The number of real roots of the equation x4+√x4+20=22 is

Question

The number of real roots of the equation x4+√x4+20=22 is (A) 4 (B) 2 (C) 0 (D) 1

Tardigrade Admin · Accepted Answer

x4+√x4+20=22 or x4-22=-√x4+20 Put x4 = y and square both the sides. (y-22)2=y+20 y2+484-44y=y+20 y2+45y+464=0 y2-29y-16y+464=0 (y-29)(y-16)=0 y=16,29 ∴ x4=16.29 or x=±2,±2.31

Q. The number of real roots of the equation $x^{4} + x^{4} + 20 = 22$ is

4

2

0

1

$x^{4} + x^{4} + 20 = 22$
or $x^{4} - 22 = - x^{4} + 20$
Put $x^{4} = y$ and square both the sides.
$(y - 22)^{2} = y + 20$
$y^{2} + 484 - 44 y = y + 20$
$y^{2} + 45 y + 464 = 0$
$y^{2} - 29 y - 16 y + 464 = 0$
$(y - 29) (y - 16) = 0$
$y = 16, 29$
$∴ x^{4} = 16.29$ or x $= \pm 2, \pm 2.31$

Q. The number of real roots of the equation x4+x4+20​=22 is

4

2

0

1

x4+x4+20​=22 orx4−22=−x4+20​ Put x4=y and square both the sides. (y−22)2=y+20 y2+484−44y=y+20 y2+45y+464=0 y2−29y−16y+464=0 (y−29)(y−16)=0 y=16,29 ∴x4=16.29 or x=±2,±2.31

Q. The number of real roots of the equation $x^{4} + x^{4} + 20 = 22$ is

$x^{4} + x^{4} + 20 = 22$
or $x^{4} - 22 = - x^{4} + 20$
Put $x^{4} = y$ and square both the sides.
$(y - 22)^{2} = y + 20$
$y^{2} + 484 - 44 y = y + 20$
$y^{2} + 45 y + 464 = 0$
$y^{2} - 29 y - 16 y + 464 = 0$
$(y - 29) (y - 16) = 0$
$y = 16, 29$
$∴ x^{4} = 16.29$ or x $= \pm 2, \pm 2.31$