Question

$if \: both(x - 2)and(x - 1 \div 2)are \: the \: factor \: of \: px {?}^{2} + 5x + r \: then$
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Answer 1

Answer:

Given: f(x) = px²+5x+r and factors are x-2 and x-1/2.

Substituting x=2 in the equation, we have

→p×2²+5×2+r = 0

→p×4+10+r = 0

→4p+10+r = 0 _ _ _ _ _(i)

Substituting x=1/2 in the equation, we have

→p×(1/2)²+5×(1/2)+r = 0

→p×1/4+5/2+r = 0

→p/4+5/2+r = 0

Making the denominators same...

→(p+10+4r)/4 = 0

→p+10+4r = 0×4

→p+10+4r = 0 _ _ _ _ _(ii)

On constituting both the equations, we get

→4p+10+r = p+10+4r

→4p-p+10-10 = 4r-r

→3p = 3r

.°. p = r

I hope this answer helps you...