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Question 30755: 2x squared minus 3x squared minus 10x plus 15
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! if it is solving
2x squared minus 3x squared minus 10x plus 15 = 0
Then 2x^2-3x^2-10x+15 = 0
- x^2-10x+15 =0 ----(1)
(x^2+10x-15)=0 (dividing by (-1)
x^2+10x = 15
The LHS resembles a^2+2ab with a= x and b = 5.
We need b^2 to complete the square.
Therefore adding b^2 = 5^2 = 25 to both the sides
x^2+10x+25 = 15+25
(x+5)^2 = 40
Taking sqrt
(x+5) = +sqrt(40) or -sqrt(40) or
x = -5+sqrt(40) or x = -5-sqrt(40)
That is x = -5+ 2sqrt(10) or x = -5-2sqrt(10)
Answer: x = -5+ 2sqrt(10) or x = -5-2sqrt(10)
Verification:putting x = -5+ 2sqrt(10) in (1)
LHS = -[-5+ 2sqrt(10) ]^2-10[-5+ 2sqrt(10)]+15
=-[25+40-20sqrt(10)]+50-20sqrt(10)+15
= -65+20sqrt(10)+50-20sqrt(10)+15
=0
=RHs
Since surds occur in conjuagate pairs
there is no need to test the validity of the second value
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