# Thread: Linear Quadratic Systems Help

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 Originally Posted by MacstersUndead 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 Originally Posted by MacstersUndead 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 Originally Posted by MacstersUndead 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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