p and q are the roots of x² − 2x + A = 0, and r and s are the roots of x² − 18x + B = 0. If p

Question

p and q are the roots of x² − 2x + A = 0, and r and s are the roots of x² − 18x + B = 0. If p < q < r < s form an arithmetic progression, find A + B.

Accepted Answer

74. Let the common difference be d. p + q = 2p + d = 2 and r + s = 2p + 5d = 18, so d = 4, p = −1. The terms are −1, 3, 7, 11, giving A = pq = −3 and B = rs = 77. A + B = 74.

$p$ and $q$ are the roots of $x^{2} - 2 x + A = 0$ , and $r$ and $s$ are the roots of $x^{2} - 18 x + B = 0$ . If $p < q < r < s$ form an arithmetic progression, find $A + B$ .

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p and q are the roots of x2−2x+A=0, and r and s are the roots of x2−18x+B=0. If p<q<r<s form an arithmetic progression, find A+B.

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Master this topic in 30 minutes a day.

More Algebra practice

$p$ and $q$ are the roots of $x^{2} - 2 x + A = 0$ , and $r$ and $s$ are the roots of $x^{2} - 18 x + B = 0$ . If $p < q < r < s$ form an arithmetic progression, find $A + B$ .