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If the squared difference of the zeros of the quadratic polynomial f(x)=x^2+px+45 is equal to 144 find the value of p

Posted by Akshat Rai 6 years, 7 months ago

CBSE > Class 10 > Mathematics

4 answers

Abhay Gupta 6 years, 7 months ago

P= 4

0Thank You

Abhay Gupta 6 years, 7 months ago

P= -4

0Thank You

Yogita Ingle 6 years, 7 months ago

Let α,β are the roots of the quadratic polynomial f(x) = x²+px+45 then

α + β = -p ---------(1) and αβ = 45

Given (α - β)² = 144

∴ (α + β)² – 4αβ = 144

⇒ (– p)² – 4 × 45 = 144 [Using (1)]

⇒ p² – 180 = 144

⇒ p² = 144 + 180 = 324

⇒ p = ± 18

Thus, the value of p is ± 18.

1Thank You

Akshat Rai 6 years, 7 months ago

I want yhe solutions

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CBSE > Class 10 > Mathematics