## adnanchowdhury 3 years ago How do I show the real roots in for this graph? http://d.pr/i/COS9

1. dpaInc

be truthful to yourself.

|dw:1334914514106:dw| This is my drawing.

3. Lumenaire

Real roots means x-axis intercepts

So is it 2 and where line y= x+16 intercepts the x axis?

5. Lumenaire

6. Lumenaire

and yes, it is 2 and -16

Are you sure? because when x=0, y=16?

8. Lumenaire

does not make a difference, you want y = 0

9. Lumenaire

and y = 0. when x = -16

10. Lumenaire

sorry, just read the question again, the roots are the solutions to the question, which means the intercepts between the 2 graphs: |dw:1334914789804:dw|

11. Lumenaire

clearly one of the roots is at x = 0 and the only other root is where the line extends to meet the curve

okay, so there are 3 roots?

13. Lumenaire

(x-2)^4 = x+16 so x^4 -8x^3+24x^2-32x+16 = x-16

14. Lumenaire

16 cancels, and you can factor out x, so one solution is at x = 0

15. Lumenaire

now, x^3 -8x^2+24x-32 = 0

16. Lumenaire

this is a third order polynomial with one x-intercept, so the solution to this equation, and x = 0, are the only 2 roots

and so how do i show this on the graph?

18. Lumenaire

|dw:1334915153788:dw|

19. Lumenaire

like that :-)