Solve for x in the Triangle with Sides x, x+1, and x+2

Q: How do I know which side is the hypotenuse?

The hypotenuse is always the longest side in a right triangle and is opposite the right angle. In this problem, since x > 0, we have x+2 > x+1 > x, so x+2 is the hypotenuse.

Q: What if I can't factor the quadratic?

If factoring is difficult, you can always use the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $. For $ x^2 - 2x - 3 = 0 $, this gives the same answers!

Right Triangle Applications with Consecutive Side Lengths

Calculate x according to the figure shown below below.

$x>0$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Use the Pythagorean theorem in the triangle

00:08 Substitute appropriate values and solve for X

00:14 Open parentheses properly

00:22 Simplify where possible

00:28 Collect terms and arrange the equation

00:46 Find the possible solutions for X

00:52 X must be greater than 0, therefore this solution is not logical

00:57 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

Calculate x according to the figure shown below below.

$x>0$

Step-by-step solution

To find $x$ in the given triangle, let's apply the Pythagorean Theorem. The squared lengths of the triangle's legs and hypotenuse are related by this equation:

(x+1)^2 + x^2 = (x+2)^2

First, expand each term:

$(x+1)^2 = x^2 + 2x + 1$
$x^2 = x^2$
$(x+2)^2 = x^2 + 4x + 4$

Plug these into the Pythagorean Theorem equation:

(x^2 + 2x + 1) + x^2 = x^2 + 4x + 4

Combine like terms:

2x^2 + 2x + 1 = x^2 + 4x + 4

Rearrange the equation to isolate terms on one side:

2x^2 + 2x + 1 - x^2 - 4x - 4 = 0

Simplify to get a quadratic equation:

x^2 - 2x - 3 = 0

Now, solve for $x$ using factoring. Look for two numbers that multiply to $-3$ and add to $-2$ . These numbers are $-3$ and $1$ :

(x - 3)(x + 1) = 0

Set each factor equal to zero:

$x - 3 = 0 \Rightarrow x = 3$
$x + 1 = 0 \Rightarrow x = -1$

Given the condition $x > 0$ , the valid solution is:

$x = 3$

Final Answer

$x=3$

Key Points to Remember

Essential concepts to master this topic

Pythagorean Theorem: For right triangles, $a^2 + b^2 = c^2$ where c is hypotenuse
Technique: Expand $(x+1)^2 + x^2 = (x+2)^2$ to get $x^2 - 2x - 3 = 0$
Check: Substitute x=3: $4^2 + 3^2 = 5^2$ gives 16 + 9 = 25 ✓

Common Mistakes

Avoid these frequent errors

Using the wrong side as hypotenuse
Don't assume the longest expression is automatically the hypotenuse = wrong equation setup! Students often write $x^2 + (x+2)^2 = (x+1)^2$ which gives impossible negative solutions. Always identify the hypotenuse as the side opposite the right angle in the diagram.

Practice Quiz

Test your knowledge with interactive questions

Look at the triangle in the diagram. How long is side AB?

FAQ

Everything you need to know about this question

How do I know which side is the hypotenuse?

The hypotenuse is always the longest side in a right triangle and is opposite the right angle. In this problem, since x > 0, we have x+2 > x+1 > x, so x+2 is the hypotenuse.

Why did we get two solutions but only use x = 3?

The quadratic equation gives x = 3 and x = -1, but the problem states $x > 0$ . Since side lengths must be positive, we reject x = -1 and use only x = 3.

What if I can't factor the quadratic?

If factoring is difficult, you can always use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . For $x^2 - 2x - 3 = 0$ , this gives the same answers!

How do I check if my triangle sides actually work?

Substitute your x-value back: if x = 3, the sides are 3, 4, and 5. Check: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ ✓. This confirms it's a valid right triangle!

Can I solve this without expanding all the squares?

While expanding is the most reliable method, you might recognize this as a Pythagorean triple pattern. However, expanding ensures you don't miss any solutions and gives you practice with algebraic manipulation.

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Solve for x in the Triangle with Sides x, x+1, and x+2

Right Triangle Applications with Consecutive Side Lengths

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

How do I know which side is the hypotenuse?

Why did we get two solutions but only use x = 3?

What if I can't factor the quadratic?

How do I check if my triangle sides actually work?

Can I solve this without expanding all the squares?

🌟 Unlock Your Math Potential

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