Solve the Quadratic Inequality: Where f(x)=-2x²+4x-6 Is Greater Than 0

Question

The following function is graphed below:

f(x)=2x2+4x6 f(x)=-2x^2+4x-6

For which values of x is

f(x)>0 true?

Video Solution

Solution Steps

00:00 Find the domain of positivity of the function
00:03 The domain of positivity is above the X-axis
00:06 The entire function is below the X-axis
00:09 Therefore it is never positive
00:12 And this is the solution to the question

Step-by-Step Solution

The given function is f(x)=2x2+4x6 f(x) = -2x^2 + 4x - 6 .

First, we find the roots of the equation f(x)=0 f(x) = 0 , which are the critical points where the function can change its sign.

Using the quadratic formula x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=2 a = -2 , b=4 b = 4 , and c=6 c = -6 :

Calculate the discriminant Δ=b24ac=424(2)(6)=1648=32\Delta = b^2 - 4ac = 4^2 - 4(-2)(-6) = 16 - 48 = -32.

The discriminant is negative, indicating there are no real roots.

Because the parabola opens downwards and there are no x-intercepts (real roots), the function does not cross the x-axis.

Hence, for all real numbers, f(x)0 f(x) \leq 0 .

Therefore, there are no values of x x for which f(x)>0 f(x) > 0 .

Thus, the solution is: No answer.

Answer

No answer