha^4 + \beta^4$

2. anonymous

@Nnesha @nincompoop @paki

3. anonymous

@Luigi0210 @campbell_st

4. campbell_st

well $\alpha^4 + \beta^4= (\alpha^2 + \beta^2)^2 - 2 \alpha^2 \beta^2$ so that is probably a start

5. campbell_st

so apply the same reasoning to deal with $(\alpha^2 + \beta^2)^2$ so that it is in terms of $\alpha + \beta$

6. campbell_st

ok... so $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$ so calculate this and then substitute it

7. campbell_st

and in the original equation $2\alpha^2 beat^2 = 2(\alpha \beta)^2$

8. anonymous

so for that I got the answer = 1

9. campbell_st

well that makes sense so you have $(1)^2 - 2(\alpha \beta)^2$

10. anonymous

a medal and i will help

11. anonymous

so, that equals = -7/2

12. campbell_st

that's it