Let - - be two roots of the equation x2-20 14x-5 12 - 0- Then - 8 - - 8 is equal to

X^2+(20)^ 14x+(5)^ 12 = 0 ∴α + β = (20)^ 14, α . β = nbsp;(5)^ 12 α ^8 + β ^8 = (α ^4 + β ^4)^2 2 α ^4 β ^4 ={(α ^2 + β ^2)^2 2 α ^2β ^2}^2 2 α ^4 β ^4 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Relations between Roots and Coefficients : Higher Order Equations

Let α, β be t...

Question

Let $α, β$ be two roots of the equation $x^{2} + (20)^{\frac{1}{4}} x + (5)^{\frac{1}{2}} = 0$ . Then $α^{8} + β^{8}$ is equal to

160

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

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

100

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

Open in App

Solution

The correct option is C 50
$x^{2} + (20)^{\frac{1}{4}} x + (5)^{\frac{1}{2}} = 0$
$∴ α + β = - (20)^{\frac{1}{4}}, α . β = (5)^{\frac{1}{2}}$
$α^{8} + β^{8} = (α^{4} + β^{4})^{2} - 2 α^{4} β^{4}$
$= {(α^{2} + β^{2})^{2} - 2 α^{2} β^{2}}^{2} - 2 α^{4} β^{4}$
$= {[{(α + β)^{2} - 2 α β}^{2} - 2 α^{2} β^{2}]}^{2} - 2 α^{4} β^{4}$
$= {⎡ ⎢ ⎢ ⎣ {⎧ ⎪ ⎨ ⎪ ⎩ 20^{\frac{1}{2}} - {2.5}^{\frac{1}{2}} ⎫ ⎪ ⎬ ⎪ ⎭}^{2} - 2.5 ⎤ ⎥ ⎥ ⎦}^{2} - {2.5}^{2}$
$= (0 - 10)^{2} - 50$
$= 50$

Suggest Corrections

124

Similar questions

Q. Let

α, β

be the roots of

x^{2} + b x + 1 = 0

. Then find the equation whose roots are

- (α + \frac{1}{β})

and

- (β + \frac{1}{α})

Q. Let

α, β

be the roots of

x^{2} + 3 x + 5 = 0

then the equation whose roots are

- \frac{1}{α}

and

- \frac{1}{β}

is :

Q. Let

α

and

β

be nonzero real roots of the quadratic equation

x^{2} + a x + b = 0

and

α + β, α - β, - α + β

and

- α - β

be the roots of the equation

x^{4} + a x^{3} + c x^{2} + d x + e = 0

Then which of the following statement is false?

Q. Let

α

and

β

be roots of the equation

x^{2} - x + 1 = 0

. Then the value of

α + β + α^{2} + β^{2} + \dots + α^{100} + β^{100}

Q. Let

α, β

be two roots of the quadratic equation

x^{2} + 4 x + 1 = 0

then find the value of

α^{2} + β^{2}

Join BYJU'S Learning Program

Let α,β be two roots of the equation x2+(20)14x+(5)12=0. Then α8+β8 is equal to

The correct option is C 50x2+(20)14x+(5)12=0 ∴α+β=−(20)14,α.β=(5)12 α8+β8=(α4+β4)2−2α4β4 ={(α2+β2)2−2α2β2}2−2α4β4 =[{(α+β)2−2αβ}2−2α2β2]2−2α4β4 =⎡⎢ ⎢⎣⎧⎪⎨⎪⎩2012−2.512⎫⎪⎬⎪⎭2−2.5⎤⎥ ⎥⎦2−2.52 =(0−10)2−50 =50

Let $α, β$ be two roots of the equation $x^{2} + (20)^{\frac{1}{4}} x + (5)^{\frac{1}{2}} = 0$ . Then $α^{8} + β^{8}$ is equal to