and
x− 3 = 0giving us x=−5 or x= 3. We can substitute these two values for x into either of the original
functions to get the y-values. Let’s use the first one:
y=− 2 x^2 − 3 x− 4For x=−5, we have
y=− 2 × (−5)^2 − 3 × (−5)− 4
=− 2 × 25 − (−15)− 4
=− 50 + 15 − 4
=− 35 − 4
=− 39
For x= 3, we have
y=− 2 × 32 − 3 × 3 − 4
=− 2 × 9 − 9 − 4
=− 18 − 9 − 4
=− 27 − 4
=− 31
The two solutions of the system are therefore (x,y)= (−5,−39) and (x,y)= (3, −31).
Finally, we check
Even though it may seem redundant, we should check both solutions in the original equations
to remove all doubt about their correctness. Let’s begin by plugging (−5,−39) into the first
equation, paying careful attention to the signs:
4 x^2 + 6 x+ 2 y+ 8 = 0
4 × (−5)^2 + 6 × (−5)+ 2 × (−39)+ 8 = 0
4 × 25 + (−30)+ (−78)+ 8 = 0
100 + (−30)+ (−78)+ 8 = 0
70 + (−78)+ 8 = 0
− 8 + 8 = 0
0 = 0
Next, we check (3, −31) in that equation:
4 x^2 + 6 x+ 2 y+ 8 = 0
4 × 32 + 6 × 3 + 2 × (−31)+ 8 = 0
Two Quadratics 453