If 2 and ½ are the zeros of px 2+5x+r, then
(a) p = r = 2
(b) p = r = - 2
(c) p = 2, r= -2
(d) p = -2, r= 2
(b) p = r = - 2
Explanation:
At First we need to verify that in from which numbers is zeros of px 2+5x+r
p (x) = px 2+5x+r
(i) p (2) = px 2+5x+r
= `p(2)^{2}+5(2)+r=0`
= 4p + 10 + r
= 4p + r = -10
p (x) = px 2+5x+r
(ii) p (`1/2`) = px 2+5x+r
= p`(1/2)^{2}`+5(`1/2`)+r=0
= `"p"/2` + `5/2` + r = 0
∴ (`1/2`) is the the zeros of px 2+5x+r
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