If \(x_1,x_2\) are the roots of the equation \[ x^2-x-3=0 \] Find the value of \[ A=7x_1^5+19x_2^4 \]

Solution:

Since \(x_1,x_2\) are roots of

\[ x^2-x-3=0 \]

we have:

\[ x^2=x+3 \]

For any root \(x\):

\[ x^3=x(x^2)=x(x+3)=x^2+3x=4x+3 \]

\[ x^4=x(4x+3)=4x^2+3x=7x+12 \]

\[ x^5=x(7x+12)=7x^2+12x=19x+21 \]

Therefore:

\[ x_1^5=19x_1+21 \]

and

\[ x_2^4=7x_2+12 \]

Substitute into \(A\):

\[ A=7(19x_1+21)+19(7x_2+12) \]

\[ A=133x_1+147+133x_2+228 \]

\[ A=133(x_1+x_2)+375 \]

From the quadratic equation:

\[ x^2-x-3=0 \]

the sum of roots is:

\[ x_1+x_2=1 \]

Hence:

\[ A=133(1)+375 \]

\[ \boxed{A=508} \]

Therefore, the required value is:

\[ \boxed{508} \]

Solution

• Credit this problem to Mr. Lim Phalkun and Mr. Sen Piseth Who wrote this book. And, I just picked up one problem from that book to share all of you.