Question

If the squared difference of the zeroes of x² + px + 45 is equal to 216, find the value of p .​

Answer 1

$\sf{\huge{\underline{\green{Given :-}}}}$

• The squared difference of the zeroes of x² + px + 45 is equal to 216 .

$\sf{\huge{\underline{\green{To\:Find :-}}}}$

• The value of p .

$\sf{\huge{\underline{\green{Answer :-}}}}$

Let α & β , be the roots of polynomial f(x).

Such that,

f(x) = x² + px + 45

Consider,

α + β = -p

αβ = 45

The squared difference of the zeroes of x² + px + 45 is equal to 216 .

• ∴ ( α + β )² – 4αβ = 144 ------(1)

Putting values in equation 1,

➝ (– p)² – 4×45 = 144

➝ p² – 180 = 144

➝ p² = 144 + 180

➝ p² = 324

➝ p = √324

➝ p = √ 18 × 18

➝ p = ± 18

The value of p is ±18 .