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

Consider a right-angled triangle, AB = 8 cm and AC = 6 cm.
Calculate the length of side BC.

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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