9709 Pure 1 · 1.1 Quadratics · Session 3 of 3 · 60 min
Solving Through to Roots
Two problems that look nothing like quadratics, and turn into quadratics anyway.
Do Now: one simultaneous pair, cold5 min
The Intersection Lab: what the algebra looks like4 min
Direct instruction: substitution, then the nickname trick17 min
Worked example: wkshtF Q1, with the safety check7 min
Practice: wkshtF, tap each question to check18 min
Recap and exit ticket5 min
Do Now · 5 min
Solve the pair.
\(x + y = 5\)
\(x^2 + y^2 = 17\)
rearrange the linear \(y = 5 - x\)
put it into the quadratic \(x^2 + (5 - x)^2 = 17\)
expand and collect \(2x^2 - 10x + 8 = 0\)
divide both sides by 2 \(x^2 - 5x + 4 = 0\)
\(x = 1\) or \(x = 4\), so the answers are \((1, 4)\) and \((4, 1)\)
That last quadratic, \(x^2 - 5x + 4 = 0\), is the same one that ran through Session 2.
The Intersection Lab · 4 min
Two answers, because there are two crossing points.
\(x + y = 5\) is a straight line. \(x^2 + y^2 = 17\) is a circle. Solving them together asks one question: where do they touch? Drag the slider to slide the line across the circle.
The discriminant is back. \(\Delta > 0\) means the line cuts the circle twice. \(\Delta = 0\) means it just touches, which is the tangency you met in Session 1. \(\Delta < 0\) means it misses, and the algebra says the same thing by refusing to give you a real answer.
Direct instruction · 1 of 6
One linear + one quadratic: substitute, solve, pair up.
Substitute the linear equation into the quadratic, so only one letter is left.
Solve the quadratic you get. Usually two \(x\)-values.
Back-substitute each one. Every \(x\) needs its partner \(y\). The answer is a pair of coordinates: an \(x\)-value and its matching \(y\)-value.
Stopping at "\(x = 1\) or \(x = 4\)" is half an answer. Those are just the two places on the picture, read off the \(x\)-axis. You still owe the height of each one: \(y = 5 - 1 = 4\), and \(y = 5 - 4 = 1\). The answer is \((1, 4)\) and \((4, 1)\).
Why not use elimination, the way you would with two linear equations? Elimination scales one or both equations so a variable's coefficient matches the other equation, then adds or subtracts to cancel that variable out.
try it: multiply the linear equation by 3 \(3 \times (x + y = 5) \to 3x + 3y = 15\): still only \(x\) and \(y\) to the first power, nothing to cancel against \(x^2 + y^2 = 17\)
Multiplying the linear equation by any constant keeps it linear. It never grows an \(x^2\) or a \(y^2\) to cancel, so no multiple of it lines up with the quadratic. Substitution is the only route, which is the word the syllabus uses.
Direct instruction · 2 of 6 · worked
All of it, on one pair.
linear \(2x + y = 7\) quadratic \(x^2 + xy = 10\)
1 · rearrange the linear for \(y\) \(y = 7 - 2x\)
2 · substitute into the quadratic \(x^2 + x(7 - 2x) = 10\)
3 · expand the bracket \(x^2 + 7x - 2x^2 = 10\)
4 · everything on one side \(-x^2 + 7x - 10 = 0\)
5 · multiply by \(-1\) so \(x^2\) is positive \(x^2 - 7x + 10 = 0\)
6 · factor and solve \((x - 2)(x - 5) = 0 \to x = 2\) or \(x = 5\)
7 · each \(x\) back into \(y = 7 - 2x\) \(x = 2 \to y = 3\) · \(x = 5 \to y = -3\)
\((2, 3)\) and \((5, -3)\)
check both in both originals \(2(2) + 3 = 7\) ✓, \(2^2 + 2(3) = 10\) ✓ · \(2(5) + (-3) = 7\) ✓, \(5^2 + 5(-3) = 10\) ✓
Why each \(x\) keeps its own \(y\): the \(y\) came out of that \(x\) through \(y = 7 - 2x\), so they travel together. Swap them and it breaks. Try \((2, -3)\): \(2(2) + (-3) = 1\), and the linear equation wanted 7.
Direct instruction · 3 of 6
The nickname trick.
You already know how to solve anything shaped like \(u^2 + bu + c = 0\). So when an equation is too ugly to solve, the move is to make it that shape.
Find the piece that keeps showing up. Cover it with a single letter, \(u\). Solve the easy equation you are left with. Then uncover it and finish the job.
A nickname is not an answer. If a friend is called Red, "Red" is not a person: you still have to go find the person. Same here. Landing on \(u = 4\) does not end the question, it starts the last step.
Direct instruction · 4 of 6
Spot the repeated piece.
In each one, the same thing appears twice: once squared, once on its own. That repeated thing is what gets the nickname.
\(x^4 - 5x^2 + 4 = 0\) (\(x^4\) is \((x^2)^2\), so let \(u = x^2\))
\(x - 4\sqrt{x} + 3 = 0\) (\(x\) is \((\sqrt{x})^2\), so let \(u = \sqrt{x}\))
\(\tan^2 x - \tan x - 1 = 0\) (let \(u = \tan x\))
Standard form first, every time. That third one is usually printed as \(\tan^2 x = 1 + \tan x\), and nothing works until everything is on one side.
The part practice sets usually skip is what comes after: swapping back to \(x\) once \(u\) is known. Getting "\(u = 1\)" is not the finish line, it is one more equation to solve.
Direct instruction · 5 of 6
The full arc, once.
\(x^4 - 5x^2 + 4 = 0\)
let \(u = x^2\) \(u^2 - 5u + 4 = 0 \to (u - 1)(u - 4) = 0 \to u = 1\) or \(4\)
undo the nickname \(x^2 = 1 \to x = \pm 1\) · \(x^2 = 4 \to x = \pm 2\)
Four roots: \(x = -2, -1, 1, 2\)
Two \(u\)-values turned into four \(x\)-values, because each \(x^2 = \text{number}\) question has a positive and a negative answer. Dropping the \(\pm\) is how a correct method still loses half the marks.
This one had four real roots. A degree-four equation does not always get four. You will see one in a moment that gets two.
Direct instruction · 6 of 6
Some answers for \(u\) are impossible.
Before you undo the nickname, ask what that nickname is allowed to be, and why.
\(u = x^2\) cannot be negative, since squaring a real number never gives a negative result
\(u = \sqrt{x}\) cannot be negative either, since the \(\sqrt{\ }\) symbol always means the non-negative root
\(u = \cos x\) lives between \(-1\) and \(1\), the whole range cosine ever takes
\(u = x^3\) can be anything, negatives included, since every real number has exactly one real cube root
Check the nickname you actually used. Do not assume negative means reject: it depends which of the four you're using.
A \(u\)-value outside its range has no solution. Write down why you're rejecting it: that line is what earns the mark. The safety check is always the same, put your answer back into the original equation and see if it holds.
Try it: pick a nickname, pick a value, and watch what comes back
Worked example · 7 min · wkshtF Q1
\(x^4 - 3x^2 - 4 = 0\)
let \(u = x^2\) \(u^2 - 3u - 4 = 0 \to (u - 4)(u + 1) = 0 \to u = 4\) or \(u = -1\)
check each \(u\) against its range \(u = x^2\), so \(u\) cannot be negative
\(x^2 = 4 \to x = \pm 2\) · \(x^2 = -1 \to\) no real solution, reject
Same \(u = x^2\) substitution as the arc before it, and this time one branch really does get rejected. Writing down why \(u = -1\) is rejected earns the mark. An answer of \(x = \pm 2\) with no reason given looks like a guess.
Practice · together · 9 min · wkshtF
Your turn. Tap a question to check it.
Q2 \(x - 4\sqrt{x} + 3 = 0\)
\(u = \sqrt{x}\) \(u^2 - 4u + 3 = 0 \to (u - 1)(u - 3) = 0 \to u = 1\) or \(3\)
both are \(\ge 0\), so both survive: \(\sqrt{x} = 1 \to x = 1\) · \(\sqrt{x} = 3 \to x = 9\)
\(u = \cos x\) \(u^2 - 2u - 8 = 0 \to (u - 4)(u + 2) = 0 \to u = 4\) or \(-2\)
both fall outside \(-1 \le \cos x \le 1\), so reject both
No solutions. That is the answer.
check \(\cos x\) can never reach \(4\) or \(-2\), since nothing satisfies \(-1 \le \cos x \le 1\) there, so there is nothing left to substitute back
Stop at the values of \(\sin x\) and \(\cos x\). Solving for the angle comes in 1.5, so this answer is complete.
Recap · 2 min
Both halves of today were the same move.
Neither problem today was a quadratic when you met it. Both became one, and you already knew what to do from there.
two equations, two letters substitute the linear one in → an ordinary quadratic in \(x\)
one ugly equation nickname the repeated piece → an ordinary quadratic in \(u\)
And both have a last step people skip. On a pair, every \(x\) owes you its \(y\). On a disguise, every \(u\) owes you its \(x\), and some \(u\)-values owe you nothing at all because they were impossible.
Exit ticket · 5 min
Last check, five minutes, on Schoology.
Solve \(x^6 + 7x^3 = 8\). Hint: what's the nickname, and what is it allowed to be?
A cubic hiding inside a quadratic like this is a real Cambridge exam pattern, not just a practice-sheet one.
Before you leave, press Submit. An attempt left open scores nothing.
Homework: Transum "Hidden Quadratics," Levels 1 and 2 (free and auto-marked).
Self-check with full worked solutions: Paul's Online Notes, 6 problems. tutorial.math.lamar.edu/problems/alg/reducibletoquadratic.aspx