Question

If a, ß are the roots of the equation 8x² – 3x +27 =0, then
the value of (a²/ß + ß²/a)⅓ is​

Answer 1

EXPLANATION.

α,β are the roots of the equation.

⇒ F(x) = 8x² - 3x + 27 = 0.

As we know that,

Sum of zeroes of a quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-3)/8 = 3/8.

Products of zeroes of a quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 27/8.

To find the value of.

(1) = (α²/β + β²/α)^1/3.

⇒ (α²/β)^1/3 + (β²/α)^1/3.

⇒ (α³)^1/3 + (β³)^1/3/(αβ)^1/3.

⇒ (α + β)/(αβ)^1/3.

⇒ (3/8)/(27/8)^1/3.

⇒ (3/8)/(3/2)³^1/3.

⇒ (3/8)/3/2.

⇒ 3/8 X 2/3.

⇒ 1/4.

                                                                                                                       

MORE INFORMATION.

Conjugate roots.

If D < 0.

One root = α + iβ.

Other root = α - iβ.

If D > 0.

One root = α + √β.

Other root = α - √β.

Answer 2

Answer:

1/4

Step-by-step explanation:

$\rm{Question:}$

If α, β are the roots of the equation 8x² – 3x +27 =0, then

the value of (α²/β + β²/α)^⅓ is​

$\rm{Solution:}$

• α+β = -b/a = 3/8

• αβ = c/a = 27/8

Therefore,

(α²/β + β²/α)^⅓ = [(a^3+b^3)/(αβ)]^⅓

α+β/(αβ)^⅓

$\implies$ (3/8)/(27/8)^⅓

$\implies$ 2/8

$\implies$ 1/4