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 Po is trying to solve the following equation by completing the square: $$49x^2+56x-64 = 0.$$He successfully rewrites the above equation in the following form: $$(ax + b)^2 = c,$$where $a$, $b$, and $c$ are integers and $a > 0$. What is the value of $a + b + c$?

 

Thanks

WhichWitchIsWhich  Oct 22, 2017

Best Answer 

 #5
avatar+5241 
+1

(x - \(\frac47\))2  = \(\frac{80}{49}\)

 

On this one...to get  a ,  b , and  c  as integers, we can multiply both sides by  72

 

72 * (x - \(\frac47\))2  = 72 * \(\frac{80}{49}\)

 

( 7 * (x - \(\frac47\)) )2  =  49 * \(\frac{80}{49}\)

 

(7x - 4)2  =  80

 

And...    7 + (-4) + 80  =  83      smiley

hectictar  Oct 22, 2017
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6+0 Answers

 #1
avatar+26329 
+1

Follow the procedure shown at: https://web2.0calc.com/questions/if-we-express-3x-2-6x-2-in-the-form-a-x-h-2-k-what-is-a-h-k

Alan  Oct 22, 2017
 #2
avatar+272 
+1

Thanks really but I'm not sure I understand the answer there... Can you expound please? thanks so much!

WhichWitchIsWhich  Oct 22, 2017
 #3
avatar+78632 
+2

49 x^2+ 56x- 64 = 0       divide through by 49

 

x^2 +  56/49 - 64/49  = 0

 

x^2 -  8/7 - 64/49  = 0       add 64/49 to both sides

 

x^2 - 8/7  =   64/49

 

Take 1/2 of 8/7 = 8/14= 4/7......square it  =  16/49   add it to both sides

 

x^2 - 8/7 + 16/49  =   64/49 + 16/49

 

x^2 - 8/7 +  16/49  =  80 / 49         factor the left side

 

(x - 4/7)^2  = 80/49    

 

So     a = 1, b = -4/7  and c = 80/49

 

And   a + b + c  =  1 - 4/7 + 80/49 =   [ 49 - 28 + 80] / 49  =  101 / 49

 

 

cool cool cool

CPhill  Oct 22, 2017
 #4
avatar+272 
+2

Thanks this really helped!

WhichWitchIsWhich  Oct 22, 2017
 #5
avatar+5241 
+1
Best Answer

(x - \(\frac47\))2  = \(\frac{80}{49}\)

 

On this one...to get  a ,  b , and  c  as integers, we can multiply both sides by  72

 

72 * (x - \(\frac47\))2  = 72 * \(\frac{80}{49}\)

 

( 7 * (x - \(\frac47\)) )2  =  49 * \(\frac{80}{49}\)

 

(7x - 4)2  =  80

 

And...    7 + (-4) + 80  =  83      smiley

hectictar  Oct 22, 2017
 #6
avatar+78632 
+1

Thanks, hectictar...I forgot that a, b, c had to be integers......

 

cool cool cool

CPhill  Oct 22, 2017

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