Solve for X: 3x+7+3x+5-x = 3(x-3)(x+3)+5x Linear-Quadratic Equation

Q: Why did we get both positive and negative square roots?

When solving $ x^2 = 13 $, we need both solutions because both $ (\sqrt{13})^2 = 13 $ and $ (-\sqrt{13})^2 = 13 $. Always include the ± symbol when taking square roots!

Q: How do I know when to use the difference of squares pattern?

Look for the form (a-b)(a+b) where you have the same terms with opposite signs. In our problem, $ (x-3)(x+3) $ fits this pattern perfectly!

Q: Can I leave the answer as ±√13 or do I need decimals?

Keep it as $ ±\sqrt{13} $! This exact form is more precise than decimals like ±3.606. Exact answers are preferred in mathematics unless specifically asked for approximations.

Q: Why did this linear-looking equation become quadratic?

The right side contained $ (x-3)(x+3) $ which expands to $ x^2 - 9 $. Even though the left side was linear, the hidden quadratic term on the right made this a quadratic equation!

Quadratic Equations with Difference of Squares

Solve the following equation:

_{^{$3x+7+3x+5-x=3(x-3)(x+3)+5x$}}

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:03 Collect terms

00:23 Use abbreviated multiplication formulas

00:40 Reduce what's possible

00:47 Divide by 3

01:02 Arrange the exercise so that 0 is on one side of the equation

01:15 Break down 13 into square root of 13 squared

01:18 Again use abbreviated multiplication formulas to find solutions

01:26 Find the two possible solutions

01:41 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 equation:

_{^{$3x+7+3x+5-x=3(x-3)(x+3)+5x$}}

Step-by-step solution

To solve this problem, let's follow these steps:

Step 1: Simplify the left-hand side of the equation:
Combine like terms: $3x + 3x - x + 7 + 5 = 5x + 12$ .
Step 2: Expand and simplify the right-hand side:
Apply the difference of squares: $3(x-3)(x+3) = 3(x^2 - 9)$ .
This simplifies to $3x^2 - 27$ .
Then add $5x$ : $3x^2 - 27 + 5x$ .
Step 3: Equate both sides and solve for $x$ :
We have $5x + 12 = 3x^2 + 5x - 27$ .
Subtract $5x$ and 12 from both sides: $0 = 3x^2 - 39$ .
This simplifies to $3x^2 = 39$ .
Divide by 3: $x^2 = 13$ .
Taking the square root of both sides, we find $x = \pm\sqrt{13}$ .

Therefore, the solution to the equation is $x = \pm\sqrt{13}$ .

Final Answer

$±\sqrt{13}$

Key Points to Remember

Essential concepts to master this topic

Rule: Combine like terms first on both sides before expanding
Technique: Use difference of squares: $(x-3)(x+3) = x^2 - 9$
Check: Substitute $x = \pm\sqrt{13}$ back: both sides equal 39 ✓

Common Mistakes

Avoid these frequent errors

Expanding (x-3)(x+3) term by term instead of using difference of squares
Don't multiply (x-3)(x+3) as x² + 3x - 3x - 9 = longer work with more error risk! This wastes time and increases mistakes. Always recognize the pattern (a-b)(a+b) = a² - b² for faster, accurate solutions.

Practice Quiz

Test your knowledge with interactive questions

Solve:

$$ (2+x)(2-x)=0 $$

FAQ

Everything you need to know about this question

Why did we get both positive and negative square roots?

When solving $x^2 = 13$ , we need both solutions because both $(\sqrt{13})^2 = 13$ and $(-\sqrt{13})^2 = 13$ . Always include the ± symbol when taking square roots!

How do I know when to use the difference of squares pattern?

Look for the form (a-b)(a+b) where you have the same terms with opposite signs. In our problem, $(x-3)(x+3)$ fits this pattern perfectly!

Can I leave the answer as ±√13 or do I need decimals?

Keep it as $±\sqrt{13}$ ! This exact form is more precise than decimals like ±3.606. Exact answers are preferred in mathematics unless specifically asked for approximations.

What if I made an algebra mistake and got x² = 8 instead?

Double-check your combining like terms! The left side should be $5x + 12$ and right side $3x^2 + 5x - 27$ . After subtracting $5x$ from both sides, you get $3x^2 = 39$ .

Why did this linear-looking equation become quadratic?

The right side contained $(x-3)(x+3)$ which expands to $x^2 - 9$ . Even though the left side was linear, the hidden quadratic term on the right made this a quadratic equation!

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