Quadratic formulas

**Maccabhoy** · May 13th 2009, 01:50 AM

Would like a bit of help on a maths quiz. No 1...I know what I am supposed to do, factorising, and putting a,b,c into the x = b+/- sqrt(b^2 etc etc formula... but im getting the wrong answer. So maybe my factorising is wrong... Can someone show me a step by step process for this question.

And No 2, Im a bit lost, so any help would be appreciated.

1)Give the turning point of the following quadratic and state whether it is a maximum or minimum.

y = (−8+9x) (−4x−5)

2) If a^2 = b^2 + 3 and a − b = 6, what is a + b? Please give an exact answer, using fractions if necessary; no decimals.

**SENTINEL4** · May 13th 2009, 02:06 AM

Originally Posted by Maccabhoy

2) If a^2 = b^2 + 3 and a − b = 6, what is a + b? Please give an exact answer, using fractions if necessary; no decimals.

a-b=6 $\Rightarrow$ a=b+6 put this to $a^2=b^2+3$ and you have

$(b+6)^2=b^2+3$ $\Rightarrow$ $b^2+12b+36=b^2+3$ $\Rightarrow$ $12b+36=3$ $\Rightarrow$ $12b=3-36$ $\Rightarrow$ $12b=-33$ $\Rightarrow$ $b=\frac{-33}{12}$

Now replace $b=\frac{-33}{12}$ back to $a=b+6$ and you find a

$a=\frac{-33}{12}+6$ $\Rightarrow$ $a=\frac{-33}{12}+\frac{72}{12}$ $\Rightarrow$ $a=\frac{39}{12}$

So now

$a+b=\frac{39}{12}+(\frac{-33}{12})=\frac{39}{12}-\frac{33}{12}=\frac{6}{12}=\frac{1}{2}$

**yeongil** · May 13th 2009, 02:20 AM

Here's another way to do #2:

If $a^{2} = b^{2} + 3$
then $a^{2} - b^{2} = 3$ .
LHS is a difference of 2 squares. Factor into $(a - b)(a + b) = 3$ .
Replace (a - b) with 6: $6(a + b) = 3$ .
Divide both sides by 6 to get $a + b = \frac{1}{2}$ .

01

**Maccabhoy** · May 13th 2009, 02:27 AM

Thanks guys. Much appreciated.

Any ideas on No 1?

**Maccabhoy** · May 13th 2009, 03:02 AM

Just a couple more...

Give the turning point of the following quadratic and state whether it is a maximum or minimum.
1) y = −4 x^2−9 x−7

and Find all solutions of the equation:
2) x^3−5 X^2 +4 x=0

**SENTINEL4** · May 13th 2009, 03:22 AM

Originally Posted by Maccabhoy

and Find all solutions of the equation:
2) x^3−5 X^2 +4 x=0

$x^3-5x^2+4x=0$ $\Rightarrow$ $x(x^2-5x+4)=0$
Now it would be x=0 or $x^2-5x+4=0$

By solving the $x^2-5x+4=0$ ypu find x=1 , x=4

So the solutions are : x=0 , x=1 , x=4

**Maccabhoy** · May 13th 2009, 04:15 AM

Originally Posted by SENTINEL4

$x^3-5x^2+4x=0$ $\Rightarrow$ $x(x^2-5x+4)=0$
Now it would be x=0 or $x^2-5x+4=0$

By solving the $x^2-5x+4=0$ ypu find x=1 , x=4

So the solutions are : x=0 , x=1 , x=4

Thanks for that!

Would you have an idea about my original No 1 question...

1)Give the turning point of the following quadratic and state whether it is a maximum or minimum.

y = (−8+9x) (−4x−5)

I was solving it by factorisation, but am getting a wrong answer...

Is this wrong... y = -36x^2 - 12x + 40.... and then working from there..

**SENTINEL4** · May 13th 2009, 04:45 AM

Originally Posted by Maccabhoy

1)Give the turning point of the following quadratic and state whether it is a maximum or minimum.

y = (−8+9x) (−4x−5)

Well i don't remember well, but i think that turning point has something to do with the second derivative.

**TheMasterMind** · May 13th 2009, 05:50 AM

Originally Posted by SENTINEL4

Well i don't remember well, but i think that turning point has something to do with the second derivative.

y = (−8+9x) (−4x−5)

simple foil gives:

y=-36x^2-13x+40

complete the square, this will give you the quadratic in vertex form, which is the turning point.

although after the foil, it is evident that it is a negative x value, this a maximum because the parabola is opening down

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