Need help with a quadratic worksheet?

Question

first of all im not asking you to do my homework. Im doing IB High Level Math and am struggling so i asked the teacher for some practice work. if you could show your work so i can understand the steps to solve the problems that would be great

1. Find the range of values for p such that 3x^2+3px+p^2=1 has real roots.

2. Find the exact solutions to the equation √(3x-2) - √(x-2)=1

3.solve the inequality (x^2-2x-10)/(x+1) >= 2

Answer 1

For the first one, you need to look at the discriminant - b^2 - 4ac. The discriminant tells you if you have real roots, double roots, or imaginary roots. First, let's see what we have in terms of p:

3x^2 + 3px + p^2 = 1
3x^2 + 3px + (p^2 - 1) = 0

a = 3
b = 3p
c = p^2 - 1

discriminant = b^2 - 4ac
discriminant = (3p)^2 - 4(3)(p^2 - 1)
discriminant = 9p^2 - 12p^2 + 12
discriminant = -3p^2 + 12

For a quadratic equation to have real roots, the discriminant has to be greater than or equal to 0. So,

0 <= -3p^2 + 12
3p^2 <= 12
p^2 <= 4
p <= 2 and p >= -2 (to account for the square root part of the problem)

So p can be anything from -2 to 2, inclusive. If you plug in values just to check, you'll see that at p = -2 or p = 2, you end up with double roots, and anything outside this range gives you imaginary roots.

For your second problem, it gets tricky with the square roots in there, but the best way to get rid of them is to square both sides of your equation until they go away:

√(3x-2) - √(x-2) = 1
√(3x-2) = 1 + √(x-2)
3x - 2 = 1 + 2√(x-2) + x - 2 <---use FOIL to get here
2x - 1 = 2√(x-2)
x - 1/2 = √(x-2)
x^2 - x + 1/4 = x - 2
x^2 - 2x + 9/4 = 0

When looking at this discriminant, we find that it is less than 0, which means that there are no real solutions to this equation.

For the last one, solve it like you would a normal equation (i.e. ignore the inequality for now - you can leave it in there, but it doesn't affect the way you solve the equation unless you multiply or divide by a negative number):

(x^2 -2x - 10) / (x + 1) >= 2
x^2 - 2x - 10 >= 2x + 2
x^2 - 4x - 12 >= 0
(x - 6)(x + 2) >= 0
x >= 6 or x >= -2

Two answers! When you graph them on a number line, the final answer has to encompass both ranges. These two overlap at x >= 6, so that's your answer. If you try to plug in x = 0 (which is >= -2) as a test point, you get -10/1 >= 2, which doesn't work. But if you plug in 1,000,000, you get something positive in the numerator (I don't care what the exact number is at this point) divided by something positive, but much smaller in the denominator (the numerator is much larger because of the squaring), and that fraction is much larger than 2.