If the squared difference of the zeros of the quadratic polynomial f(x) = x2 +px + 45 is equal to 144, find the value of p.

The given quadratic polynomial is f(x) = x2 + px + 45.

Let ∝ and β be the zeroes of the given quadratic polynomial.

∴∝ + β = – p and ∝β = 45           ...(1)

Given, (∝ – β)2 = 144

∴ (∝ + β)2 – 4∝β = 144

⇒ (– p)2 – 4 × 45 = 144               [Using (1)]

p2 – 180 = 144

p2 = 144 + 180 = 324

Thus, the value of p is ± 18.

  • 132

square of sum of zeros = p2
Product  of Zeros = 45
Squared difference of zeros = 144
Now

Squared difference + Square of sum = 4(product)
144+p2 =4*45
p2 =180 -144

p2 = 36

p = +6 or -6

  • -40

Why r u writing +6 or --6????  But ur ans... izzz... wrng!!!  :(

  • -22
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