Solve the Quadratic Equation: 3x² + 7 = 2x² + 9

Q: Why do I get two answers for this quadratic equation?

Every quadratic equation has two solutions because when you square both a positive and negative number, you get the same result. Since $ (\sqrt{2})^2 = 2 $ and $ (-\sqrt{2})^2 = 2 $, both values work!

Q: Should I leave my answer as ±√2 or convert to decimals?

Keep it as $ ±\sqrt{2} $! This is the exact answer. Converting to decimals like ±1.414... gives you an approximation that's less precise.

Q: What if I get x² = negative number?

If you get something like $ x^2 = -5 $, there are no real solutions because you can't take the square root of a negative number in real numbers.

Q: Do I need to check both solutions?

Always check both! Substitute $ +\sqrt{2} $ and $ -\sqrt{2} $ back into the original equation. Both should make the equation true.

Quadratic Equations with Square Root Solutions

Solve the following exercise

$x\cdot3\cdot x+7=2x^2+9$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Calculate the products

00:09 Isolate X

00:35 Extract the root

00:39 When extracting a root there are always 2 solutions (positive, negative)

00:42 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

Solve the following exercise

$x\cdot3\cdot x+7=2x^2+9$

Step-by-step solution

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

Simplify the given expression.
Rearrange into standard quadratic form.
Solve using applicable method.

Let's begin the process:
1. Simplify the left-hand side: $x \cdot 3 \cdot x + 7 = 3x^2 + 7$ .

2. Set up the equation by balancing: $3x^2 + 7 = 2x^2 + 9$ .

3. Rearrange the terms to form a quadratic equation: $3x^2 - 2x^2 + 7 - 9 = 0$ .

This simplifies to: $x^2 - 2 = 0$ .

4. Solve for $x$ :
By adding 2 to both sides, we have: $x^2 = 2$ .
Take the square root of both sides: $x = \pm\sqrt{2}$ .

Therefore, the solution to the problem is $\pm\sqrt{2}$ .

Final Answer

$±\sqrt{2}$

Key Points to Remember

Essential concepts to master this topic

Simplification: Collect like terms to isolate the x² coefficient
Technique: From $x^2 = 2$ , take square root of both sides
Check: Substitute $±\sqrt{2}$ back: $(±\sqrt{2})^2 = 2$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting the ± symbol when taking square roots
Don't write just $x = \sqrt{2}$ = missing half the solutions! The square root operation produces both positive and negative values since both $(\sqrt{2})^2 = 2$ and $(-\sqrt{2})^2 = 2$ . Always include ± when solving $x^2 = n$ .

Practice Quiz

Test your knowledge with interactive questions

Solve the following exercise:

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

FAQ

Everything you need to know about this question

Why do I get two answers for this quadratic equation?

Every quadratic equation has two solutions because when you square both a positive and negative number, you get the same result. Since $(\sqrt{2})^2 = 2$ and $(-\sqrt{2})^2 = 2$ , both values work!

Should I leave my answer as ±√2 or convert to decimals?

Keep it as $±\sqrt{2}$ ! This is the exact answer. Converting to decimals like ±1.414... gives you an approximation that's less precise.

How do I know when to use the square root method?

Use the square root method when your equation simplifies to $x^2 = n$ (no x term). If you have $ax^2 + bx + c = 0$ with a b term, use factoring or the quadratic formula instead.

What if I get x² = negative number?

If you get something like $x^2 = -5$ , there are no real solutions because you can't take the square root of a negative number in real numbers.

Do I need to check both solutions?

Always check both! Substitute $+\sqrt{2}$ and $-\sqrt{2}$ back into the original equation. Both should make the equation true.

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