anonymous one year ago 3x^2-9x-12=0 by completing the square with explanation if possible.

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1. anonymous

divide the whole thing by 3

2. anonymous

x^2-3x-4=0

3. anonymous

$3(x^2-3x-4) = 0 \rightarrow 3(x^2-3x)=4$

4. anonymous

to complete the square...you take the middle term and divide by 2 and square it

5. anonymous

$3(x^2-3x+(\frac{ 3 }{ 2 })^2)=4+(\frac{ 3 }{ 2 })^2$

6. anonymous

looks like i forgot to x3 on the other end

7. anonymous

$3(x^2-3x+(\frac{ 3 }{ 2 })^2=12+3(\frac{ 3 }{ 2 })^2$

8. anonymous

its 3 not 4

9. anonymous

and also completing the square method

10. anonymous

you should get $3(x^2-3x+(\frac{ 9 }{ 4 })=18\frac{ 3 }{ 4 }$

11. anonymous

now you can factorize the$3(x^2-3x+\frac{ 9 }{ 4 })$part

12. anonymous

you catching on?

13. anonymous

yes, thank you very much

14. anonymous

you can factorize that on your own?

15. anonymous

yes, i was having trouble getting to that point though. appreciate your help!

16. anonymous

yw

17. mathstudent55

Start with your equation: $$3x^2-9x-12=0$$ Notice that there is a common factor of 3 for all terms, so divide both sides by 3: $$x^2-3x-4=0$$ Now add 4 to both sides. $$x^2-3x=4$$ The term that needs to be added to the left sides is obtained this way. Take half of the x-term coefficient and square it. Half of -3 is $$-\dfrac{3}{2}$$. The square of that is $$\dfrac{9}{4}$$ Now we add $$\dfrac{9}{4}$$ to both sides $$x^2-3x + \dfrac{9}{4}=4 + \dfrac{9}{4}$$ We rewrite the left sides as the square of a binomial and we add the numbers on the right side. $$\left( x - \dfrac{3}{2} \right)^2 = \dfrac{16}{4} + \dfrac{9}{4}$$ $$\left( x - \dfrac{3}{2} \right)^2 = \dfrac{25}{4}$$ Now we apply the rule: If $$x^2 = k$$, then $$x = \pm \sqrt k$$ to get: $$x - \dfrac{3}{2} = \pm \sqrt{\dfrac{25}{4}}$$ $$x - \dfrac{3}{2} = \pm \dfrac{5}{2}$$ We separate the equation into two parts: $$x - \dfrac{3}{2} = \dfrac{5}{2}$$ or $$x - \dfrac{3}{2} = -\dfrac{5}{2}$$ Add $$\dfrac{3}{2}$$ to both sides in both equations: $$x = \dfrac{8}{2}$$ or $$x = -\dfrac{2}{2}$$ Reduce the fractions: $$x = 4$$ or $$x = -1$$

18. mathstudent55

The step of completing the square by adding the square of half of the x-term coefficient to both sides only works if the coefficient of the x^2 term is 1. After we divided the equation by 3 on both sides, we did have a coefficient of 1 for the x^2 term, so we were able to complete the square that way.

i agree with you