SAT original mathematics module 1 question 15

Kabadosh, we were told that:
(ax + 2)(bx + 7) = 15x² +cx +14 ——— *
a + b = 8 ——— (i)
and to to find the two possible values of 'c'.

Let's solve:
The first thing we do is expand equation *, (ax + 2)(bx + 7).

(ax + 2)(bx + 7) = ax(bx +7) +2(bx +7);
= ax(bx) +7(ax) +2(bx) +2(+7);
= abx² +7ax +2bx +14;
(ax + 2)(bx + 7) = abx² +7ax +2bx +14;

REM: (ax + 2)(bx + 7) = 15x² +cx +1 and (ax + 2)(bx + 7) = abx² +7ax +2bx +14;

Therefore,
abx² +7ax +2bx +14 = 15x² +cx +1;
Compare the LHS to the RHS
x² : ab = 15 ——— (ii)
x. : 7a +2b = c ——— (iii)

From (i) and (ii)
a + b = 8; therefore a = 8 - b
ab = 15; therefore (8 -b)b = 15
8b -b² = 15;
b² -8b +15 = 0;
b² -5b -3b +15 = 0;
b(b -5) -3(b -5) = 0
(b -3) (b -5) = 0
b -3 = 0; or b -5 = 0;
b = 3; or b = 5;

From a +b = 8; a = 8 -b.
If A: b = 3:
a = 8 -3 = 5. (a,b) = (5,3).

If B: b = 5:
a = 8 -5 = 3. (a,b) = (3,5).

From (iii):
c = 7a +2b;
If A: (a,b) = (5,3). a = 5; b = 3.
c = 7(5) +2(3);
c = 35 + 6;
c = 41;

If B: (a,b) = (3,5). a = 3; b = 5.
c = 7(3) +2(5);
c = 21 + 10;
c = 31;

The two possible solutions for 'c' are: 41 and 31

Option D

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