SOLUTION: Hi, Can you please help me? . Quadratic Equations ( You solve this problem same as quadratic equation) . Solve {{{ x - 3 (sqrt(x)) - 4 = 0 }}} . I replaced {{{ sqrt(x) }}} wi

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Question 160328: Hi, Can you please help me?
.
Quadratic Equations ( You solve this problem same as quadratic equation)
.
Solve +x+-+3+%28sqrt%28x%29%29+-+4+=+0+
.
I replaced +sqrt%28x%29+ with "u", and solved like a normal quadratic equation
.
I found out that +u+=+4+ and +u+=+%28-1%29+
.
since +u+=+sqrt+%28x%29+, I found that +x+=+u%5E2+
.
If we replace the "u" with "4" and (-1), it means +x+=+4%5E2+ and +u+=+%28-1%29%5E2+, +x+=+16+ and +x+=+1+
.
I found the answers, but when I looked at the answers in the book, the book only had +x+=+16+, why wouldn't +x+=+1+, or did the book forget to mention +x+=+1+
.
Thanks ahead of time, Levi

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
It's a very good idea to use proofs for ALL answers.
x - 3 (sqrt(x)) - 4 = 0
Let x=1:
Proof:
x-3sqrtx-4=0
1-3sqrt1-4=0
1-3*1-4=0
1-3-4=0
1-7=0
-6=0 not a valid proof.
Let x=-1
-1-3sqrt-1-4=0
1-3*i-4=0
-3-3i=0 not a valid proof.
Proof for x=16:
16-3sqrt16-4=0
16-3*4-4=0
16-12-4=0
16-16=0
0=0 a valid proof.