Simplify: 3(x + 4) − 2x
√a asks "what number, multiplied by itself, gives a?" Every positive number actually has two square roots — a positive one and a negative one — but the radical symbol √ by itself always means the positive (principal) root.
| √100 | = 10, because 10 · 10 = 100 |
| All square roots of 36 | ±6, because both 6·6 = 36 and (−6)·(−6) = 36 |
√a · √b = √(a·b), and likewise √(a·b) = √a · √b. This is the key tool for both multiplying "unfriendly" roots and for simplifying roots of non-perfect squares.
| √2 · √8 | = √16 = 4 (multiply first, then simplify) |
| √48 | = √16 · √3 = 4√3 (16 is the largest perfect-square factor of 48) |
| −√75 | = −√25 · √3 = −5√3 |
√a asks 'what non-negative number squares to a?' Simplifying means smuggling perfect squares out from under the radical: split the number into (biggest perfect square) × (leftover), root the square, and leave the rest inside. The radical is a grouping symbol — whatever stays inside stays together.
| Simplify: √200 | √100 · √2 = 10√2. |
| Simplify: 3√18 | 3 · √9 · √2 = 3 · 3√2 = 9√2. The outside 3 multiplies with what comes out. |
Simplify √28.
A rational number can be written as a ratio of two integers and has a terminating or repeating decimal. An irrational number cannot, and its decimal never terminates or repeats. Square roots of non-perfect squares are always irrational.
| 5/4 | Rational — terminates at 1.25 |
| √7 | Irrational — 7 is not a perfect square, decimal never repeats |
| 7 + √20 | Irrational (rational 7 plus irrational √20) |
| √8 · √18 | = √144 = 12 — Rational! Two irrationals multiplied can land on a perfect square. |
| √11 · √11 | = √121 = 11 — Rational, since a number times itself under one root is just that radicand. |
A square root symbol doesn't automatically mean irrational — check whether the radicand is a perfect square first.
| 4 − √9 | = 4 − 3 = 1, which is rational (√9 simplifies away) |
Rational numbers can be written as a fraction of integers — their decimals stop or repeat. Irrational numbers can't — their decimals run forever without pattern. √n is rational only when n is a perfect square. Rational + irrational = irrational; the messiness never cancels out of a sum.
| Classify: √36, √12, 0.75, π | √36 = 6 rational; √12 irrational; 0.75 = 3/4 rational; π irrational. |
| Is 2 + √3 rational? | No — if it were, subtracting 2 would make √3 rational, a contradiction. |
Why is √(−4) not a real number?
f(x) = √x has domain x ≥ 0 and range y ≥ 0 (you can never input a negative x or get a negative output). The graph is always increasing and looks like "half a parabola" lying on its side.
| y = √(x + 4) + 2 | Shifted 4 units LEFT, 2 units UP. Domain: x ≥ −4. Range: y ≥ 2. |
| y = √(x − 1) − 4 | Shifted 1 unit RIGHT, 4 units DOWN. Domain: x ≥ 1. Range: y ≥ −4. |
The expression under the radical must be ≥ 0. Set it up as an inequality and solve for x.
| f(x) = √(x − 8) | x − 8 ≥ 0 → x ≥ 8 is the domain |
y = √x starts at the origin and rises slowly — half a parabola on its side. Shifting follows the universal rules: inside the root moves horizontally (opposite the sign), outside moves vertically (as written). The starting point (the graph's 'corner') travels with the shifts.
| Describe f(x) = √(x − 4) + 1. | Right 4, up 1 — starting point moves from (0,0) to (4, 1). |
| What's the domain of f(x) = √(x + 6)? | x + 6 ≥ 0 → x ≥ −6. Shifting left 6 also slides the domain left 6. |
Solve: x² = 16
Squaring is not reversible in one direction — taking the square root of both sides always introduces a ± symbol, because both the positive and negative root square back to the same number.
| x² = 20 | x = ±√20 = ±2√5 (simplest radical form) |
| (x − 2)² = 25 | Already isolated — take the square root of both sides |
| x − 2 = ±5 | x = 2 + 5 = 7 or x = 2 − 5 = −3 |
| 2(x + 5)² − 50 = 150 | Add 50, divide by 2: (x+5)² = 100 → x + 5 = ±10 → x = −15 or x = 5 |
| (x − 3)² + 10 = 38 | (x−3)² = 28 → x − 3 = ±√28 = ±2√7 → x = 3 ± 2√7 |
Equations with squares or roots solve by inverse operations, with two safety rules: square-rooting both sides of x² = k needs a ± (two numbers square to k); and after squaring both sides of a root equation, check your answer in the original — squaring can create impostor solutions.
| Solve: 2x² − 5 = 45 | x² = 25 → x = ±5. Forgetting the ± loses half the answer. |
| Solve: √(2x + 1) = 7 | 2x + 1 = 49 → x = 24. Check: √49 = 7 ✓. |
Multiply using an area model: (x + 3)(x + 7)
(x + b)² always expands to x² + 2bx + b². The linear coefficient is always TWICE b, and the constant is always b SQUARED — this relationship is the entire engine behind completing the square.
| (x + 6)² | = x² + 6x + 6x + 36 = x² + 12x + 36 |
| x² + 10x + 25 | Half of 10 is 5, and 5² = 25 ✓ → (x + 5)² |
| x² + 8x + ____ | Half of 8 is 4, 4² = 16 → x² + 8x + 16 = (x + 4)² |
| x² − 20x + ____ | Half of −20 is −10, (−10)² = 100 → x² − 20x + 100 = (x − 10)² |
The geometric meaning of completing the square: x² + bx is an incomplete square tile — an x-by-x square plus a b-by-x strip. Split the strip in half, wrap it around two sides, and the missing corner is (b/2)². Adding that corner literally 'completes the square.'
| What completes x² + 14x? | Corner = (14/2)² = 49 → x² + 14x + 49 = (x + 7)². |
| Solve by picture-logic: x² + 2x = 8 | Corner = 1: (x + 1)² = 9 → x + 1 = ±3 → x = 2, −4. |
Complete the square: x² − 6x + ____
Completing the square works on EVERY quadratic — even ones that don't factor nicely. Zeroes found by factoring are always rational; if a quadratic can't be factored, completing the square reveals irrational zeroes (or shows there are none at all).
| x² + 6x + 2 = 0 | (x + 3)² − 7 = 0 → x = −3 ± √7 — irrational, so factoring could never have found this |
| f(x) = 2x² − 4x − 16 | Factor 2 from the x-terms: 2(x² − 2x) − 16 → 2(x−1)² − 2 − 16 = 2(x−1)² − 18 |
| 2(x−1)² − 18 = 0 → (x−1)² = 9 | x − 1 = ±3 → x = 4 or x = −2 |
Completing the square finds zeros even when factoring fails — that's its superpower. Isolate the x-terms, add the corner to both sides, collapse, square-root with ±. Irrational answers like x = 3 ± √5 are normal here; leave them in radical form unless asked to round.
| Zeros of f(x) = x² − 6x + 4? | x² − 6x = −4 → (x − 3)² = 5 → x = 3 ± √5. |
| Why not factor? | No integer pair multiplies to 4 and adds to −6 — the roots are irrational, invisible to integer factoring. |
Identify a, b, and c in: 2x² − 4x + 1 = 0
For ax² + bx + c = 0, the zeroes are x = (−b ± √(b² − 4ac)) / (2a). This comes directly from completing the square on the general equation — it just packages that same process into one algorithm.
| x² + 6x + 3 = 0 | a=1, b=6, c=3 |
| x = (−6 ± √(36−12)) / 2 = (−6 ± √24)/2 | √24 = 2√6, so x = (−6 ± 2√6)/2 = −3 ± √6 |
| x² + 4x − 6 = −x − 2 | Add x, add 2 to both sides → x² + 5x − 4 = 0, THEN identify a=1,b=5,c=−4 |
The expression under the radical, b² − 4ac, is called the discriminant. If it's negative, the square root isn't a real number, so the quadratic has NO real zeroes — the parabola never crosses the x-axis.
| x² + 2x + 8 = 0 | b² − 4ac = 4 − 32 = −28 → no real solutions; the parabola misses the x-axis entirely |
When a problem gives messy, real-world numbers (like a height model), round your final answers rather than expressing them in radical form.
| −16t² + 20t + 60 = 0 (height model) | t = (−20 ± √4240) / −32 → t ≈ 2.7 seconds (reject the negative time) |
The quadratic formula is completing the square done once, in general, forever: x = (−b ± √(b² − 4ac)) / 2a. It works on every quadratic. Ritual: write the equation as ax² + bx + c = 0, list a, b, c with their signs, substitute with parentheses, simplify the discriminant first.
| Solve: x² − 4x − 3 = 0 | a=1, b=−4, c=−3: x = (4 ± √(16+12))/2 = (4 ± 2√7)/2 = 2 ± √7. |
| Solve: 3x² + 2x − 8 = 0 | x = (−2 ± √(4 + 96))/6 = (−2 ± 10)/6 → x = 4/3, −2. |
List the three methods you know for solving a quadratic equation.
All three methods require the equation to equal zero first. Factoring is fastest when the trinomial factors nicely. Completing the Square and the Quadratic Formula always work, even for irrational or non-real zeroes.
| x² + 5x − 12 = 8x − 2 | Rearrange: x² − 3x − 10 = 0 → factors to (x−5)(x+2)=0 → x = 5, −2 |
| f(x) = (x − h)² + k, vertex (2,4) | Setting (x−2)²+4 = 0 gives (x−2)² = −4, which has no real solution |
| Graphically: | The parabola's vertex is above the x-axis and opens upward, so it never crosses — no real zeroes, confirming the algebra |
| P = −0.03T² + 25T − 3600 (popcorn popping model) | Set P = 0, use the Quadratic Formula: T ≈ 185°F or T ≈ 648°F — the two temperatures giving 0% popping |
Zeroes found any of the three ways should always match the x-intercepts when the function is graphed — this is the fastest way to catch an arithmetic error.
The discriminant b² − 4ac is the crystal ball under the radical: positive → two real roots (graph crosses twice); zero → one repeated root (vertex kisses the x-axis); negative → no real roots (graph never touches). Perfect-square discriminant → rational roots, factorable.
| Describe the roots of 2x² + x + 3 = 0. | D = 1 − 24 = −23 < 0 → no real roots; the parabola floats above the axis. |
| For what k does x² + kx + 9 = 0 have exactly one root? | D = k² − 36 = 0 → k = ±6. |
By hand: solve x² − 4x − 5 = 0 by factoring.
This is the calculator authorized for this course. Use these steps to verify ANY answer from this unit — a simplified radical, a completed-square vertex, or a quadratic-formula solution — the same way you did in Unit 8.
Calculator root-checking closes the loop: graph the function, use the zero-finder with bounds around each crossing, and compare to your algebra. Irrational answers compare as decimals — 3 ± √5 ≈ 5.236 and 0.764. Mismatch means recheck signs in a, b, c first; that's where most formula errors hide.
| Algebra gave x = 2 ± √3; calculator shows zeros at 3.732 and 0.268. | 2 + 1.732 = 3.732 ✓ and 2 − 1.732 = 0.268 ✓ — same answers in decimal costume. |
| solve() returns 'false' | The equation has no real solution — consistent with a negative discriminant, not a calculator error. |
These problems pull from earlier units on purpose. Switching between skills feels harder in the moment, but it helps you remember longer and matches how a real exam mixes topics. Try each one on paper first, then reveal.Estos problemas provienen de unidades anteriores a propósito. Cambiar de una destreza a otra se siente más difícil en el momento, pero te ayuda a recordar por más tiempo y se parece a cómo un examen real mezcla los temas. Intenta cada uno en papel primero, luego revela la respuesta.