Determine the number of real solutions

Question

Y=x^2-7x+10

y=-6x+5

Answer 1

Hi Madison,

Since both equations begin with Y then we can set them equal to each other:

x² - 7x + 10 = -6x + 5. We can add 6x and subtract 5 from the right side to get

x² - 7x + 6x + 10 - 5 = 0. Simplifying the equation gives us

x² - x + 5 = 0

This is a quadratic equation with three terms. The coefficients of each term are a = 1, b = -1, c = 5.

To find the number of real solutions we calculate what’s called the discriminant: b² - 4ac.

b² - 4ac =

(-1)² - 4(1)(5) =

1 - 20 =

-19

Whenever the discriminant is negative that means we have zero real solutions.

Please let me know if I can be more helpful, thanks!

Joe