Unravel the Equation: Solving √(x+1) × √(x+2) = x+3

Q: Why do I need to square both sides of the equation?

Squaring both sides eliminates the square root symbols, making the equation easier to solve. Since $ \sqrt{x+1} \times \sqrt{x+2} = \sqrt{(x+1)(x+2)} $, squaring gives us (x+1)(x+2) on the left side.

Q: What are domain restrictions and why do they matter?

Domain restrictions ensure the expressions under square roots are non-negative. For $ \sqrt{x+1} $, we need x+1 ≥ 0, so x ≥ -1. For $ \sqrt{x+2} $, we need x+2 ≥ 0, so x ≥ -2. The solution must satisfy both conditions.

Q: Can I multiply the square roots together first?

Yes! $ \sqrt{x+1} \times \sqrt{x+2} = \sqrt{(x+1)(x+2)} $, so the equation becomes $ \sqrt{(x+1)(x+2)} = x+3 $. Then square both sides to get (x+1)(x+2) = (x+3)².

Q: How do I check if x = -7/3 is actually valid?

Substitute into the domain requirements: x+1 = -7/3+1 = -4/3 $ \sqrt{x+1} $ undefined! This means x = -7/3 is an extraneous solution introduced by squaring.

Q: What should I do if my algebraic solution doesn't satisfy the domain?

If your solution creates negative values under square roots, it's an extraneous solution. This means the original equation has no real solutions. Always state this clearly in your final answer.

Q: Why does squaring both sides sometimes create fake solutions?

Squaring can turn negative numbers into positive ones. For example, if a = -2 and b = 2, then a ≠ b, but a² = b² = 4. That's why we must always check our solutions in the original equation!

Radical Equations with Square Root Simplification

Solve the following equation:

$\sqrt{x+1}\times\sqrt{x+2}=x+3$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Square both sides to eliminate the radicals

00:13 Expand brackets correctly, multiply each term by each term

00:30 Use the shortened multiplication formulas to expand the brackets

00:41 Collect like terms

00:51 Simplify where possible

00:55 Isolate X

01:10 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve the following equation:

$\sqrt{x+1}\times\sqrt{x+2}=x+3$

Step-by-step solution

To solve the equation $\sqrt{x+1} \times \sqrt{x+2} = x+3$ , we will follow these steps:

Step 1: Identify the terms. Let $a = \sqrt{x+1}$ and $b = \sqrt{x+2}$ , so $a \cdot b = x + 3$ .
Step 2: Square both sides of the equation to remove the square roots: $(ab)^2 = (x+3)^2$ .
Step 3: Express using the original variables: $(x+1)(x+2) = (x+3)^2$ .
Step 4: Expand both sides:
Left side: $x^2 + 3x + 2$
Right side: $x^2 + 6x + 9$ .
Step 5: Rearrange and simplify the equation by subtracting one side from the other:
$x^2 + 3x + 2 = x^2 + 6x + 9$ becomes $0 = 3x + 7$ .
Step 6: Solve the linear equation: $3x = -7$ , hence $x = -\frac{7}{3}$ .
Step 7: Verify that the solution $x = -\frac{7}{3}$ fits the initial domain requirements for the square roots.

Therefore, the solution to the problem is $x = -\frac{7}{3}$ . This matches choice 3 in the provided answer choices.

Final Answer

$x=-\frac{7}{3}$

Key Points to Remember

Essential concepts to master this topic

Domain: Both x+1 ≥ 0 and x+2 ≥ 0 must be satisfied
Technique: Square both sides: $(\sqrt{x+1} \times \sqrt{x+2})^2 = (x+3)^2$
Check: Substitute x = -7/3: $\sqrt{-4/3} \times \sqrt{-1/3} = -7/3 + 3$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to check domain restrictions after solving
Don't just solve algebraically and accept x = -7/3 without checking domains! This creates undefined square roots since x+1 = -4/3 < 0. Always verify that x+1 ≥ 0 and x+2 ≥ 0 before accepting any solution to radical equations.

Practice Quiz

Test your knowledge with interactive questions

Choose the expression that has the same value as the following:

$$ (x+y)^2 $$

FAQ

Everything you need to know about this question

Why do I need to square both sides of the equation?

Squaring both sides eliminates the square root symbols, making the equation easier to solve. Since $\sqrt{x+1} \times \sqrt{x+2} = \sqrt{(x+1)(x+2)}$ , squaring gives us (x+1)(x+2) on the left side.

What are domain restrictions and why do they matter?

Domain restrictions ensure the expressions under square roots are non-negative. For $\sqrt{x+1}$ , we need x+1 ≥ 0, so x ≥ -1. For $\sqrt{x+2}$ , we need x+2 ≥ 0, so x ≥ -2. The solution must satisfy both conditions.

Can I multiply the square roots together first?

Yes! $\sqrt{x+1} \times \sqrt{x+2} = \sqrt{(x+1)(x+2)}$ , so the equation becomes $\sqrt{(x+1)(x+2)} = x+3$ . Then square both sides to get (x+1)(x+2) = (x+3)².

How do I check if x = -7/3 is actually valid?

Substitute into the domain requirements: x+1 = -7/3+1 = -4/3 < 0, which makes $\sqrt{x+1}$ undefined! This means x = -7/3 is an extraneous solution introduced by squaring.

What should I do if my algebraic solution doesn't satisfy the domain?

If your solution creates negative values under square roots, it's an extraneous solution. This means the original equation has no real solutions. Always state this clearly in your final answer.

Why does squaring both sides sometimes create fake solutions?

Squaring can turn negative numbers into positive ones. For example, if a = -2 and b = 2, then a ≠ b, but a² = b² = 4. That's why we must always check our solutions in the original equation!

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