If the grap of the equation y=qx2+6x+1/2 has two x-intercepts, what are the possible value of q?

Question

A. q less than or equal to 18
B. q less than 18 and q does not equal 0
C. q less than or equal to 9 and q does not equal 0
D. q less than 9 and q does not equal 0

How do you solve this?

Answer 1

The discriminant (b^2 - 4ac) of the quadratic equation is:

a) negative when there are NO x-intercepts
b) zero when there is ONE x-intercept
c) positive when there are TWO x-intercepts

Since you want two x-intercepts, you want the disciminant to be positive:

b^2 - 4ac > 0
6^2 - 4*q*1/2 > 0
36 - 2q > 0
18 - q > 0
18 > q
q < 18

Also, you have to add the condition that q is not equal to zero. Because if q=0, the equation is "0x^2 + 6x + 1/2" and that is no longer a quadratic equation -- it's just the line y=6x+1/2 which has only one x-intercept.

So the answer is (B). q must be less than 18, and also cannot equal zero.

Answer 2

Since this question is multiple choice, it's best to begin by checking and ruling out answers.
q cannot equal 0, because if it did, the equation would be y=6x + 1/2, which only has 1 x-intercept. Answer A is not correct.
Next, to find x-intercepts, set the equation equal to zero and factor. In this case, it is easier to work with whole numbers, so multiply each side by 2 giving 0=2qx^2 + 12x + 1.
When this is factored, we will have 0=(ax+1)(bx+1), where a times b will equal 2q, and a+b=12.
To get the largest possible number for q, a and b will both equal 6. Since ab=2q, 2q=36, and q=18. Therefore, q must be less than 18.

Answer 3

When the discriminant (b^2-4ac) of a quadratic is positive, there are 2 roots or x-intercepts.

So in this case, the discriminant is: 6^2-4q(1/2) =36-2q.
We need to make sure that: 36-2q>0

36-2q>0
18-q>0
18>q
q<18

Also, note that q is the leading coefficient, so if it were to be 0, then we would not have a quadratic, but rather a linear equation. So that tells us the answer is B.