Answer + Step-by-step explanation:
Recall That the number of solution
of a quadratic equation ax² + bx + c = 0
depends on the discriminant b² - 4ac :
if b² - 4ac > 0 , the equation has two distinct solutions.
if b² - 4ac = 0 , the equation has only one solution.
if b² - 4ac < 0 , the equation has no solutions.
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System 1 :
y = x² + 4x + 7 and y = 2
⇔ x² + 4x + 7 = 2
⇔ x² + 4x + 5 = 0
→ b² - 4ac = 4² - 4×1×5 = 16 - 20 = -4 < 0
Then the quadratic equation has no solutions
Therefore the system has no solutions.
System 2 :
y = x² - 2 and y = x + 5
⇔ x² - 2 = x + 5
⇔ x² - x - 7 = 0
→ b² - 4ac = (-1)² + 4×7 = 29 > 0
Then the quadratic equation has two solutions
Therefore the system has two solutions.
System 3 :
y = -x² - 3 and y = 9 + 2x
⇔ -x² - 3 = 9 + 2x
⇔ -x² - 2x - 12 = 0
→ b² - 4ac = (-2)² - 4×(-1)×(-12) = 4 - 48 = -44 < 0
Then the quadratic equation has no solutions
Therefore the system has two solutions.
System 4 :
y = -3x - 6 and y = 2x² - 7x
⇔ -3x - 6 = 2x² - 7x
⇔ 2x² - 4x + 6 = 0
→ b² - 4ac = (-4)² - 4×(2)×(6) = 16 - 48 = -32 < 0
Then the quadratic equation has no solutions
Therefore the system has two solutions.
System 5 :
y = x² and y = 10 - 8x
⇔ x² = 10 - 8x
⇔ x² + 8x - 10 = 0
→ b² - 4ac = 8² - 4×1×(-10) = 64 + 40 = 104 > 0
Then the quadratic equation has two solutions
Therefore the system has two solutions.
