# Thread: help please last few q's

1. ## help please last few q's

(7/3) ^-2 how can i find the value of this as an integer.

Im not sure how to sketch the graph y=9-x^2

Also..i dnt know if this will go right on here..

a=3/2 b=9-17(17 in a sqr root) c=9+17(also in sqr root)
_____ ______
4 4

How can i shpw that a+b+c =abc do i need to add them all and then multiply them all and it should be the same?

Thnx

2. Originally Posted by Chez_
(7/3) ^-2 how can i find the value of this as an integer.
This makes no sense. Can you provide the original question, in full, and then show your work and give your thoughts on the solution?

Im not sure how to sketch the graph y=9-x^2
Are you required to be sure? You should be learning. You may experience periods of wonder and absence of clarity. Complete the square. Find the zeros. Find the intercepts. Find the axis of symmetry. Identify the curve.

Also..i dnt know if this will go right on here..

a=3/2 b=9-17(17 in a sqr root) c=9+17(also in sqr root)
_____ ______
4 4
It didn't. HTML ignores extra spaces. Learn just a hair of LaTex and you'll be there. Think about the Rational Root Theorem. Try to imagine what it has to do with those irrational roots. If a cubic expression has a factor (3x-2), can you write a general quadratic expression with the information from the other two zeros?

How can i shpw that a+b+c =abc do i need to add them all and then multiply them all and it should be the same?
You can't, but you may wish to play with (a+b+c)^2 and see what comes out.

3. Originally Posted by Chez_
(7/3) ^-2 how can i find the value of this as an integer.

Im not sure how to sketch the graph y=9-x^2

Also..i dnt know if this will go right on here..

a=3/2 b=9-17(17 in a sqr root) c=9+17(also in sqr root)
_____ ______
4 4

How can i shpw that a+b+c =abc do i need to add them all and then multiply them all and it should be the same?

Thnx

This is just the indices rules. If negative power, then just flip the fraction, then just solve normally how you would:
$(\frac{7}{3})^{-2}$
$(\frac{3}{7})^2$
$\frac{9}{49}$

To draw a quadratic, observe the shape and intercepts.
$y=9-x^2$
Intecepts:
Crosses $y$ when $x=0$ $\implies y=9$
Crosses $x$ when $y=0$ $\implies x = +/- \sqrt 9 = +/- 3$
Shape: $-ve \therefore$ 'n' shaped.

Yes, your method of working out $a+b+c = abc$ is correct. Test both and show they are same. In your current question, it doesn't:

$a+b+c = abc$
$a = \frac{3}{2}, b=9- \sqrt17, c=9+\sqrt17$
$abc= (\frac{3}{2})(9- \sqrt17)(9+\sqrt17) = 96$
$a + b + c = (\frac{3}{2})+(9- \sqrt17)+(9+\sqrt17) = 19.5$

EDIT: But, in the exam, they could ask one that does work e.g: $a=1, b=2, c=3$.

4. (7/3) ^-2 how can i find the value of this as an integer.
Problem with this is that the result is not an integer. The result will be a real number that is rational. A number is rational if it can be displayed as a fraction of integers.