Solve x^2 - 7x > -6
We want to find the values of x for which this inequality is 0. Let's first fine the values of x for which x2- 7x = -6.Rearranging, x2 - 7x + 6 = 0. Now we're looking for the values of x for which x2 - 7x + 6 > 0 which is the same as our first inequality.Solve this quadratic using an appropriate method.x2 -x -6x + 6 = 0x(x-1) -6(x-1) = 0(x-6)(x-1) = 0x = 6 or x = 1At these values, x2 - 5x + 6 = 0. Look at what happens as we increase or decrease our values of x by creating a timeline. Is it positive or negative?First look at x = 1. Try a value of x less than 1. Try x = 0(0)2 - 7(0) + 6 = 6 which is greater than 0. This is what we want. So we know a possible answer to our question is x < 1So when x < 1, x2 - 7x + 6 > 0Try a value in between x = 1 and x = 6. Try 2(2)2 -7(2) + 6 = 2 - 14 + 6 = -6So when x > 1 and x < 6, x2 - 7x + 6 < 0. So we don't want any of the values of x in between 1 and 6. Because x2 - 7x + 6 is a quadratic and quadratics change sign when they cross the x -axis at y = 0, we now know that it will be positive, greater than 0, when x > 6. We can check this by trying x = 7.(7)2 -7(7) + 6 = 6 because 7(7) = 72 so 72 - 7(7) = 0Therefore, x2 - 7x + 6 > 0 and x2- 5x > -6 are true when x < 1 or x > 6.
