## Dodo1 2 years ago 3-sqrt(x)/9-x. how do you break down?

1. UnkleRhaukus

$\frac{3-\sqrt x}{9-x}$ factorize the denominator use $(a+b)(a-b)=a^2-ab+ba-b^2=a^2-b^2$ cancel common factors

2. Dodo1

yes i know that but how do you that for sqrt?

3. UnkleRhaukus

use $$x=\sqrt x^2$$

4. Dodo1

mmm so 3-sqet(x)=(x^2-3^9)?

5. UnkleRhaukus

your looking at the numerator (bottom bit of the fraction) you should be looking at the denominator ( the bottom bit of the fraction) $9-x=(\cdots)(\cdots)$

6. Dodo1

9-x...... sqert(3)-(x)??

7. UnkleRhaukus

$9-x=3^2-\sqrt x^2=(\cdots)(\cdots)$

8. Dodo1

(3-sq$\left( 3-\sqrt{x} \right)$

9. Dodo1

(3-sq$\left( 3-\sqrt{x} \right)$(3-sq$\left( 3-\sqrt{x} \right)$?

10. Dodo1

$\left( 3-\sqrt{x} \right)$

11. Dodo1

\left( 3-sqrt{x} \right) \left( 3-sqrt{x} \right)

12. Dodo1

(3-sqrt(x)) (3-sqrt(x)

13. UnkleRhaukus

not quite$a^2-b^2=(a+b)(a-b)$ so$9-x=3^2-\sqrt x^2=$

14. Dodo1

3-sqrt(x)?

15. Dodo1

hope its right

16. UnkleRhaukus

$9-x=3^2-\sqrt x^2=(3+\sqrt x)(3-\sqrt x)$

17. UnkleRhaukus

so$\frac{3-\sqrt x}{9-x}=\frac{3-\sqrt x}{(3+\sqrt x)(3-{\sqrt x)}}$$\qquad\qquad =\frac{\cancel{3-\sqrt x}}{(3+\sqrt x)\cancel{(3-{\sqrt x)}}}$

18. Dodo1

Thanks!!

19. Dodo1

20. Dodo1

I got　(x+4)(x^2+4x+16)/(x+4)?

21. UnkleRhaukus

almost $(a^3+b^3)=(a+b)(a^2-ab+b^2)$

22. UnkleRhaukus

once you have fixed the sign error, cancel the common factor

23. Dodo1

Thank you it was very helpful!!