Question 421852: can u please help me solve this question?
1.The straight line y=3-4x does not intersect the curve y=5x^2-x+q.Find the range of values of q.
Found 2 solutions by Theo, richard1234:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! I've never seen a problem like this before, but I think I have a solution, although it may not be the preferred way of dealing with this type of problem.
If the equation of y = 5x^2 - x + q is to intersect with the line y = 3-4x, then to find the points of intersection, we would set y equal to 3-4x and solve for x.
accordingly, we set 5x^2 - x + q = 3 - 4x
if we subtract 3 from both sides of the equation and we add 4x to both sides of the equation, we get:
5x^2 + 3x - 3 + q = 0
we can rearrange the terms to get:
5x^2 + 3x - (q+3) = 0
the standard form of a quadratic equation is ax^2 + bx + c = 0
in our equation:
a = 5
b = 3
c = (q+3)
we find the piece parts of the quadratic formula as follows:
2a = 10
-b = -3
4ac = 4*5*(q-3) = 20q - 60
b^2 = 9
b^2 - 4ac = 9 - 20q + 60- = 69-20q
the discriminant is equal to 69 - 20q
in order for there to be real roots to this equation, the discriminant has to be >= 0
set 69-20q >= 0 and solve for q
add 20q to both sides of this equation to get 69 >= 20q
divide both sides of this equation by 20 to get 69/20 >= q
commute this equation to get q <= 69/20
simplify this to get q <= 3.45
if q <= 3.45, then we WILL get an intersection of the graph of the equation y = 3 - 4x with the graph of the equation y = 5x^2 - x + q
conversely, if q > 3.45, then we WILL NOT get an intersection of the graph of the equation y = 3 - 4x with the graph of the equation y = 5x^2 - x + q
the following graph sets q equal to 3.45
the following graph sets q equal to 4 which is greater than 3.45
the following graph sets q equal to 3 which is less than 3.45
you can see that when q <= 3.45, the 2 graphs intersect, and you can see that when q > 3.45, the 2 graphs do not intersect.
in order for there to be a real solution to the intersection of the graph of the 2 equations, the roots of the resulting quadratic equation had to be real.
in order for them to be real, the discriminant had to be >= 0.
that led to the solution.
Answer by richard1234(7193)  (Show Source):
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