What can be the value of p for which one root of the quadratic equation (p^{2} – 5p + 3)x^{2} + (3p - 1)x + 2 = 0 is twice as large as the other?

Option 2 : 2/3

RBI Grade B 2020: Full Mock Test

9364

200 Questions
200 Marks
120 Mins

**Calculation:**

Let m, 2m be the roots of the given equation.

∴ sum of the roots:

⇒ m + 2m

⇒ 3m

[∴ -b/a]

⇒ (1 – 3p)/(p^{2} – 5p + 3) ----(i)

∴ product of the roots:

⇒ m(2m)

⇒ 2m^{2}

[∴ c/a]

⇒ 2/(p^{2} – 5p + 3) ----(ii)

By (i) and (ii) we have

⇒ 9m^{2}/2m^{2} = (1 – 3p)^{2}/( p^{2} – 5p + 3)^{2} × (p^{2} – 5p + 3)/2

⇒ 9/2 = (1 – 3p)^{2}/( p^{2} – 5p + 3) × 1/2

⇒ 9(p^{2} – 5p + 3) = (1 – 3p)^{2}

⇒ 9p^{2} – 45p + 27 = 1 – 9p^{2} – 6p

⇒ 39p = 26

⇒ p = 26/39

⇒ p = 2/3

**∴ the value of p is 2/3.**

__Key Points__

Sum of Roots = -b/a

Product of roots = c/a