SOLUTION: Hi I need help with these radical equation questions. Thanks in advance :)
Solve each equation and check for extraneous solutions:
1. [sqrt(2x^2 - 1)]=x
A.[sqrt(1/3)]
Algebra.Com
Question 950712: Hi I need help with these radical equation questions. Thanks in advance :)
Solve each equation and check for extraneous solutions:
1. [sqrt(2x^2 - 1)]=x
A.[sqrt(1/3)]
B.{-1,1}
C.{-1/2,1}
D.1
E.0
F.none of the above
2.[sqrt(2x^2 + 5x + 6)]=x
A.no solution
B.{-2,-3}
C.{2,3}
D.-2
E.3
F.0
3. [sqrt(x + 3)] = 2[SQRT(x)]
A.no real solution
B.3
C.{-3,3}
D.0
E.3/2
F.1
4.[sqrt(2x^2 + x - 12)] = x
A.[sqrt(6)]
B.3
C.no solution
D.[sqrt(12)]
E.2
F.{-2,2}
5.[sqrt(2x^2 + 6x + 4)] = x + 1
A.1
B.{-3, -1}
C.{-3/2, -1}
D.no solution
E.-1
F.-3
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
1.
.......square both sides
solutions:
=>
=>
answer is:
B.{-1,1}
2.
solutions:
=>
=>
answer:
B.{-2,-3}
3.
answer:
F.1
since , =>=>
so, these lines intersect at (,)
4.
solutions:
=>
=>
answer is:
B.3
5.
solutions:
=>
=>
answer:
B.{-3, -1}
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