Question

Solve the equation2x^2-3x-5√(2x^2-3x+2)+6=0
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ANSWER = {1,1/2,-2,7/2}

Answer 1

anybody there

Answer 2

\[equation:- 2x^2-3x-5\sqrt{2x^2-3x+2}+6=0\]

Answer 3

\[2x^2-3x-5\sqrt{2x^2-3x+2}+6=0\]

Answer 4

OK

Answer 5

then how will i get the answer

Answer 6

\[2 x^2-3 x+6-5 \sqrt{2 x^2-3 x+2}=0 \]\[2 x^2-3 x+6=5 \sqrt{2 x^2-3 x+2} \]\[(2 x^2-3 x+6)^2=(5 Sqrt[2 x^2-3 x+2])^2 \]\[4 x^4-12 x^3-17 x^2+39 x-14=0 \]\[(x-1) (x+2) (2 x-7) (2 x-1)=0 \]

Answer 7

The Book says that the answer is {1,1/2,-2,7/2}

Answer 8

hello anybody there? help plz

Answer 9

If x-1=0, then x=1
If x+2=0 then x=-2
If 2x-7=0 then x= 7/2
if 2x-1=0 then x=1/2