## kanigayan Group Title can some one tell me how to get factors i mean method 3b^2-20b-12 2 years ago 2 years ago

1. fontez

x=−b±(√b2−4ac)/2a

2. kanigayan

aaaaa what's that ? i don't think that's the way

3. amitlpu91

this will give u two value for x..........& say it x1,x2...... now factor is (x-x1)(x-x2)...ok

4. kanigayan

|dw:1330616930509:dw|

5. fontez

what??!! no no!

6. kanigayan

(b-6) (3b-2) am i right ?

7. ash2326

we have $3b^2-20b-12$ standard quadratic equation $ax^2+bx+c=0$ we know its solution is given as $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ here a=3 b=-20 and c=-12 so $x=\frac{-(-20)\pm \sqrt{(-20)^2-4\times 3\times -12}}{2\times 3}$ we get $x=\frac{20\pm \sqrt{400+144}}{6}$ now we get $x=\frac{20\pm \sqrt{400+144}}{6}$ we get $x=\frac{20\pm \sqrt{544}}{6}$ $544= 17\times 16\times 2$ so $x=\frac{20\pm \sqrt{16\times 34}}{6}$ we get $x=\frac{20\pm 4\sqrt{ 34}}{6}$ let's divide by 2 $x=\frac{10\pm \sqrt{34}}{3}$ so the two factors are $(x-(\frac{10+ \sqrt{34}}{3}))(x-(\frac{10- \sqrt{34}}{3}))$

8. ash2326

Sorry it'd be $(x-(\frac{10+ 2\sqrt{34}}{3}))(x-(\frac{10 -2\sqrt{34}}{3}))$ I missed the 2:(

9. kanigayan

ah that's too advance i guess but right factors (b-6)(3b-2)

10. ash2326

No, these are not the factors $(b-6)(3b-2)= 3b^2-20b+12$ but in your question you have $3b^2-20b-12$

11. kanigayan

yah :) do u know how to get (b-6)(3b-2)