Need help on quadratics problem.

Question

If $(ax+2)(bx+7)=15x^2+cx+14$ for all values of $x$, and $a+b=8$, what are the two possible values for $c$?
$\quad$ A)$\ $ $3$ and $5$
$\quad$ B)$\ $ $6$ and $35$
$\quad$ C)$\ $ $10$ and $21$
$\quad$ D)$\ $ $31$ and $41$

Although the question asks for $2$ possible values of $c$, aren't there $4$ in total: $31$, $41$, $45$, and $27$?

Answer 1

I think you have found $a,b \in \{3,5\}$ and your "extra solutions" correspond to $a=b=3$ and $a=b=5$. However, these do not satisfy $ab=15$ or $a+b=8$ - you have to go back to the original conditions and check these are satisfied.

Answer 2

$ab=15$ and $a+b=8$ has exactly two solutions. You can solve it by substitution and you will get a quadratic equation.

Each $(a,b)$ correponds to one $c$.

Hence there are at most $2$ possible values for $c$.

Answer 3

Option-d is correct.

Solution:

$a+b=8$

$\left(ax+2\right)\left(bx+7\right) = 15x^{2} +cx + 14$

$\rightarrow abx^{2}+\left(7a+2b\right)x+14=15x^{2}+cx+14$

Comparing coefficients we get,

$ab=15$ and $7a+2b=c$

Now we create a new quadratic equation whose roots are $a$ and $b$.

$x^{2}-\left(a+b\right)x + ab = 0$

$\rightarrow x^{2} - 8x + 15 =0$ $-(1)$

By factorising the above expression, we get,

$\rightarrow \left( x-3\right) \left( x-5\right) = 0 $

$\rightarrow x=3 $ or $ x=5 $

Therefore $a= 3$, $b=5$ or $a=5$, $b=3 $

Putting the possible values of $a$ and $b$ in the given equation, we get,

$ 7a+2b= c$

$\rightarrow c= 31 $ or $c=41$

Here $(a,b)$ is a unique duplet for every real solution in $(1)$ ,hence there can be only two real solutions for $c$.

Answer 4

You have $ab= 15$ and $a+b=8$. Given that this will lead to a quadratic in either $a$ or $b$, you know that you can expect at most two distinct values for each variable. You can (of course) solve the quadratic, but since this is a multi-choice question, try to work more efficiently. Inspection will almost immediately show that $a= 3, b= 5$ is a solution pair (if you work with the prime factorisation of $15$ to check for integer solutions first). By symmetry ($a$ and $b$ are transposable), it should now be obvious that $a=5, b=3$ (the values reversed) is another solution pair. Since we've found two values for each of $a$ and $b$, we're done looking. Trying out just one set in $c = 7a + 2b$ will immediately zero the choice to just option $D$.

I know this is a verbose answer, but I just wanted to illustrate the thought process. Once you get used to thinking like this, it'll take literally seconds with only mental computation.