If one root of the equation 4x 2 −2x+(β−4)=0 be the reciprocal of the other, then find β

Question :If one root of the equation 4x 2 −2x+(β−4)=0 be the reciprocal of the other, then find β

The correct answer :Let the roots of the quadratic equation be α and 1/α.

By Vieta’s formulas, we know that the sum of the roots of a quadratic equation of the form ax^2 + bx + c = 0 is given by -b/a, and the product of the roots is given by c/a.

So, we have:

α + (1/α) = 2β/4 = β/2 (1) α(1/α) = 4-β/4 = 1-β/2 (2)

Multiplying equation (1) by α, we get:

α^2 + 1 = (β/2)α

Multiplying equation (2) by 4/α, we get:

4 + β = 8/α – 2β/α

Substituting α = 1/α from equation (1), we get:

4 + β = 8α – 2βα 4 + β = 8(α – α^2)

Using the identity α – α^2 = (α – 1/2)^2 – 1/4, we get:

4 + β = 8[(α – 1/2)^2 – 1/4] 4 + β = 8(α – 1/2)^2 – 2

Substituting β/2 for α + 1/α from equation (1), we get:

4 + β = 8(β/2 – 1/2)^2 – 2 4 + β = 2(β – 1)^2 – 2 4 + β = 2(β^2 – 2β + 1) – 2 4 + β = 2β^2 – 4β + 2 2β^2 – 5β + 2 = 0

Using the quadratic formula, we get:

β = [5 ± sqrt(25 – 16)]/4 β = [5 ± 3]/4

Therefore, β can be either 2 or 1.

So, the possible values of β are 1 and 2.