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Mathematics
Robertson
16 April, 08:09
X^2+10x-7 / x^2+10x+25=0
how do i solve for x?
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Zoey Burton
16 April, 10:03
0
You just have to solve the next equation:
x²+10x-7=0
Because, if the numerator is equal to "0", then the next expression
(x²+10x-7) / (x²+10x+25) = 0
Therefore:
x²+10x-7=0
x=[-10⁺₋√ (100+28) ]/2 = (-10⁺₋√128) / 2 = (-10⁺₋8√2) / 2=-5⁺₋4√2
We have two possible solutions:
x₁=-5-4√2
x₂=-5+4√2
Now, we have to check these possible solutions.
Remember the denominator cannot be equal to "0".
if x=-5-4√2
[ (-5-4√2) ²+10 (-5-4√2) - 7] / [ (-5-4√2) ²+10 (-5-4√2) + 25]=0/32=0
Therefore: x=-5-4√2 is a solution.
if x=-5+4√2
[ (-5+4√2) ²+10 (-5+4√2) - 7] / [ (-5+4√2) ²+10 (-5+4√2) + 25]=0/32=0
Therefore: x=-5+4√√2 is a solution.
Answer: we have two solutions for x:
x₁=-5-4√2
x₂=-5+4√2
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Ross Chase
16 April, 10:59
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Factorize both the numerator and denominator then equate to zero ... you'll get X
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