SOLUTION: Find the real roots of each equation. If there are none state such.
16x^2 + 7x/7 = 4x+7/4
I tried finding common denominator
4(16x^2+7x)-7(4x+7)=0 all over 28
64x^2+28x-28
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: Find the real roots of each equation. If there are none state such.
16x^2 + 7x/7 = 4x+7/4
I tried finding common denominator
4(16x^2+7x)-7(4x+7)=0 all over 28
64x^2+28x-28
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Question 545610: Find the real roots of each equation. If there are none state such.
16x^2 + 7x/7 = 4x+7/4
I tried finding common denominator
4(16x^2+7x)-7(4x+7)=0 all over 28
64x^2+28x-28x-49=0 /28
(8x-7) (8x+7) all over 28=0
not sure what to do next.
Thank you! Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! (16x^2 + 7x)/7 = 4x+7/4
multiply by 28
4*16x^2+28x=28x+49
64x^2-49=0
(8x+7)(8x-7)=0
either (8x+7)=0 OR (8x-7)=0
8x+7=0
8x=-7
x=-7/8
8x-7=0
8x=7
x=7/8
