1
Look for the halved exponent.
If the middle power is exactly half the first, the equation is quadratic in disguise.
| Equation | Let u be | Becomes |
|---|
| x⁴ − 13x² + 36 = 0 | x² | u² − 13u + 36 = 0 |
|---|
| x − 5√x + 6 = 0 | √x | u² − 5u + 6 = 0 |
|---|
| x⁶ − 7x³ − 8 = 0 | x³ | u² − 7u − 8 = 0 |
|---|
| (x+1)² − 5(x+1) + 6 = 0 | x + 1 | u² − 5u + 6 = 0 |
|---|
| 1x² − 4x + 3 = 0 | 1x | u² − 4u + 3 = 0 |
|---|
The pattern is always the same three parts: something squared, that same something,
and a number. Name the something u and the equation you are left with is one you can
already solve.
2
Substitute, then solve the quadratic.
let u = x², so x⁴ = u²
because (x²)² = x⁴
u² − 13u + 36 = 0
an ordinary quadratic
(u − 4)(u − 9) = 0
×36, +(−13)
u = 4 or u = 9
not the answer yet
u is not x
Stopping at 4 and 9 answers a question about u. The problem asked for x, and there
is one more line to go — this is the single most common way marks are lost here.
3
Go back to x — and expect four answers.
x = −3, −2, 2, 3
four solutions
A quartic can have up to four roots, and each value of u splits into two. Two answers
to a fourth-degree equation is a warning sign.
4
The radical one: the same move.
x is (√x)², so u = √x turns it into a quadratic too.
u² − 5u + 6 = 0
with u = √x
(u − 2)(u − 3) = 0
u = 2 or 3
√x = 2 → x = 4
square both sides
x = ±2, ±3 · x = 4 or 9
Reject a negative u here
√x is never negative, so a solution like u = −2 must be thrown away — it produces no
real x. Squaring can also invent roots, which is why every answer from a radical
equation gets substituted back.
5
When a value of u leads nowhere.
x⁴ + 3x² − 4 = 0
u² + 3u − 4 = 0
u = 1 or u = −4
u = 1
x² = 1
x = ±1 ✓
u = −4
x² = −4
no real x ✗
So this quartic has only two real roots. Each u is a candidate, not a guarantee — check
that it survives the trip back.
The bracket version needs no unsquaring
For (x+1)² − 5(x+1) + 6 = 0 with u = x + 1, the roots u = 2 and u = 3 give
x + 1 = 2 and x + 1 = 3 — so x = 1 and x = 2. One subtraction each, and no ± to
worry about.
6
Variations: a leading 4, u = 1x, a fractional exponent.
Solve 4x⁴ − 13x² + 9 = 0.
4u² − 13u + 9 = 0, u = x²
ac = 36: pair −4 and −9
(4u − 9)(u − 1) = 0 → u = 94 or 1
a fraction for u
x = ±32 or ±1
x = 32: 4(8116) − 13(94) + 9 = 0 ✓
Different: with a leading coefficient the u’s can be fractions, and the ± still applies to each.
Solve 2x² − 1x − 1 = 0.
2u² − u − 1 = 0, u = 1x
1x² = u²
(2u + 1)(u − 1) = 0 → u = −½ or 1
—
x = −2 or 1
flip each u; x = −2: ½ + ½ − 1 = 0 ✓
Different: the way back is a reciprocal, not a root — no ±, and x = 0 is banned from the start.
u = −½ is not x = −½
Stopping at u gives −½ and 1, and −½ fails: 8 + 2 − 1 = 9, not 0. The
substitution was u = 1x, so each u has to be turned upside down.
Solve x2/3 − 5x1/3 + 4 = 0.
u² − 5u + 4 = 0, u = x1/3
23 is twice 13
(u − 1)(u − 4) = 0 → 3√x = 1 or 4
—
x = 1 or 64
cube each; 64: 16 − 20 + 4 = 0 ✓
Different: the halved power is a fraction; undo it with a cube, so one x per u.
7
Cubes have no ±, and u can be a whole expression.
With u = x³ the way back is a cube root, and a real number has only one real cube root. Solve x6 − 26x³ − 27 = 0.
u² − 26u − 27 = 0 → (u − 27)(u + 1) = 0
u = x³
x³ = 27 → x = 3
not ±3: (−3)³ = −27
x³ = −1 → x = −1
a negative u is fine for a cube
x = 3 or x = −1 — two real roots, sum 2
When the repeated piece is x² − 2x, call the whole of it u. Solve (x² − 2x)² − 11(x² − 2x) + 24 = 0.
u² − 11u + 24 = 0 → u = 3 or u = 8
u = x² − 2x
x² − 2x = 3 → (x − 3)(x + 1) = 0
x = 3 or −1
x² − 2x = 8 → (x − 4)(x + 2) = 0
x = 4 or −2
x = −2, −1, 3, 4
each u gives its own quadratic
Sum of the u's is not the sum of the x's
Asked for the sum of the solutions, adding u = 27 and u = −1 (or 3 and 8) answers
the wrong question. Go back to x first: here 3 + (−1) = 2, and −2 − 1 + 3 + 4 = 4.
On the exam
These appear as “how many real solutions”, “sum (or product) of the real
solutions”, or a plain solve. Spot the halved power, solve for u, then go back to
x with the right inverse: ± for a square, one value for a cube, reject a
negative √x or an impossible x² = −4.
✓
Check: every root, in the original.
x = −3
81 − 117 + 36 = 0
✓
x = 2
16 − 52 + 36 = 0
✓
x = 9
9 − 5(3) + 6 = 0
✓ √9 = 3
x = 4
4 − 5(2) + 6 = 0
✓
Remember the flow
Spot the halved power · Let u = the smaller piece · Solve · Return to x
Then test each root — squaring and reciprocals both invent solutions.
