Solve x²-49=0 vs x²+49=0: Finding the Largest X-Value

Q: What does it mean when we get imaginary solutions?

When $ x^2 = -49 $, we need the square root of a negative number. This gives us imaginary solutions like $ x = \pm 7i $, where i represents $ \sqrt{-1} $.

Q: How do I compare real and imaginary numbers?

You can only compare real numbers directly. Since $ x^2 + 49 = 0 $ has only imaginary solutions, we ignore them and find the largest real solution from $ x^2 - 49 = 0 $.

Q: Why does x² - 49 = 0 have real solutions but x² + 49 = 0 doesn't?

In $ x^2 - 49 = 0 $, we get $ x^2 = 49 $ (positive), so real solutions exist. In $ x^2 + 49 = 0 $, we get $ x^2 = -49 $ (negative), requiring imaginary numbers.

Q: Should I always write both positive and negative solutions?

Yes! When solving $ x^2 = k $ where k > 0, always write $ x = \pm\sqrt{k} $. Both solutions are equally valid and important.

Q: What if the problem asks for the largest x-value and I have imaginary solutions?

Focus on the real solutions only. Imaginary numbers can't be compared with real numbers on a number line, so find the largest real value from your solutions.

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the largest X

00:03 First let's find X from section 1

00:08 Let's isolate X

00:13 Take the root to find X

00:23 These are the possible solutions for X from section 1

00:26 Now let's calculate X from section 2

00:32 Let's isolate X

00:41 Any number squared is always greater than 0

00:46 Minus 49 is less than 0 therefore section 2 has no solution

00:49 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Solve each equation separately and find which x is the largest.

$x^2-49=0$
$x^2+49=0$

2

Step-by-step solution

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

Step 1: Solve the first equation $x^2 - 49 = 0$ .
Step 2: Solve the second equation $x^2 + 49 = 0$ .
Step 3: Compare the $x$ values to determine which is largest.

Now, let's work through each step:

Step 1: Solve $x^2 - 49 = 0$ .

Add 49 to both sides to isolate the square term: $x^2 = 49$ .

Take the square root of both sides to find $x$ :
$x = \pm \sqrt{49}$
$x = \pm 7$ .

So, the solutions are $x = 7$ and $x = -7$ .

Step 2: Solve $x^2 + 49 = 0$ .

Subtract 49 from both sides to isolate the square term: $x^2 = -49$ .

Take the square root of both sides considering complex numbers:
$x = \pm \sqrt{-49}$
Using imaginary numbers: $x = \pm 7i$ .

These solutions are non-real complex numbers.

Step 3: Compare the solutions.

For the first equation, the real solutions are $x = 7$ and $x = -7$ .

Since the second equation only has imaginary solutions, the largest real $x$ is from the first equation: $x = 7$ .

Therefore, the largest $x$ is 7, in the first equation.

3

Final Answer

Equation 1

Key Points to Remember

Essential concepts to master this topic

Rule: Set equation equal to zero and solve for x
Technique: $x^2 - 49 = 0$ gives $x = \pm 7$ , $x^2 + 49 = 0$ gives $x = \pm 7i$
Check: Substitute back: $7^2 - 49 = 0$ works ✓

Common Mistakes

Avoid these frequent errors

Ignoring the ± when taking square roots
Don't write just x = 7 when solving $x^2 = 49$ = missing half the solutions! Quadratic equations typically have two solutions. Always include both positive and negative roots: $x = \pm 7$ .

Practice Quiz

Test your knowledge with interactive questions

Solve the following equation:

$$ 2x^2-8=x^2+4 $$

FAQ

Everything you need to know about this question

What does it mean when we get imaginary solutions?

+

When $x^2 = -49$ , we need the square root of a negative number. This gives us imaginary solutions like $x = \pm 7i$ , where i represents $\sqrt{-1}$ .

How do I compare real and imaginary numbers?

+

You can only compare real numbers directly. Since $x^2 + 49 = 0$ has only imaginary solutions, we ignore them and find the largest real solution from $x^2 - 49 = 0$ .

Why does x² - 49 = 0 have real solutions but x² + 49 = 0 doesn't?

+

In $x^2 - 49 = 0$ , we get $x^2 = 49$ (positive), so real solutions exist. In $x^2 + 49 = 0$ , we get $x^2 = -49$ (negative), requiring imaginary numbers.

Should I always write both positive and negative solutions?

+

Yes! When solving $x^2 = k$ where k > 0, always write $x = \pm\sqrt{k}$ . Both solutions are equally valid and important.

What if the problem asks for the largest x-value and I have imaginary solutions?

+

Focus on the real solutions only. Imaginary numbers can't be compared with real numbers on a number line, so find the largest real value from your solutions.

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