Question

The number of real roots of the equation |x² + 4x +3|+ 2x +5=0 are​

Answer 1

Answer:

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Step-by-step explanation:

The given equation is x  

2/3

+x  

1/3

−2=0

Put x  

1/3

=y, then y  

2

+y−2=0

⇒(y−1)(y+2)=0⇒y=1 or y=−2

⇒x  

1/3

=1 or x  

1/3

=−2

∴x=(1)  

3

=1 or x=(−2)  

3

=−8

Hence, the real root of the given equation are 1 & −8.

Answer 2

Answer:

x²+4x+3| = x² + 4x +3 when

x² + 4x +3 >= 0

x(x+3) >=0

i.e when x <= -3 or when x >= 0.....(1)

So the equation becomes

x² + 4x +3 + 2x +5 = 0

Which gives

x = - 4 and x = - 2

But we can't take x = - 2 as it not satisfies equation 1

So x = -4

Now |x² + 4x + 3| = -(x² + 4x + 3)

When x² + 4x + 3 <0

i.e when

-3 < x < 0.......(2)

So the equation becomes

-(x² + 4x + 3) + 2x+5 = 0

-x² -2x +2 = 0

x = √3 - 1 and x = -(√3+1)

But only x = - (√3+1) satisfy equation 2 so there are two answers

x = - 4 and x= - (√3 +1)

Step-by-step explanation:

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