If one root of the equation $a x^{2} + b x + c = 0$ is equal to $n^{t h}$ power of the other root, show that ${(a^{n} c)}^{\frac{1}{n + 1}} + {(a c^{n})}^{\frac{1}{n + 1}} + b = 0$

Open in App

Solution

Let the roots of the equation $a x^{2} + b x + c = 0$ be $α, α^{n}$

Product of roots $α^{n + 1} = \frac{c}{a}$

$⟹ α = {(\frac{c}{a})}^{\frac{1}{n + 1}}$

Sum of roots $α + α^{n} = \frac{- b}{a}$

$⟹ {(\frac{c}{a})}^{\frac{1}{n + 1}} + {(\frac{c}{a})}^{\frac{n}{n + 1}} = \frac{- b}{a}$

$⟹ a {(\frac{c}{a})}^{\frac{1}{n + 1}} + a {(\frac{c}{a})}^{\frac{n}{n + 1}} + b = 0$

$∴ {(a^{n} c)}^{\frac{1}{n + 1}} + {(a c^{n})}^{\frac{1}{n + 1}} + b = 0$

Suggest Corrections

Similar questions

Q. If one root of the quadratic equation

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

is equal to the

n^{t h}

power of the other, then show that

{({a c}^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b = 0

Q. If one root of the equation

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

is equal to the

n^{t h}

power of the other, then

{(a c^{n})}^{1 / n + 1} + {(a^{n} c)}^{1 / n + 1} + b

is equal to

Q. If one root of

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

is equal to

n^{t h}

power of the other, then

(a c^{n})^{\frac{1}{(n + 1)}} + (a^{n} c)^{\frac{1}{(n + 1)}}

is equal to

If one root of the quadratic equation $a x^{2} + b x + c = 0$ is equal to the n^th power of the other root, then the value of $(a c^{n})^{\frac{1}{n + 1}}$ + $(a^{n} c)^{\frac{1}{n + 1}}$

If one root of the quadratic equation $a x^{2}$ +bx+c=0 is equal to the nth power of the other root, then the value of $(a c^{n})^{\frac{1}{n + 1}} + (a^{n} c)^{\frac{1}{n + 1}} =$

Join BYJU'S Learning Program

If one root of the equation ax2+bx+c=0 is equal to nth power of the other root, show that (anc)1n+1+(acn)1n+1+b=0

Let the roots of the equation ax2+bx+c=0 be α,αnProduct of roots αn+1=ca⟹α=(ca)1n+1Sum of roots α+αn=−ba⟹(ca)1n+1+(ca)nn+1=−ba⟹a(ca)1n+1+a(ca)nn+1+b=0∴(anc)1n+1+(acn)1n+1+b=0

If one root of the equation $a x^{2} + b x + c = 0$ is equal to $n^{t h}$ power of the other root, show that ${(a^{n} c)}^{\frac{1}{n + 1}} + {(a c^{n})}^{\frac{1}{n + 1}} + b = 0$