A system of equations with one linear equation and one quadratic equation has "zero or one or two" solutions.
Consider a system of equations with one linear equation and one quadratic equation as
Case (1): y = x² + 4x + 2 ...(1)
and x + y = 2 ...(2)
By the substitution method:
from (2), y = 2 - x; Substituting in (1)
⇒ 2 - x = x² + 4x + 2
⇒ x² + 4x + 2 + x - 2 = 0
⇒ x² + 5x = 0
⇒ x(x + 5) = 0
∴ x = 0 and x = -5
If x = 0, then y = 2 - 0 = 2
If x = -5, then y = 2 + 5 = 7
So, the system has two solutions at (0, 2) and (-5, 7).
Case (2): y = x² + 4x + 2 ...(1)
and x - y = 2 ...(2)
By the substitution method:
from (2), y = x - 2; Substituting in (1)
⇒ x - 2 = x² + 4x + 2
⇒ x² + 4x + 2 + 2 - x = 0
⇒ x² + 3x + 4 = 0
we know that b² - 4ac will decide the nature of roots.
So, (3)² - 4(1)(4) = 9 - 16 = -7 < 0
Since the determinant is less than 0, the roots are imaginary. Hence the system has no solution. That means the line and the parabola do not meet.
Case (3): y = x² + 4x + 2 ...(1)
and y = x - 1/4 ...(2)
Substituting (2) in (1), we get
⇒ x - 1/4 = x² + 4x + 2
⇒ x² + 4x + 2 - x + 1/4 = 0
⇒ x² + 3x + 9/4 = 0
⇒ 4x² + 12x + 9 = 0
⇒ (2x + 3)² = 0
⇒ 2x + 3 = 0
∴ x = -3/2
If x = -3/2 then y = -3/4 - 1/4 = -7/4
So, the system has one solution at (-3/2, -7/4).
Therefore, the given type of system has "zero or one or two" solutions.
Learn more about the system of equations here:
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