Which parabola will have one real solution with the line y = x – 5?

y = x2 + x – 4
y = x2 + 2x – 1
y = x2 + 6x + 9
y = x2 + 7x + 4

For each parabola, you have to do a system of equation with the line y = x - 5 and find which of them has one real solution.

For that you can solve the systems and find the roots, but you can also use the rule that to have one real solution the discriminant of the quadratiic equation (b^2 - 4ac) has to be zero.

1)

y = x - 5
y = x^2 + x - 4

=> x - 5 = x^2 + x - 4

=> x^2 + 1 = 0

It is easy to tell, by simple inspection, that this equation has not real solutions.

2)

y = x -5
y = x^2 + 2x - 1

=> x - 5 = x^2 + 2x  - 1

=> x^2 + x + 4 = 0

discriminant = b^2 - 4ac = 1^2 - 4(1)(4) = 1 - 16 = - 15

A negative discriminant means that there are not real solutions.
 
3)

y = x - 5
y = x^2 + 6x + 9

x - 5 = x^2 + 6x + 9

=> x^2 + 5x + 14 = 0

=> b^2 - 4ac = 5^2 - 4(1)(14) = 25 - 56 = - 31 => no real solutions

 
4)

y = x - 5
y = x^2 + 7x + 4

x - 5 = x^2 + 7x + 4

=> x^2 + 6x + 9 = 0

=> b^2 - 4ac = 6^2 - 4(1)(9) = 36 - 36 = 0 => the system has one real solution.

By the way, that solution is easy to find because you can factor the equation as: (x + 3)^2 = 0 => x = - 3