1. ## Linear Quadratic Systems Help

Determine the restrictions on the y-intercept so that y = 3x^2 + 6x - 1 intersects with a line with slope 2 in more than one place.
I equate the problem like how it was shown in my lesson
and made an equation for slope 2
y = 3x^2 +6x -1 (Quadratic Equation)
y = 2x + b (Linear Equation)
3x^2 + 6x -1 = 2x + b
3x^2 + 6x - 2x -1 -b
3x^2 + 4x -1 - b
And I equate the problem and I get
3x^2 + 4x - 1 - b
I'm not sure if i'm able to use this equation in the quadratic formula since there is a,b,c and d in this equation.

2. ## Re: Linear Quadratic Systems Help

Since 1 and b are constants with respect to x, you can group them in brackets, like so
3x^2 + 4x + (-1-b) = 0

Can you finish?

3. ## Re: Linear Quadratic Systems Help

Since 1 and b are constants with respect to x, you can group them in brackets, like so
3x^2 + 4x + (-1-b) = 0

Can you finish?
Would that make it 1b?

4. ## Re: Linear Quadratic Systems Help

No.
Consider the polynomial ax^2 + bx + c

In the polynomial 3x^2 + 4x + (-1-b)
a = 3
b = 4
c = -1 - b

5. ## Re: Linear Quadratic Systems Help

Ok i get it now, so I can use a, b and c into the quadratic formula then?

6. ## Re: Linear Quadratic Systems Help

That's correct. I'll leave the rest to you.

7. ## Re: Linear Quadratic Systems Help

That's correct. I'll leave the rest to you.
Thank you! If you have the time would you mind checking my answer when i'm done?

Sure.

9. ## Re: Linear Quadratic Systems Help

When i solve it i get

x = - 2 +/- sqrt(7 + 3b) / 3 as my answer

10. ## Re: Linear Quadratic Systems Help

That is the correct solution for x, but the question is asking for the restriction of b (the y-intercept).

Recall the following.
D, the determinant, is the expression under the square root. ie. b^2 - 4ac

If D > 0, then there are two solutions.
If D = 0, then there is a repeated solution
If D < 0, then there are no real number solutions

Since the question asks for the restriction on b when the parabola intersects the line in more than one place, D > 0
D = 28 + 12b
D > 0 -> D/4 > 0 -> 7 + 3b > 0

Isolate b to determine the restriction on b.

11. ## Re: Linear Quadratic Systems Help

That is the correct solution for x, but the question is asking for the restriction of b (the y-intercept).

Recall the following.
D, the determinant, is the expression under the square root. ie. b^2 - 4ac

If D > 0, then there are two solutions.
If D = 0, then there is a repeated solution
If D < 0, then there are no real number solutions

Since the question asks for the restriction on b when the parabola intersects the line in more than one place, D > 0
D = 28 + 12b
D > 0 -> D/4 > 0 -> 7 + 3b > 0

Isolate b to determine the restriction on b.
quick question can i use D = 7 + 3b or does it have to be D = 28 + 12b

12. ## Re: Linear Quadratic Systems Help

By definition, D = 28 + 12b.
However, you can divide D by a positive number as this will not change the sign of the inequality. The most common case you will find while studying quadratics is D/4. D/4 in this case is 7+3b.

Starting at either D or D/4 doesn't change the solution for isolating b. In fact, starting at D, you'll divide by 4 on both sides anyway
D = 28 + 12b > 0
Dividing both sides by 4 yields D/4 = 7 + 3b > 0

13. ## Re: Linear Quadratic Systems Help

As my final answer i got if b = 7/-3 then the line intersects with the parabola twice. since x can have 2 possible values there are 2 values of x that satisfies the equation

14. ## Re: Linear Quadratic Systems Help

Close. You need the inequality. Solving it yields b > -7/3. This means that if b > -7/3, then the parabola will intersect the line twice.

15. ## Re: Linear Quadratic Systems Help

My bad that's what I meant b > 7/-3. Also do I have to solve it for D = 0 and D > 0 as well or can I just stop at D < 0

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