Q. Let p(x)=x²-5x+a and q(x)=x²-3x+b, where a and b are +ve integers. Suppose HCF [p(x),q(x)]=x-1 and k(x)=LCM [p(x),q(x)]. If the leading coefficient of the highest degree term of k(x) is 1, then sun of the roots of (x-1)+k(x) is
pls solve with proper soln
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Answered by
3
Answer:
p(x)=x
2
−5x+a
q(x)=x
2
−3x+b
Given that HCF(p(x),q(x))=x−1, the other factors of p(x) and q(x) become (x−4) and (x−2) respectively.
∴p(x)=(x−1)(x−4)
q(x)=(x−1)(x−2)
⇒k(x)=LCM(p(x),q(x))=(x−1)(x−2)(x−4)
⇒(x−1)+k(x)=(x−1)+(x−1)(x−2)(x−4)
=(x−1)(x
2
−6x+8+1)
=(x−1)(x
2
−6x+9)
=x
3
−7x
2
+15x−9
So, by applying the relation between the coefficients and the roots , we get sum of roots=7.
Sum of roots is thus 7,
Answered by
2
p(x)=x
2
−5x+a
q(x)=x
2
−3x+b
Given that HCF(p(x),q(x))=x−1, the other factors of p(x) and q(x) become (x−4) and (x−2) respectively.
∴p(x)=(x−1)(x−4)
q(x)=(x−1)(x−2)
⇒k(x)=LCM(p(x),q(x))=(x−1)(x−2)(x−4)
⇒(x−1)+k(x)=(x−1)+(x−1)(x−2)(x−4)
=(x−1)(x
2
−6x+8+1)
=(x−1)(x
2
−6x+9)
=x
3
−7x
2
+15x−9
MAIN ANSWER ➡️ 7
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