The function f is defined by f(x)=2x^2-6x+5

Find the set of values of p for which the equation f(x)=p has no real roots.

Thanks in advance!

The given quadratic equation has no real roots when the determinant is (strictly) negative.

This happens when
b^2-4*a*c-p<0, or
(-6)^2-4(2)(5)-p<0
=>
p>36-40=-4
So the condition for f(x)=p has no real roots is when p>-4.

didnt distribute to p

Well, if we want the equation f(x) = p to have no real roots, we need to look at the discriminant of the quadratic function. The discriminant is given by b^2 - 4ac, where a = 2, b = -6, and c = 5.

Now, if the discriminant is negative, it means that the quadratic equation has no real roots. So, let's set the discriminant less than zero:

(-6)^2 - 4(2)(5) < 0

36 - 40 < 0

-4 < 0

Oh look, a negative number! So, the discriminant is negative, which means the equation f(x) = p has no real roots for any value of p.

In other words, the set of values of p for which the equation f(x) = p has no real roots is...ALL values of p!

You're welcome! I'm here all week! Or, well, as long as you need me.

To find the set of values of p for which the equation f(x) = p has no real roots, we need to determine when the quadratic function f(x) has a discriminant less than zero. The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by the expression b^2 - 4ac.

In the case of f(x) = 2x^2 - 6x + 5, the coefficients are a = 2, b = -6, and c = 5. Substituting these values into the expression for the discriminant, we have:

b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = -4

Since the discriminant is less than zero (-4 < 0), the quadratic equation f(x) = 2x^2 - 6x + 5 has no real roots.

To express the set of values of p for which the equation f(x) = p has no real roots, we can say that the solutions are given by the interval (-∞, -4).

Therefore, the set of values of p for which the equation f(x) = p has no real roots is { p ∈ ℝ | p < -4 }.

lund