Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
766
3
avatar

How would I find the roots of 9x^2-6x+1=64, without using the quadratic formula?

 

So far I have, 9x^2 - 6x - 63 = 0, how do I return it to it's factored form? 

 Aug 16, 2015

Best Answer 

 #1
avatar+118696 
+8

9x26x+1=649x26x63=03(3x22x21)=03x22x21=0

 

You need 2 numbers that multiply to -3*21=-63    and add to -2

That is really easy.  

One has to be neg and the other positive (since they mult to a neg)

The bigger one is the neg one because that add to a neg.

there absolute value is only 2 apart.

So

 

they are  -9 and +7

write -2x as -9x+7x

 

3x22x21=03x29x+7x21=0$nowfactorinpairs$3x(x3)+7(x3)=0(3x+7)(x3)=03x+7=0orx3=0x=7/3orx=+3$slightcorrectionherethanksAlan:)$

.
 Aug 16, 2015
 #1
avatar+118696 
+8
Best Answer

9x26x+1=649x26x63=03(3x22x21)=03x22x21=0

 

You need 2 numbers that multiply to -3*21=-63    and add to -2

That is really easy.  

One has to be neg and the other positive (since they mult to a neg)

The bigger one is the neg one because that add to a neg.

there absolute value is only 2 apart.

So

 

they are  -9 and +7

write -2x as -9x+7x

 

3x22x21=03x29x+7x21=0$nowfactorinpairs$3x(x3)+7(x3)=0(3x+7)(x3)=03x+7=0orx3=0x=7/3orx=+3$slightcorrectionherethanksAlan:)$

Melody Aug 16, 2015
 #2
avatar+21 
+5

Thanks for the help I was really having trouble :)

 Aug 16, 2015
 #3
avatar+118696 
0

You are welcome Purpe.

I am glad that you have signed up.

Welcome to the forum :)

 Aug 16, 2015

3 Online Users

avatar