Number of real solutions of the equation x4+8 x2+16=4 x2-12 x+9 is equal to -

Question

Number of real solutions of the equation x4+8 x2+16=4 x2-12 x+9 is equal to - (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

(x2+4)2=(2 x-3)2 ⇒ x2+4=±(2 x-3) ⇒ x2+2 x+1=0 text or underbracex2-2 x+7 text D < 0=0 ⇒ (x+1)2=0 text or text No solution ⇒ x=-1 Have only one solution.

Q. Number of real solutions of the equation $x^{4} + 8 x^{2} + 16 = 4 x^{2} - 12 x + 9$ is equal to -

1

2

3

4

$(x^{2} + 4)^{2} = (2 x - 3)^{2}$
$\Rightarrow x^{2} + 4 = \pm (2 x - 3)$
$\Rightarrow x^{2} + 2 x + 1 = 0 or D < 0 x^{2} - 2 x + 7 = 0$
$\Rightarrow (x + 1)^{2} = 0 or No solution$
$\Rightarrow x = - 1$
Have only one solution.

Q. Number of real solutions of the equation x4+8x2+16=4x2−12x+9 is equal to -

1

2

3

4

(x2+4)2=(2x−3)2 ⇒x2+4=±(2x−3) ⇒x2+2x+1=0 or D <0x2−2x+7​​=0 ⇒(x+1)2=0 or No solution ⇒x=−1 Have only one solution.

Q. Number of real solutions of the equation $x^{4} + 8 x^{2} + 16 = 4 x^{2} - 12 x + 9$ is equal to -

$(x^{2} + 4)^{2} = (2 x - 3)^{2}$
$\Rightarrow x^{2} + 4 = \pm (2 x - 3)$
$\Rightarrow x^{2} + 2 x + 1 = 0 or D < 0 x^{2} - 2 x + 7 = 0$
$\Rightarrow (x + 1)^{2} = 0 or No solution$
$\Rightarrow x = - 1$
Have only one solution.