Question

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

Answer 1

EXPLANATION.

One roots of the quadratic equation.

(a² - 5a + 3)x² + (3a - 1)x + 2 = 0. is twice larger than other.

As we know that,

Let one roots be = α.

Other roots be = β = 2α.

Sum of the zeroes of quadratic polynomial.

⇒ α + β = - b/a.

⇒ α + 2α = - (3a - 1)/(a² - 5a + 3).

⇒ 3α = - (3a - 1)/(a² - 5a + 3). - - - - - (1).

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

⇒ (α)(2α) = (2)/(a² - 5a + 3).

⇒ 2α² = (2)/(a² - 5a + 3). - - - - - (2).

Squaring equation (1), we get.

⇒ (3α)² = - (3a - 1²)/(a² - 5a + 3)². - - - - - (1).

⇒ 9α² = - (3a - 1)/(a² - 5a + 3)². - - - - - (1).

Divide equation (1) & (2), we get.

⇒ 9α²/2α² = -(3a - 1)²/(a² - 5a + 3)² x (a² - 5a + 3)/(2).

⇒ 9/2 = -(3a - 1)²/(2)(a² - 5a + 3).

⇒ 9 = -(3a - 1)²/(a² - 5a + 3).

⇒ 9(a² - 5a + 3) = - (3a - 1)².

⇒ 9a² - 45a + 27 = 9a² + 1 - 6a.

⇒ 9a² - 45a + 27 - 9a² - 1 + 6a = 0.

⇒ - 45a + 6a + 27 - 1 = 0.

⇒ - 39a + 26 = 0.

⇒ 39a = 26.

⇒ a = 26/39.

⇒ a = 2/3.

Option [A] is correct answer.

Answer 2

Answer:

Option (a) 2/3

Step-by-step explanation:

Since , One root of the quadratic equation

(a² - 5a + 3 ) x² + (3a - 1) x + 2 = 0 is twice as large as the other , Then let their roots be α and 2α

∵ sum of roots of a quadratic polynomial =α+β = cofficient of x / cofficient of x² = - b/a

• ∴ 3α = - (3a - 1 )/ (a² - 5a + 3 )

• α = - (3α - 1 ) / 3(a² - 5a + 3 ____(1)

∵ product of roots of a quadratic polynomila = αβ constant term / cofficient of x² = c/a

• ∴ 2α² = 2/(a²-5a+3)

• α² = 1/(a²-5a+3) _______(2)

Put value of α from eqution (1) into the equation (2)

We get,

• →{- (3α-1) / 9(a² - 5a + 3) }²= 1 / a² - 5α + 3

• → (3α - 1) ² / 9(a² - 5a + 3)² = 1/a² - 5a + 3

• → (3a-1)² = 9( a²-5a +3)

• → 9a² - 6a + 1 = 9a² - 45a + 27

• → 45 - 6a = 27 - 1

• → a = 26/39

• → 2/3 Answer