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Question

The real solutions of x2+4x+3+2x+5=0 is/are

A
13
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B
1+3
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C
2
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D
4
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Solution

The correct options are
A 13
D 4
The Given equation is, x2+4x+3+2x+5=0
Now there can be two cases.
Case 1:
x2+4x+30(x+1)(x+3)0
x(,3][1,)(1)
Then given equation becomes,
x2+4x+3+2x+5=0
x2+6x+8=0
(x+4)(x+2)=0x=4,2
But x=2 does not satisfy (1), hence rejected
x=4 is the solution.

Case 2 :
x2+4x+3<0
(x+1)(x+3)<0
x(3,1)(2)
Then given equation becomes,
(x2+4x+3)+2x+5=0
x22x+2=0x2+2x2=0
x=2±4+82x=1+3,13
Out of which x=13 is solution.
Combining the two cases we get the solutions of given equation as x=4,13

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