Solving y = x²: Finding Points Where y = -6

Q: Why can't x² equal a negative number?

When you square any real number, you're multiplying it by itself. Whether the number is positive or negative, the result is always positive! For example: $ 3^2 = 9 $ and $ (-3)^2 = 9 $.

Q: What does the graph of y = x² look like?

The graph is a U-shaped curve called a parabola that opens upward. It never goes below the x-axis, which means y is never negative. The lowest point is at $ (0,0) $.

Q: Are there any values of x that make y = x² negative?

No, absolutely not! For any real number x, the value of $ x^2 $ is always zero or positive. This is a fundamental property of squaring real numbers.

Q: What if I need to solve x² = -6 using complex numbers?

In complex numbers, $ x = \pm i\sqrt{6} $ where $ i = \sqrt{-1} $. However, this problem asks about real solutions, so the answer is still no real solutions exist.

Q: How can I tell if a quadratic equation has real solutions?

For $ x^2 = k $, real solutions exist only when $ k \geq 0 $. If k is negative (like -6), there are no real solutions.

Quadratic Functions with Negative Outputs

Given the function:

$y=x^2$

Is there a point for ? $y=-6$ ?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Does the point exist?

00:03 We'll substitute appropriate values according to the given data, and solve to find the point

00:08 Any number squared will always equal a positive number

00:11 Therefore, the point does not exist

00:15 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

Given the function:

$y=x^2$

Is there a point for ? $y=-6$ ?

Step-by-step solution

To solve this problem, we'll follow these steps:

Step 1: Identify the given function and required y-value.
Step 2: Attempt to solve $x^2 = -6$ for $x$ .
Step 3: Conclude based on the results of the equation.

Let's work through each step:

Step 1: The function we are dealing with is $y = x^2$ , and we need to find $x$ such that $y = -6$ .

Step 2: Substitute $-6$ for $y$ , which gives us:

$x^2 = -6$

Step 3: To solve $x^2 = -6$ , consider whether it is possible for a real number squared to equal a negative number. A critical point here is that the square of any real number is non-negative. Therefore, there is no real value of $x$ that satisfies $x^2 = -6$ .

Thus, there is no point where $y = -6$ for the function $y = x^2$ .

The correct answer is No.

Final Answer

Key Points to Remember

Essential concepts to master this topic

Domain Rule: Real numbers squared are always non-negative (≥ 0)
Technique: Set x² = -6 and recognize impossibility for real solutions
Check: Verify any real number squared gives positive result ✓

Common Mistakes

Avoid these frequent errors

Assuming negative outputs are always possible
Don't think every y-value has corresponding x-values for y = x² = wrong solutions! The function y = x² only produces non-negative outputs because squaring any real number gives a positive result or zero. Always remember that x² ≥ 0 for all real numbers x.

Practice Quiz

Test your knowledge with interactive questions

Complete:

The missing value of the function point:

$$ f(x)=x^2 $$

$$ f(?)=16 $$

FAQ

Everything you need to know about this question

Why can't x² equal a negative number?

When you square any real number, you're multiplying it by itself. Whether the number is positive or negative, the result is always positive! For example: $3^2 = 9$ and $(-3)^2 = 9$ .

What does the graph of y = x² look like?

The graph is a U-shaped curve called a parabola that opens upward. It never goes below the x-axis, which means y is never negative. The lowest point is at $(0,0)$ .

Are there any values of x that make y = x² negative?

No, absolutely not! For any real number x, the value of $x^2$ is always zero or positive. This is a fundamental property of squaring real numbers.

What if I need to solve x² = -6 using complex numbers?

In complex numbers, $x = \pm i\sqrt{6}$ where $i = \sqrt{-1}$ . However, this problem asks about real solutions, so the answer is still no real solutions exist.

How can I tell if a quadratic equation has real solutions?

For $x^2 = k$ , real solutions exist only when $k \geq 0$ . If k is negative (like -6), there are no real solutions.

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