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mariannu
Solve equation by factoring. 4) x^2 - 15x=-56 3) a^2- 64=0 I just want to know how you to move numbers to the other side of the equal sign.
first you have to move all numbers to one side: x^2-15x+56=0
@mariannu for a, if you were to add 56 (or subtract -56) from each side, you would get what dave said, it's really just that easy.
then you are going to find common factors: (x- 8)(x-7 )=0 then set each side to 0: x-8=0 and x-7=0 solve for x on both sides, and x=8 and x=7
but this is solving each equation by factorin I am solving quadratics. not that.
Or my bad Xd your doing exactly what it is? XD
hahaha i was gonna say, you asked for factoring, which is one way to solve quadratics haha do you need help solving the second one?
Because the other one I did It was 5v^2=-30v-40 XD and I gott v=+- jsqaure root of -14
and I didn't have to move it to the other side. What is the easiet way to solve a^2 -64=0 and x^2-15x=-56
I just don't knwo how to solve these type of quadratrics : p XD accept the one that I have the answer for I know how to do that type of problem
the easiest way to solve x^2-15=-56 is how i outlined it above, realizing you have to move -56 to the other side is honestly the trickiest part. and a^2-64=0 is the same thing the first basic rule to solving quadratics is setting it to zero. Ill walk you through the second problem you have and explain what i am doing with each step
a^2-64=0 --------1) already set to 0 so we are good there (a- )(a+ )=0 -------2) since you know that a is squared break it up, and because 64 is negative you know one will be positive and one will be negative. 64/8=8 -----------3) testing for the factor (i chose 8 first because 8^2 is 64) (a+8)(a-8)=0-------4) plug in the factor (a+8)=0 and (a-8)=0 ----5)set both sides to 0 a=0-8 and a=0-8---6)solve both new found equations a=-8 ---------------7)-8 is the answer
Wouldn't it be like this? |dw:1366267980891:dw| the answer/
Since it has a negative? XD Because the other one that I had the answer to was switched to a positive because of imaginery solution
i do not believe it would be under a square root. i do not remember my rules to imaginary numbers that well tho. and it wouldnt be plus minus because there shouldnt be a square root,
Well then ok I will just ask my teacher tomorrow morning if its correect then thank u though Xd and one more I just want to know if this is correct |dw:1366268265757:dw||dw:1366268307044:dw|
It suppose to be a -4 k
not by my work, x should equal 4 and negative 8
Well thank you though. I appreciate it. ^.^ Going to study mre for the test. XD
haha no problem. i am not sure exactly what your teacher is looking for, so i guess dont quote me exactly if i am not doing it in the form your teacher wants
Well its ok. You still helped me at least understand it ^.^ though.