What is the largest number c such that 2x^2 + 5x + c = x^2 - 4x has at least one real solution? Express your answer as a common fraction.
+105 You should actually try to do these problems, they would help you a lot in the long run, because quadratics help in many math competitions...
But anyways, I'll still help:
We have:
\(2x^2 + 5x + c = x^2 - 4x\)
When subtracting the x^2 from both sides we get:
\(x^2 + 5x + c = - 4x\)
Add 4x to both sides:
\(x^2+9x + c = 0\)
To have at least one real solution, and the largest... (Hmm, I actually don't know how to do this, but I have intuition!)
I just want you to think about this example for a moment (Solving by intuition again!)
93 * 95 = 8835
94 * 94 = 8836
Do you see something intriguing here? The closer the numbers are together, the greater their product is! (Try it!)
So, we would need two numbers that are close together, possibly equal, and then we will get the greatest product...
We find that these two numbers would have to add to 9, and if we want them to be equal, we use 4.5!
So 4.5 * 4.5 = 20.25, so our constant "c" would be 20.25!
We have to write this as a fraction thought, so:
20 1/4 = 81/4
(Hope my explanation was good, and if there was anything wrong please notify me!)
+399 This problem is pretty easy if you know about discriminants.
Our equation is:
\(2x^2+5x+c=x^2-4x\)
First, move everything to the left side
\(x^2+9x+c=0\)
The discrimnant is the \(b^2-4ac\) (when the equation is \(ax^2+bx+c=0\)) part of the quadratic formula, basically
If the discriminant is greater than zero, there is 2 real solutions
If the discriminant is zero, there is one solution, a double root!
If the discriminant is less then zero, there is no real solutions
Using this we get: \(9^2-4c\) as the discrimant, which simplifies to \(81 - 4c\)
We want to get c as big as possible without this expression being negative, hence we want this to equal zero.
Solving \(81-4c=0\)
\(4c=81\)
\(c=81/4\)
