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andreev551 [17]
3 years ago
12

I don't remember what it is that I'm doing, as I did the lessons before spring break. Help?

Mathematics
1 answer:
blsea [12.9K]3 years ago
6 0
Extraneous solutions, is answers that we get because of squaring both sides of the radical equation, but in reality, they are not going to be the solutions of  the given equation.
(√(4x+41))²=(x+5)²
4x+41=x²+10x+25
x²+6x-16=0
(x-2)(x+8)=0
x1=2 , x2=-8,
And now we must to check them by substitution into initial equation

√(4x+41)=x+5
1)   x=2,   √(4*2+41)=2+5, √49=7, 7=7 true
2) x=-8,     √(4*(-8)+41)=-8+5,   √9 =-3    false,
so an extraneous solution x=-8
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<h3>How to determine the equivalent fraction?</h3>

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Can someone help me with this problem?
dlinn [17]

Answer

7 or 4

because do you see how v is lined 2 I don't remember if you times or divide .I'm going to go with times.

So what is 2 x 10 since V=10

2 x 10=20

Then we are going to divide so what is 20 divided by 5 since e=5

20 divided by 5=4

Next our last step what is 4 + 3=7 and there's are answer.

The answer is divide then this would be the answer.

so V(10) divided by 2=5  divided by e(5) =1 + 3=4

Hope this helps have a great day

8 0
3 years ago
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