Finding Where x² - 4x + 5 > 0: Solving the Quadratic Inequality

Q: Why can't I just set the quadratic equal to zero and solve?

Setting it equal to zero finds where the function crosses the x-axis, but this quadratic never crosses! Since $ x^2 - 4x + 5 = 0 $ has no real solutions, the parabola stays entirely above the x-axis.

Q: How do I know if a quadratic is always positive or always negative?

Check the discriminant $ b^2 - 4ac $. If it's negative and $ a > 0 $, the parabola opens up and stays positive. If \( a , it opens down and stays negative.

Q: What does completing the square actually tell me?

Completing the square gives you vertex form $ (x-h)^2 + k $, where $ (h,k) $ is the vertex. The $ k $ value tells you the minimum (if opens up) or maximum (if opens down) value of the function.

Q: Why is the minimum value 1 and not 0?

Because $ (x-2)^2 $ equals 0 at minimum (when x = 2), but we add 1 to get $ (x-2)^2 + 1 $. So the smallest possible value is 0 + 1 = 1.

Q: Could this function ever be negative?

Never! Since $ (x-2)^2 \geq 0 $ for all real x, we have $ (x-2)^2 + 1 \geq 1 > 0 $. The function is always at least 1, so it's always positive.

Quadratic Inequalities with Vertex Form

Look at the following function:

$y=x^2-4x+5$

Determine for which values of $x$ the following is true:

$f\left(x\right)>0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$y=x^2-4x+5$

Determine for which values of $x$ the following is true:

$f\left(x\right)>0$

Step-by-step solution

The given function is $y = x^2 - 4x + 5$ . To find where this function is positive, we'll first analyze the properties of this quadratic.

Let's start by completing the square. We have:

y = x^2 - 4x + 5

To complete the square, take the coefficient of $x$ , which is $-4$ , halve it to get $-2$ , and then square it to get $4$ . Add and subtract this inside the expression:

y = (x^2 - 4x + 4) + 1 = (x-2)^2 + 1

Now, the expression is in vertex form $y = (x-2)^2 + 1$ , which indicates a parabola with a vertex at $(2, 1)$ and opens upwards. The vertex is the minimum point of the function.

Since the minimum value of $y$ is 1 (when $x = 2$ ), and the parabola opens upwards, the function is positive for all real $x$ , because $(x-2)^2 \geq 0$ for any real number $x$ , making $(x-2)^2 + 1 > 0$ .

Therefore, the answer is that the function is positive for all values of $x$ .

In conclusion, the correct choice is:

The function is positive for all values of $x$ .

Final Answer

The function is positive for all values of $x$ .

Key Points to Remember

Essential concepts to master this topic

Rule: Complete the square to find vertex and determine sign
Technique: Convert $x^2 - 4x + 5$ to $(x-2)^2 + 1$
Check: Minimum value is 1 when x = 2, so function is always positive ✓

Common Mistakes

Avoid these frequent errors

Trying to factor instead of completing the square
Don't attempt to factor $x^2 - 4x + 5$ = no real roots found! This quadratic doesn't factor with real numbers because the discriminant is negative. Always complete the square or use the discriminant to analyze when the quadratic has no real roots.

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

Why can't I just set the quadratic equal to zero and solve?

Setting it equal to zero finds where the function crosses the x-axis, but this quadratic never crosses! Since $x^2 - 4x + 5 = 0$ has no real solutions, the parabola stays entirely above the x-axis.

How do I know if a quadratic is always positive or always negative?

Check the discriminant $b^2 - 4ac$ . If it's negative and $a > 0$ , the parabola opens up and stays positive. If $a < 0$ , it opens down and stays negative.

What does completing the square actually tell me?

Completing the square gives you vertex form $(x-h)^2 + k$ , where $(h,k)$ is the vertex. The $k$ value tells you the minimum (if opens up) or maximum (if opens down) value of the function.

Why is the minimum value 1 and not 0?

Because $(x-2)^2$ equals 0 at minimum (when x = 2), but we add 1 to get $(x-2)^2 + 1$ . So the smallest possible value is 0 + 1 = 1.

Could this function ever be negative?

Never! Since $(x-2)^2 \geq 0$ for all real x, we have $(x-2)^2 + 1 \geq 1 > 0$ . The function is always at least 1, so it's always positive.

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Finding Where x² - 4x + 5 > 0: Solving the Quadratic Inequality

Quadratic Inequalities with Vertex Form

❤️ Continue Your Math Journey!

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why can't I just set the quadratic equal to zero and solve?

How do I know if a quadratic is always positive or always negative?

What does completing the square actually tell me?

Why is the minimum value 1 and not 0?

Could this function ever be negative?

🌟 Unlock Your Math Potential

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