Question

14. If alpha and beta are zeroes of the polynomial x ^ 2 - p(x + 1) + c such that (alpha + 1)(beta + 1) = 0 , then find the value of c....pls reply fast

Answer 1

Step-by-step explanation:

Join oa <br /><br /><br /><br />

1. Pa = (pn-an) , pb = (pn+bn ) <br /><br /><br /><br />

pa.Pb = (pn-an)(pn+bn) <br /><br /><br /><br />

on perpendicular ab so an = bn <br /><br /><br /><br />

pa.Pb = (pn-an)(pn+an) = pn^2-an^2 <br /><br /><br /><br />

2. Pn^2-an^2 = <br /><br /><br /><br />

in ∆ona <br /><br /><br /><br />

oa^2 = on^2+an^2 <br /><br /><br /><br />

an^2 = oa^2-on^2 <br /><br /><br /><br />

pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 <br /><br /><br /><br />

in ∆ onp <br /><br /><br /><br />

op^2 = on^2+pn^2 <br /><br /><br /><br />

so. Op^2- oa^2 <br /><br /><br /><br />

oa= ot <br /><br /><br /><br />

pn^2-an^2= op^2-ot^2 <br /><br /><br /><br />

3. From 1 & 2 <br /><br /><br /><br />

pa.Pb = op^2-ot^2 <br /><br /><br /><br />

in ∆ otp <br /><br /><br /><br />

op^2 = pt^2+ot^2 <br /><br /><br /><br />

op^2-ot^2= pt^2 <br /><br /><br /><br />

so pa.Pb = pt^2 <br /><br /><br /><br />

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