Find the value of p, for which one zero of the quadratic polynomial

Q) Find the value of p, for which one zero of the quadratic polynomial p x ² – 14 x + 8 is 6 times the other.

(Q 24 – 30/2/2 – CBSE 2026 Question Paper)

Ans:

Step 1: Let the roots be α and β.

It is given that one zero is 6 times the other,

∴ α = 6 β ………….. (i)

Step 2: Let’s compare the given polynomial with standard polynomial form a x ² + bx + c, we get:

a = p, b = – 14, c = 8

Step 3: Now sum of roots, α + β = – $\frac{b}{a} = - \frac{- 14}{p}$

By substituting value from equation (i), we get:

α + β = $\frac{14}{p}$

∴ 6 β + β = $\frac{14}{p}$

∴ 7 β = $\frac{14}{p}$

∴ β = $\frac{2}{p}$ ………. (ii)

Step 4: Now product of roots, α . β = $\frac{c}{a} = \frac{8}{p}$

By substituting value of α from equation (i), we get:

α . β = $\frac{8}{p}$

∴ 6 β . β = $\frac{8}{p}$

∴ 3 β2 = $\frac{4}{p}$

By substituting value of β from equation (ii), we get:

∴ 3 $(\frac{2}{p})^2 = \frac{4}{p}$

∴ 3 $(\frac{4}{p^2}) = \frac{4}{p}$

∴ 3 $(\frac{1}{p^2}) = \frac{1}{p}$

∴ $(\frac{3}{p^2}) = \frac{1}{p}$

∴ 3 p = p 2

∴ p 2 – 3 p = 0

∴ p (p – 3) = 0

∴ p = 0 and p = 3

Step 5: Here we reject p = 0 (leading coefficient ≠ 0,

∴ p = 3

Therefore the value of p is 3.

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