The number of real roots of the equation $| x |^{6 / 5} - 26 | x |^{3 / 5} - 27 = 0$ is

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $2$
${| x |}^{6 / 5} - 26 {| x |}^{3 / 5} - 27 = 0.$
${| x |}^{(3 / 5)^{2}} - 26 {| x |}^{3 / 5} - 27 = 0$
Let ${| x |}^{3 / 5} = t .$
$t^{2} - 26 t - 27 = 0.$
$t^{2} - 27 t + t - 27 = 0.$
$t (t - 27) + 1 (t - 27) = 0.$
$∴ t = 27; t = - 1 .$

Suggest Corrections

Similar questions

Q. The product of real roots of the equation

| x |^{\frac{6}{5}} - 26 | x |^{\frac{3}{5}} - 27 = 0

Q. The number of real roots of the equation

(x + 3)^{2} + (x + 1)^{2} + (x - 5)^{2} + (x - 6)^{2} = 0

Q. How many distinct real roots are possible for the below equation?

x^{6} - 26 x^{3} - 27 = 0

Find the number of positive real roots for equation 3 $x^{5}$ - 4 $x^{4}$ - 42 $x^{3}$ + 56 $x^{2}$ + 27x - 36 = 0. If all the roots of above given equation are real.

Q. For the equation

x^{8} + 6 x^{7} - 5 x^{4} - 3 x^{2} + 2 x - 5 = 0

, determine the maximum number of real roots possible.

Join BYJU'S Learning Program

The number of real roots of the equation |x|6/5−26|x|3/5−27=0 is

The correct option is C 2|x|6/5−26|x|3/5−27=0.|x|(3/5)2−26|x|3/5−27=0Let |x|3/5=t.t2−26t−27=0.t2−27t+t−27=0.t(t−27)+1(t−27)=0.∴t=27;t=−1.

The number of real roots of the equation $| x |^{6 / 5} - 26 | x |^{3 / 5} - 27 = 0$ is

The correct option is C $2$
${| x |}^{6 / 5} - 26 {| x |}^{3 / 5} - 27 = 0.$
${| x |}^{(3 / 5)^{2}} - 26 {| x |}^{3 / 5} - 27 = 0$
Let ${| x |}^{3 / 5} = t .$
$t^{2} - 26 t - 27 = 0.$
$t^{2} - 27 t + t - 27 = 0.$
$t (t - 27) + 1 (t - 27) = 0.$
$∴ t = 27; t = - 1 .$