Math formulas in context.

Check the conditions first

• The triangle must have a right angle. The two sides meeting at that angle are the legs a and b.
• c is the hypotenuse, opposite the right angle and longer than either leg. Use positive side lengths in the same unit.
• To test whether a triangle is right-angled, put its longest side in the c position. The converse says that equality guarantees a right angle.

Why it works

1. Arrange four identical right triangles inside a square of side a + b, with their hypotenuses surrounding a central square of side c. The acute angles of each triangle add to 90°, so the central corners are right angles.

2. Count the same area as the four triangles plus the central square.

   $$(a+b)^2=4\left(\frac{ab}{2}\right)+c^2$$

3. Expand both sides and cancel the matching 2ab terms.

   $$a^2+2ab+b^2=2ab+c^2$$

Worked example, step by step

A right triangle has legs 3 cm and 4 cm. Find its hypotenuse.

1. Label the unknown hypotenuse c. Both given lengths are legs, so add their squares.

   $$c^2=3^2+4^2$$

2. Square each number before adding.

   $$c^2=9+16=25$$

3. Take the positive square root because c is a length. The hypotenuse is 5 cm.

   $$c=\sqrt{25}=5$$

Check your result

Substitute the result into the original relation. Also check that 5 is greater than both 3 and 4, and less than their sum 7.

Avoid these pitfalls

• Do not use a + b = c: the theorem relates squares of lengths, not the lengths themselves.
• If a leg is missing, subtract its companion’s square from c² before taking the square root.
• A picture that looks right-angled is not enough. Use a given right angle, perpendicular directions, or a justified converse test.

Check the conditions first

• First write ax² + bx + c = 0, with real coefficients and a ≠ 0. If a = 0, the equation is not quadratic.
• Let D = b² − 4ac. If D > 0 there are two distinct real roots; if D = 0 there is one repeated real root.
• If D < 0, there are no real roots. Complex numbers extend the formula to that case, but these examples work over the real numbers.

Why it works

1. Start from ax² + bx + c = 0. Multiply by 4a and move the constant term to the right.

   $$4a^2x^2+4abx=-4ac$$

2. Add b² to both sides to complete the square.

   $$(2ax+b)^2=b^2-4ac$$

3. When D ≥ 0, take both square-root branches, subtract b, and divide by 2a. This produces the formula above, including when a is negative.

   $$2ax+b=\pm\sqrt{D}$$

Worked example, step by step

Solve x² − 5x + 6 = 0 and check both answers.

1. Read the signed coefficients from the standard form.

   $$a=1,\quad b=-5,\quad c=6$$

2. Compute the discriminant. It is positive, so expect two distinct real roots.

   $$D=(-5)^2-4(1)(6)=25-24=1$$

3. Substitute into the formula, keeping the entire numerator over 2a.

   $$x=\frac{5\pm\sqrt{1}}{2}=\frac{5\pm1}{2}$$

4. Evaluate the plus and minus cases separately.

   $$x_1=3,\qquad x_2=2$$

Check your result

Both values make the original expression zero. Their sum is 5 = −b/a and their product is 6 = c/a, giving another useful check.

Avoid these pitfalls

• Read b with its sign: when b = −5, the numerator begins with −b = 5. Also, (−5)² = 25.
• Do not forget the minus branch or divide only the square-root term by 2a.
• A negative discriminant means no real solution; it does not mean the roots are negative. A zero discriminant gives the same root twice.

Check the conditions first

• For positive integer n, the rule holds at every real x. Constant functions have derivative 0; treat f(x) = 1 directly rather than writing 0 · x⁻¹ at x = 0.
• For a negative integer exponent, exclude x = 0. For any real exponent n, the rule is valid on x > 0.
• Some rational powers also allow negative inputs; check the particular function’s real domain. Behavior at x = 0 needs separate attention: √x is defined there but has no finite derivative there.

Why it works

1. For f(x) = x³, begin with the difference quotient, using h ≠ 0.

   $$\frac{(x+h)^3-x^3}{h}=3x^2+3xh+h^2$$

2. Let h approach zero. The terms containing h vanish, leaving 3x². The binomial expansion gives the same pattern for all positive integer powers.

   $$f'(x)=\lim_{h\to0}(3x^2+3xh+h^2)=3x^2$$

3. For an arbitrary real n and x > 0, write xⁿ as e to the power n ln x. Once the chain rule and the derivatives of exp and ln are known, they justify the same result.

   $$\frac{d}{dx}e^{n\ln x}=e^{n\ln x}\frac{n}{x}=nx^{n-1}$$

Worked example, step by step

For f(x) = x³, find the slope and tangent line at x = 2.

1. Multiply by the exponent, then lower that exponent by one.

   $$f'(x)=3x^{3-1}=3x^2$$

2. Evaluate the derivative at x = 2 to obtain the slope.

   $$f'(2)=3(2^2)=12$$

3. Evaluate the original function to locate the point of tangency.

   $$f(2)=8\quad\Longrightarrow\quad (2,8)$$

4. Use point-slope form. This line matches the function’s value and slope at the point.

   $$y-8=12(x-2)\quad\Longrightarrow\quad y=12x-16$$

Check your result

The tangent line gives y = 8 at x = 2. A nearby secant slope is 12.0601 when h = 0.01, approaching 12 as h approaches zero. A numerical check supports the calculation but is not a proof.

Avoid these pitfalls

• Subtract one from the exponent, not from the whole term: the derivative of x³ is 3x², not 3x³ − 1.
• A composite expression needs the chain rule. For (2x + 1)³, multiply 3(2x + 1)² by the inner derivative 2.
• The rule is for a constant exponent. It does not directly differentiate aˣ or xˣ. Also distinguish f(2), a function value, from f′(2), a slope.