Solve the Equation: (x+7)(x-7)/3 = -11-x²

Q: Why do I expand (x+7)(x-7) first?

Using the difference of squares pattern $ a^2 - b^2 = (a+b)(a-b) $ gives you $ x^2 - 49 $ quickly. This avoids FOIL and makes the equation simpler!

Q: What if I get confused by the negative signs?

Take it step by step! When you have $ -11 - x^2 $, think of it as $ -11 + (-x^2) $. This helps you see where each term goes when rearranging.

Q: How do I know when to take the square root?

When you get an equation like $ x^2 = 4 $, you need both positive and negative solutions. Remember: $ (2)^2 = 4 $ and $ (-2)^2 = 4 $!

Q: Why do I get two answers instead of one?

Quadratic equations typically have two solutions! This is because when you square a positive or negative number, you get the same result. Both $ x = 2 $ and $ x = -2 $ work.

Q: Should I check both solutions in the original equation?

Absolutely! Always substitute both $ x = 2 $ and $ x = -2 $ back into $ \frac{(x+7)(x-7)}{3} = -11 - x^2 $ to verify they both work.

Quadratic Equations with Fraction Clearing

$\frac{(x+7)(x-7)}{3}=-11-x^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:03 Let's use the shortened multiplication formulas

00:20 Let's use our exercise

00:34 Multiply by 3 to eliminate the fraction

00:55 Arrange the equation so that the right side equals 0

01:15 Collect like terms

01:25 Divide by 4

01:40 Break down 4 into 2 squared

01:47 Let's use the shortened multiplication formulas again

01:54 Find what makes each bracket equal to zero

02:06 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

$\frac{(x+7)(x-7)}{3}=-11-x^2$

Step-by-step solution

To solve the problem, begin with simplifying the left-hand side of the equation:

$(x+7)(x-7) = x^2 - 49$ .

Thus, the original equation $\frac{(x+7)(x-7)}{3} = -11 - x^2$ simplifies to:

$\frac{x^2 - 49}{3} = -11 - x^2$ .

Multiplying every term by 3 to clear the fraction, we obtain:

$x^2 - 49 = -33 - 3x^2$ .

Add $3x^2$ to both sides to consolidate $x^2$ terms on one side:

$x^2 + 3x^2 = -33 + 49$ .

This simplifies to:

$4x^2 = 16$ .

Divide by 4 on both sides:

$x^2 = 4$ .

Taking the square root of both sides provides:

$x = \pm 2$ .

Therefore, the solution to the problem is $x = \pm 2$ , corresponding to the choice labeled

±2

Final Answer

±2

Key Points to Remember

Essential concepts to master this topic

Rule: Multiply both sides by the denominator to eliminate fractions
Technique: Use difference of squares: $(x+7)(x-7) = x^2 - 49$
Check: Substitute x = ±2 back into original equation to verify ✓

Common Mistakes

Avoid these frequent errors

Forgetting to multiply all terms by 3
Don't multiply only the left side by 3 = unbalanced equation! This breaks the equality and gives wrong solutions like x = ±6. Always multiply every single term on both sides by the same number.

Practice Quiz

Test your knowledge with interactive questions

Solve:

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

FAQ

Everything you need to know about this question

Why do I expand (x+7)(x-7) first?

Using the difference of squares pattern $a^2 - b^2 = (a+b)(a-b)$ gives you $x^2 - 49$ quickly. This avoids FOIL and makes the equation simpler!

What if I get confused by the negative signs?

Take it step by step! When you have $-11 - x^2$ , think of it as $-11 + (-x^2)$ . This helps you see where each term goes when rearranging.

How do I know when to take the square root?

When you get an equation like $x^2 = 4$ , you need both positive and negative solutions. Remember: $(2)^2 = 4$ and $(-2)^2 = 4$ !

Why do I get two answers instead of one?

Quadratic equations typically have two solutions! This is because when you square a positive or negative number, you get the same result. Both $x = 2$ and $x = -2$ work.

Should I check both solutions in the original equation?

Absolutely! Always substitute both $x = 2$ and $x = -2$ back into $\frac{(x+7)(x-7)}{3} = -11 - x^2$ to verify they both work.

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