Any chance i can have a step by step demo of these. thanks.

-x + 8X - 15 = 0

6x^2 - 17x + 5 = 0

-3b^2 - 26b = -9

Thanks

2. Originally Posted by Brownhash
Any chance i can have a step by step demo of these. thanks.

-x^2 + 8X - 15 = 0

6x^2 - 17x + 5 = 0

-3b^2 - 26b = -9

Thanks
It should be squared...right?

For the first one, get rid of the negative sign in front of $\displaystyle x^2$:

$\displaystyle -(x^2-8x+15)=0 \implies (x-3)(x-5)=0 \implies {\color{red}\boxed{x=3}} \ \text{or} {\color{red}\boxed{x=5}}$.

For the second one, use the quadratic formula:

$\displaystyle x=\frac{-(-17)\pm\sqrt{(-17)^2-4(6)(5)}}{2(6)}=\frac{17\pm\sqrt{169}}{12}=\frac{1 7\pm 13}{12}\implies {\color{red}\boxed{x=\frac{5}{2}}} \ \text{or} \ {\color{red}\boxed{x=\frac{1}{3}}}$

Try doing the last one yourself. I'd recommend that you divide through by -3 to clean things up a little bit!

I hope this helped you out!

3. ## Thankyou

Thanks for the help there much appreciated.

4. Originally Posted by Brownhash
Any chance i can have a step by step demo of these. thanks.

-x + 8X - 15 = 0

6x^2 - 17x + 5 = 0

-3b^2 - 26b = -9

Thanks

Your second equation can be factored as well:

$\displaystyle 6x^2-17x+5$

My technique:
1. Multiply leading coef. times constant: 6 X 5 = 30
2. Find factors of 30 that add up to -17. That woud be -15 and -2
3. Replace -17x with these 2 factors.
4. Factor by grouping

Here goes:

$\displaystyle 6x^2-15x-2x+5=0$

$\displaystyle (6x^2-15x)-(2x+5)=0$

$\displaystyle 3x(2x-5)-1(2x-5)=0$

$\displaystyle (3x-1)(2x-5)=0$

$\displaystyle x=\frac{1}{3} \ or \ x=\frac{5}{2}$

5. Your third equation can be factored as well:

$\displaystyle -3b^2-26b=-9$

$\displaystyle -3b^2-26b+9=0$

Multiply by -1

$\displaystyle 3b^2+26b-9=0$

My technique:
1. Multiply leading coef. times constant: 3 X 9 = 27
2. Find factors of 27 that add up to 26. That woud be -1 and 27
3. Replace 26b with these 2 factors.
4. Factor by grouping

Here goes:

$\displaystyle 3b^2+27b-b-9=0$

$\displaystyle (3b^2+27b)-b-9=0$

$\displaystyle 3b(b+9)-1(b+9)=0$

$\displaystyle (3b-1)(b+9)=0$

$\displaystyle b=\frac{1}{3} \ or \ b=-9$