if both x=2 and x=1/2 are the factors of polynomial px^2+5x+r. show that p=r
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hiiii dear
ur answer
given Given, f(x) = px^2+5x+r and factors are x-2, x-1/2
Substitute x = 2 in place of equation, we get
= p*2^2+5*2+r=0
= 4p + 10 + r = 0 ------------ (i)
Substitute x = 1/2 in place of equation.
p/4 + 5/2 + r = 0
p + 10 + 4r = 0 ------------------- (ii)
On solving (i),(ii) we get
4p+r=-10 and p+4r+10=0
4p+r=p+4r
3p=3r
p = r.
Hope this helps!
ur answer
given Given, f(x) = px^2+5x+r and factors are x-2, x-1/2
Substitute x = 2 in place of equation, we get
= p*2^2+5*2+r=0
= 4p + 10 + r = 0 ------------ (i)
Substitute x = 1/2 in place of equation.
p/4 + 5/2 + r = 0
p + 10 + 4r = 0 ------------------- (ii)
On solving (i),(ii) we get
4p+r=-10 and p+4r+10=0
4p+r=p+4r
3p=3r
p = r.
Hope this helps!
HarathiKallem:
hi can u please explain how did u get the second equation
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