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

Q: How can a quadratic inequality have no solution?

When a downward-opening parabola (negative leading coefficient) has no x-intercepts, it stays below the x-axis everywhere. Since $ f(x) = -2x^2 + 4x - 6 $ never touches or crosses the x-axis, it's always negative!

Q: What does a negative discriminant tell me?

A negative discriminant \( \Delta = -32 means no real roots exist. The parabola doesn't intersect the x-axis, so it's either always positive or always negative depending on which way it opens.

Q: How do I know the parabola opens downward?

Look at the leading coefficient (the number in front of $ x^2 $). Since a = -2 , the parabola opens downward like an upside-down U.

Q: Should I still use the quadratic formula if the discriminant is negative?

No need! Once you calculate $ \Delta = b^2 - 4ac $ and find it's negative, you know there are no real solutions. The quadratic formula would give complex numbers, but we only care about real solutions for this inequality.

Q: Can I verify this by testing a point?

Absolutely! Pick any x-value like x = 0: $ f(0) = -2(0)^2 + 4(0) - 6 = -6 $. Since this gives a negative result and the function has no roots, it's negative everywhere.

Q: What if the question asked for f(x) ≤ 0 instead?

Then the answer would be all real numbers! Since $ f(x) $ is always negative (never zero or positive), every x-value satisfies $ f(x) ≤ 0 $.

Quadratic Inequalities with No Real Solutions

The following function is graphed below:

$f(x)=-2x^2+4x-6$

For which values of x is

$f(x)>0$ true?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

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 written solution

Follow each step carefully to understand the complete solution

Understand the problem

The following function is graphed below:

$f(x)=-2x^2+4x-6$

For which values of x is

$f(x)>0$ true?

Step-by-step solution

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

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

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

Calculate the discriminant $\Delta = b^2 - 4ac = 4^2 - 4(-2)(-6) = 16 - 48 = -32$ .

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, $f(x) \leq 0$ .

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

Thus, the solution is: No answer.

Final Answer

No answer

Key Points to Remember

Essential concepts to master this topic

Discriminant Rule: When $\Delta = b^2 - 4ac < 0$ , no real roots exist
Sign Analysis: Downward parabola with $\Delta = -32 < 0$ stays negative always
Verification: Test any value like $f(0) = -6 < 0$ confirms no positive values ✓

Common Mistakes

Avoid these frequent errors

Assuming quadratic inequalities always have solutions
Don't assume every inequality f(x) > 0 has an answer interval = missing "no solution" cases! When the discriminant is negative and the parabola opens downward, the function is always negative. Always check the discriminant first and consider that "no solution" might be the correct answer.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of $$ x $$ where $f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

How can a quadratic inequality have no solution?

When a downward-opening parabola (negative leading coefficient) has no x-intercepts, it stays below the x-axis everywhere. Since $f(x) = -2x^2 + 4x - 6$ never touches or crosses the x-axis, it's always negative!

What does a negative discriminant tell me?

A negative discriminant $\Delta = -32 < 0$ means no real roots exist. The parabola doesn't intersect the x-axis, so it's either always positive or always negative depending on which way it opens.

How do I know the parabola opens downward?

Look at the leading coefficient (the number in front of $x^2$ ). Since a = -2 < 0, the parabola opens downward like an upside-down U.

Should I still use the quadratic formula if the discriminant is negative?

No need! Once you calculate $\Delta = b^2 - 4ac$ and find it's negative, you know there are no real solutions. The quadratic formula would give complex numbers, but we only care about real solutions for this inequality.

Can I verify this by testing a point?

Absolutely! Pick any x-value like x = 0: $f(0) = -2(0)^2 + 4(0) - 6 = -6$ . Since this gives a negative result and the function has no roots, it's negative everywhere.

What if the question asked for f(x) ≤ 0 instead?

Then the answer would be all real numbers! Since $f(x)$ is always negative (never zero or positive), every x-value satisfies $f(x) ≤ 0$ .

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