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Chapter Review · a rapid pre-test review (speedrun)

Special Manipulations: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Equivalent equations
Two equations with the same solution set. Every reversible step gives one; an even power need not.
Radical equation
An equation hiding the variable under a root, such as 2x1=5\sqrt{2x - 1} = 5.
Extraneous solution
A candidate that solves the raised or cleared equation but fails the original, so it is discarded.
Principal square root
The nonnegative value x\sqrt{\phantom{x}} returns, so 9=3\sqrt{9} = 3 and never 3-3. An odd root is unrestricted: 83=2\sqrt[3]{-8} = -2.
Self-similar expression
An infinite expression containing a perfect copy of itself, so naming the whole xx makes the inner copy xx too.
Continued fraction
A stack of divisions running on forever, such as 1+11+11+1 + \frac{1}{1 + \frac{1}{1 + \cdots}}.
Golden ratio φ\varphi
The positive number equal to one plus its own reciprocal, φ=1+521.618\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.618.
Symmetric expression
An expression in xx and yy unchanged by swapping the letters. Both x+yx + y and xyxy pass, and so does x2+y2x^2 + y^2; xyx - y and x2yx^2 y fail.
The sum ss and the product pp
The abbreviations s=x+ys = x + y and p=xyp = xy, the two quantities blind to the swap that every symmetric polynomial is built from.
Palindromic (reciprocal) equation
A polynomial equation whose coefficient list reads the same forwards and backwards, such as 2x49x3+14x29x+2=02x^4 - 9x^3 + 14x^2 - 9x + 2 = 0.

Formulas and theorems

  • Raising both sides to a power

    An odd power gives an equivalent equation. An even power keeps every solution and can add candidates the original never had.

    Use when Real values. An even power hides the sign of its base, so an=bna^n = b^n leaves open a=ba = b or a=ba = -b; an odd power does not. After any even power, check every candidate in the original.

  • Square root of a square

    x2=x\sqrt{x^2} = \lvert x \rvert

    Use when Every real xx. Order matters: rooting first gives (x)2=x\left(\sqrt{x}\right)^2 = x exactly, but that needs x0x \ge 0.

    e.g. (7)2=7\sqrt{(-7)^2} = 7, not 7-7.

  • Taking an even root of both sides

    (xh)2=k xh=k xh=±k\begin{gathered} (x - h)^2 = k \\ \Longrightarrow\ \lvert x - h \rvert = \sqrt{k} \\ \Longrightarrow\ x - h = \pm\sqrt{k} \end{gathered}

    Use when k0k \ge 0 for real solutions. The absolute value forces the two signs, and both are genuine: rooting undoes a square rather than performing one, so nothing extraneous appears.

    e.g. (x+2)2=9(x + 2)^2 = 9 gives x=1x = 1 or x=5x = -5.

  • Multiplying or dividing by an expression that can vanish

    Multiplying both sides by an expression that can be zero adds solutions; dividing by one loses them.

    Use when Whenever the multiplier or divisor holds the variable. After multiplying, test every candidate in the ORIGINAL and discard any it leaves undefined. Never divide by a variable expression: factor instead, since x2=5xx^2 = 5x divided by xx drops x=0x = 0.

  • Self-similar towers

    x=c+x  x2xc=0x=cx  x(xc)=0\begin{gathered} x = \sqrt{c + x} \ \Longrightarrow \ x^2 - x - c = 0 \\ x = \sqrt{cx} \ \Longrightarrow \ x(x - c) = 0 \end{gathered}

    Use when c>0c > 0. Every layer is strictly positive, so keep the strictly positive root, not merely a nonnegative one: the additive tower has only one nonnegative root, while x(xc)=0x(x - c) = 0 offers two, and the value there is x=cx = c, never 00.

  • Continued fraction

    x=a+1xx2ax1=0x = a + \frac{1}{x} \quad\Longrightarrow\quad x^2 - ax - 1 = 0

    Use when a>0a > 0 and x0x \neq 0. The parts are all positive, so keep the positive root. With a=1a = 1 this is φ\varphi.

    e.g. a=2a = 2 gives x=1+22.414x = 1 + \sqrt{2} \approx 2.414, rejecting 121 - \sqrt{2}.

  • Repeating decimal as a shifted copy

    10kx=B+xx=B10k110^k x = B + x \quad\Longrightarrow\quad x = \frac{B}{10^k - 1}

    Use when x=0.Bx = 0.\overline{B} with a kk-digit block starting right after the decimal point. The power of ten must match the block length, or the tail will not line up. Convergence is never in question here: a decimal already names a point on the number line.

    e.g. x=0.27x = 0.\overline{27}: 100x=27+x100x = 27 + x, 99x=2799x = 27, x=311x = \tfrac{3}{11}.

  • Infinite geometric series, self-similar form

    S=a1+rSS=a11rS = a_1 + rS \quad\Longrightarrow\quad S = \frac{a_1}{1 - r}

    Use when r<1\lvert r \rvert < 1. Outside that range the series has no sum, so whatever the formula returns means nothing.

  • What the self-similar equation proves

    It gives the value the expression must have IF it settles on a number, and never proves that it does.

    Use when Settle that separately first: r<1\lvert r \rvert < 1 for a geometric series, or partial values that climb while held under a fixed ceiling for a tower.

  • Sum and product reductions

    x2+y2=s22px3+y3=s33ps1x+1y=sp(xy)2=s24p\begin{gathered} x^2 + y^2 = s^2 - 2p \\ x^3 + y^3 = s^3 - 3ps \\ \frac{1}{x} + \frac{1}{y} = \frac{s}{p} \\ (x - y)^2 = s^2 - 4p \end{gathered}

    Use when s=x+ys = x + y, p=xyp = xy; every symmetric polynomial in xx and yy reduces this way. The reciprocal form needs x0x \neq 0 and y0y \neq 0. Reapply x2+y2=s22px^2 + y^2 = s^2 - 2p to x2x^2 and y2y^2 for x4+y4=(s22p)22p2x^4 + y^4 = (s^2 - 2p)^2 - 2p^2.

  • A symmetric system as one quadratic

    t2st+p=0t^2 - st + p = 0

    Use when Monic, minus the sum and plus the product. Its roots are the unordered pair {x,y}\{x, y\}, so the method never says which is which. The pair is real exactly when s24ps^2 \ge 4p, since (xy)2=s24p(x - y)^2 = s^2 - 4p, and a complex conjugate pair otherwise.

  • Pairing a number with its reciprocal

    k=x+1xx2+1x2=k22x3+1x3=k33k\begin{gathered} k = x + \frac{1}{x} \\ x^2 + \frac{1}{x^2} = k^2 - 2 \\ x^3 + \frac{1}{x^3} = k^3 - 3k \end{gathered}

    Use when x0x \neq 0. These are the ss and pp identities with the product fixed at p=1p = 1. A real xx forces k2\lvert k \rvert \ge 2, since k24=(x1x)2k^2 - 4 = \left(x - \tfrac{1}{x}\right)^2.

    e.g. k=4k = 4: x2+1x2=14x^2 + \tfrac{1}{x^2} = 14 and x3+1x3=52x^3 + \tfrac{1}{x^3} = 52.

  • Axis of symmetry

    x=b2ax = -\frac{b}{2a}

    Use when a0a \neq 0, for ax2+bx+c=0ax^2 + bx + c = 0. The roots differ only by the sign before b24ac\sqrt{b^2 - 4ac}, so they always average to this line, and when b24ac0b^2 - 4ac \ge 0 they sit the same distance either side of it; the vertex rides on it.

    e.g. x26x+5=0x^2 - 6x + 5 = 0 has roots 11 and 55, mirrored about x=3x = 3.

Problem types, step by step

Solve a radical equation with one radical

  1. Isolate the radical on one side by itself.
  2. Raise both sides to the power matching the index: square a square root, cube a cube root.
  3. Solve what remains.
  4. Substitute every candidate into the ORIGINAL equation and discard the failures.

e.g. 1+2x+3=61 + \sqrt{2x + 3} = 6: 2x+3=252x + 3 = 25, x=11x = 11, and 1+25=61 + \sqrt{25} = 6 holds.

Solve an equation with two radicals

  1. Isolate one radical, moving the other across.
  2. Square both sides, expanding the binomial side in full; one radical survives.
  3. Isolate the survivor and square again.
  4. Solve, then test every candidate in the original: two squarings, two chances for an impostor.

e.g. 3x+1x+4=1\sqrt{3x + 1} - \sqrt{x + 4} = 1 gives x25x=0x^2 - 5x = 0; of x=0x = 0 and x=5x = 5, only 55 survives.

Solve by substituting for the repeated root or power

  1. Name the repeated piece: u=x1/3u = x^{1/3} for an equation in x2/3x^{2/3}, or u=xu = \sqrt{x} for one quadratic in x\sqrt{x}.
  2. Solve in the new letter; an even power of it brings a ±\pm.
  3. Discard what the substitution forbids: x\sqrt{x} is never negative, while x1/3x^{1/3} may be.
  4. Undo the substitution and check in the original.

e.g. x2/3=4x^{2/3} = 4: u=±2u = \pm 2, so x=u3=±8x = u^3 = \pm 8, and both 88 and 8-8 check.

Evaluate a self-similar expression

  1. Name the whole expression xx.
  2. Replace the copy of the whole sitting inside it by xx, turning the endless process into one finite equation.
  3. Solve it: square to clear a root, multiply through by xx to clear a fraction.
  4. Keep only the root the expression can have: a tower of positive layers is strictly positive, not merely nonnegative, and a fraction of positive parts is positive.
  5. Confirm the expression settles on a value before trusting the number.

e.g. 12+12+\sqrt{12 + \sqrt{12 + \cdots}}: x2x12=0x^2 - x - 12 = 0, so x=4x = 4 and not 3-3.

Evaluate a symmetric expression from the sum and product

  1. Read off ss and pp, recovering whichever is missing by rearranging a reduction such as x2+y2=s22px^2 + y^2 = s^2 - 2p.
  2. Rewrite the requested expression in ss and pp.
  3. Substitute the numbers; xx and yy are never needed one at a time.
  4. When the problem pairs xx with 1x\tfrac{1}{x}, use kk instead, with the product already 11.

e.g. s=5s = 5, p=3p = 3: x2+y2=256=19x^2 + y^2 = 25 - 6 = 19 and x3+y3=12545=80x^3 + y^3 = 125 - 45 = 80.

Solve a symmetric system in xx and yy

  1. Reduce whatever you are given to ss and pp.
  2. Build t2st+p=0t^2 - st + p = 0 and solve it.
  3. Report the unordered pair {x,y}\{x, y\} of its roots, then check it against both original equations.

e.g. x+y=6x + y = 6, x2+y2=20x^2 + y^2 = 20: p=8p = 8, t26t+8=0t^2 - 6t + 8 = 0, so {x,y}={2,4}\{x, y\} = \{2, 4\}.

Solve a palindromic equation

  1. Confirm x=0x = 0 is not a root, which a nonzero constant term settles.
  2. Divide every term by the middle power of xx, which is x2x^2 for a quartic.
  3. Group each term with its mirror and substitute u=x+1xu = x + \tfrac{1}{x}, using x2+1x2=u22x^2 + \tfrac{1}{x^2} = u^2 - 2.
  4. Solve the quadratic in uu.
  5. For each uu, multiply x+1x=ux + \tfrac{1}{x} = u through by xx and solve for xx.
  6. Check the list: roots come in reciprocal pairs, so rr without 1r\tfrac{1}{r} means one is missing.

e.g. 2x49x3+14x29x+2=02x^4 - 9x^3 + 14x^2 - 9x + 2 = 0 becomes 2u29u+10=02u^2 - 9u + 10 = 0, roots 22, 12\tfrac{1}{2}, 11, 11.

Exam traps

  • Trap Reporting every candidate the algebra survives: squaring x+2=x\sqrt{x + 2} = x gives x2x2=0x^2 - x - 2 = 0, answered as x=2x = 2 and x=1x = -1.

    Fix Squaring is not reversible, so test both in the ORIGINAL. At x=2x = 2: 4=2\sqrt{4} = 2, true. At x=1x = -1: 1=11\sqrt{1} = 1 \neq -1, so it is extraneous. The solution is x=2x = 2 alone.

  • Trap Squaring term by term, turning x=x2\sqrt{x} = x - 2 into x=x2+4x = x^2 + 4.

    Fix The right side is a binomial and must be expanded in full: x=x24x+4x = x^2 - 4x + 4. For the same reason a+b\sqrt{a + b} is not a+b\sqrt{a} + \sqrt{b}.

  • Trap Assuming the negative candidate, or the smaller one, is automatically the extraneous one.

    Fix Only substitution decides. In 3x+1x+4=1\sqrt{3x + 1} - \sqrt{x + 4} = 1 the smaller candidate x=0x = 0 fails, while in x2/3=4x^{2/3} = 4 both 88 and 8-8 are genuine.

  • Trap Trusting a self-similar value without asking whether the expression settles: 1+2+4+8+1 + 2 + 4 + 8 + \cdots gives S=1+2SS = 1 + 2S, hence S=1S = -1.

    Fix Every line of that algebra is right; writing S=S = was not. The running totals 1,3,7,15,1, 3, 7, 15, \ldots pass every bound, so there is no number to name. Grandi's 11+11 - 1 + 1 - \cdots fails the same way: bracketing gives 00 or 11, and the self-similar move gives 12\tfrac{1}{2}.

  • Trap Misreading the copy inside, turning 2+2+\sqrt{2 + \sqrt{2 + \cdots}} into x=2+xx = 2 + \sqrt{x} or x=2+xx = \sqrt{2} + x.

    Fix The copy must be the whole expression again. Under the outermost radical sits 22 plus another full tower, so the equation is x=2+xx = \sqrt{2 + x}.

  • Trap Slipping a sign or a factor in the reductions: x2+y2=s2+2px^2 + y^2 = s^2 + 2p, or x3+y3=s33px^3 + y^3 = s^3 - 3p.

    Fix The cross term is subtracted, and the cubic correction carries a factor of ss. With s=6s = 6, p=4p = 4: x2+y2=368=28x^2 + y^2 = 36 - 8 = 28, not 4444; and x3+y3=21672=144x^3 + y^3 = 216 - 72 = 144, not 204204.

Chapter test Questions from across the chapter