Question

# If the squared difference of the zeroes of the quadratic polynomial f(x)=x​​​​​​2+px+45 is equal to 144 find the value of p.

Solution

## we can let ' a ' and ' b ' be the roots of the quadratic polynomial f (x) = x² + px + 45  ∴ we can  write a+b= -p (sum of the roots)-(negative of coeff of x)    we can also write ab=45(product of roots)-(constant) it is given that, (a - b)² =144 ∴ (a + b)² –4ab=144 (calculate) ⇒ (– p)² –4×45=144   ⇒ p ²–180=144 ⇒ p²= 144+180 ⇒p²=324 ⇒p= ±√324 ∴p=±18  Mathematics

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