Question

equation .​Previous year IIT jee Question
Chapter :- complex number and quadratic​

Answer 1

EXPLANATION.

If α ≠ β.

⇒ α² = 5α - 3.

⇒ β² = 5β - 3.

As we know that,

We can write equation as,

α, β are the roots of the equation,

⇒ x² = 5x - 3.

⇒ x² - 5x + 3 = 0.

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = -b/a.

⇒ α + β = -(-5)/1 = 5.

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

⇒ αβ = 3/1 = 3.

The equation whose roots are α/β and β/α.

⇒ α/β + β/α.

⇒ (α² + β²)/(αβ).

⇒ [(α + β)² - 2αβ]/(αβ).

⇒ [(5)² - 2(3)]/(3).

⇒ [25 - 6]/3.

⇒ 19/3.

⇒ α + β = 19/3.

⇒ α/β x β/α = 1.

⇒ αβ = 1.

Formula of the quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (19/3)x + (1) = 0.

⇒ x² - 19x/3 + 1 = 0.

⇒ 3x² - 19x + 3 = 0.

Option [B] is correct answer.

Answer 2

Answer:

Option (B) 3x² - 19x +3

Step-by-step explanation:

Given :- α² = 5α - 3 → x² - 5x + 3 = 0

and , β² = 5β - 3 → β² - 5β +3 = 0

Those two equations shows that α and β are the roots of the equation x² - 5x + 3 = 0

∴ α + β = 5 and αβ = 3

Now, α / β + β / α = α² + β ² / αβ

• = (α + β)² - 2αβ / αβ

• = 5² - 2 *3 / 3

• = 25 - 6 / 3

• = 19/3

• α/β × β/α = 1

Thus, the equation having roots α / β and β/α is given by

• → x² - ( α/β + β/α )x + α/β × β /α = 0

• → x² - 19/3x + 1 = 0

• → 3x² - 19x + 3 = 0