Question

slove this question this is IIT 2017 question for you

Answer 1

Answer:

Values of x satisfying the giving equation are

x=-4,  -2,  -1+√3,  -1-√3

Step-by-step explanation:

case I

x²+4x+3>0

Now open modules and solve i.e.

x² + 4x +3 + 2x + 5 = 0

=> x² + 6x + 8 = 0

=> (x+4)(x+2) = 0

we get x=-4,-2

So x= -4 or -2 is a solution in case 1

Case 2

x²+4x+3<0

Open modules with -ve sign

Equation we get is

-x² - 4x -3 + 2x + 5 = 0

=> -x² -2x + 2 = 0

=> x² + 2x - 2 = 0

Solving for x we get x=[-2 ±(√(4–4(1)(-2)]/2

x=-1±√3

Hence  Values of x satisfying the giving equation are

x=-4,  -2,  -1+√3,  -1-√3

Answer 2

Answer:

Step-by-step explanation:

Formula used:

If |u|=v then u=v (or) u= -v

$|{x}^2+4x+3|+2x+5=0$

$|{x}^2+4x+3|=-(2x+5)$

$case(i):$

${x}^2+4x+3=-(2x+5)\\\\{x}^2+4x+(2x+5)+3=0$

${x}^2+6x+8=0\\\\(x+2)(x+4)=0\\x=-2,-4$

${x}^2+2x+1=2+1\\\\(x+1)^{2}=3$

case(ii)

${x}^2+4x+3=(2x+5)\\\\{x}^2+2x-2=0$

${x}^2+2x+1=2+1\\\\(x+1)^{2}=3$

$x+1=\sqrt{3},\:x+1=-\sqrt{3}\\\\x=-1+\sqrt{3}, x=-1-\sqrt{3}$