if $(ax+2)(bx+7) = 15x^2 + cx + 14$ for all values of $x$, and $a + b = 8$, what are the two possible values of $c$?
What i thought of: C would be $X's$ coefficient. Therefore $-(r+s)$ were $r, s$ are the roots would yield the value.
SAT- No calculator question. Solving for coefficient in quadratic.
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algebra-precalculus
2 Answers
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Multiplying the factors in the left hand side, you have that $$abx^2 + (7a + 2b)x + 14 = 15x^2 + cx + 14.$$ Comparing coefficients, we find that $ab = 15$ and hence $a = \frac{15}{b}$. Since $a+b = 8$, we have that $a + \frac{15}{a} = 8$. Multplying both sides by $a$ (this is possible since $a \neq 0$), we find $a^2 - 8a + 15 = 0$. Solving for $a$, we find that $a= 3$ or $a = 5$. Therfore $b = 5$ respectively $3$.
Because $7a + 2b = c$, we have that for $(a,b) = (3,5)$ that $c = 31$ and for $(a,b) = (5,3)$ that $c = 41$. So the only two possibilities for $c$ are $\{31, 41\}$.
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If $a+b=8$ and (from the leading coefficient) $ab=15$, then $$(a-b)^2=(a+b)^2-4ab = 8^2-4\cdot15=4,$$ so $a-b=\pm2$. On the other hand, $$c=2b+7a=\frac92(a+b)+\frac52(a-b)=36\pm 5.$$