Rearrange until it looks like ax² + bx + c = 0, then use what you already know.
Solve for x
x + 6x = 5
Multiply everything by x: x² + 6 = 5x, so x² − 5x + 6 = 0, which factors as (x − 2)(x − 3) = 0. Both x = 2 and 3 are allowed, since neither is 0. Check: 2 + 3 = 5 ✓ and 3 + 2 = 5 ✓ C flips the signs — factoring gives (x − 2), so the root is +2.
x(x + 4) = 21
not standard form
x² + 4x = 21
expand
x² + 4x − 21 = 0
zero on one side
(x + 7)(x − 3) = 0, x = −7 or 3
factor and solve
Setting x = 21 and x + 4 = 21 from the original is invalid. 21 can be made from endless pairs of factors, so nothing follows about either bracket.
Note the restriction first, then multiply through.
x + 8x = 6
x ≠ 0
x² + 8 = 6x
× x, every term
x² − 6x + 8 = 0
standard form
x = 2 or x = 4
neither is 0, so both stand
Multiplying the x term by x is what creates the x². That is why a rational equation so often turns quadratic.
1x + 1x − 3 = 1
x ≠ 0, x ≠ 3
(x − 3) + x = x(x − 3)
× x(x − 3)
2x − 3 = x² − 3x
expand
x² − 5x + 3 = 0
D = 25 − 12 = 13
x = 5 ± √132
≈ 4.30 or 0.70, both allowed
Neither root is 0 or 3, so both survive. When a root does match a restriction, it is extraneous and must be thrown out — the checking step is not optional.
The algebra was flawless and still produced an unusable answer. Multiplying by (x − 4) quietly assumed it was not zero.
A bus travels 120 km
10 km/h faster saves 1 hour
120s − 120s + 10 = 1
time = distance ÷ speed
120(s + 10) − 120s = s(s + 10)
× s(s + 10)
s² + 10s − 1200 = 0
1200 = 120 × 10
(s + 40)(s − 30) = 0, s = 30 km/h
reject −40
Check: 12030 = 4 hours, and 12040 = 3 hours — exactly one hour less ✓
Isolate the root, square both sides, solve the quadratic, then test each answer in the ORIGINAL.
Solve √x + 7 = x + 1.
x + 7 = (x + 1)²
square both sides
x + 7 = x² + 2x + 1
expand the bracket fully
x² + x − 6 = 0, so (x + 3)(x − 2) = 0
candidates −3 and 2
x = 2: √9 = 3 and 2 + 1 = 3
keep ✓
x = −3: √4 = 2 but −3 + 1 = −2
reject ✗
Squaring makes 2 = −2 look true, because both sides square to 4. A student who stops after factoring answers "−3 or 2". A square root is never negative, so the only solution is x = 2.
If one part is the square of another, call the smaller part u, solve for u, then go back to x.
Solve x⁴ − 13x² + 36 = 0.
u² − 13u + 36 = 0
let u = x², so x⁴ = u²
(u − 4)(u − 9) = 0
u = 4 or u = 9
x² = 4 or x² = 9
put x² back
x = ±2 or x = ±3
four solutions; check: 16 − 52 + 36 = 0 ✓
Solve x + √x − 6 = 0.
u² + u − 6 = 0
let u = √x, so x = u²
(u + 3)(u − 2) = 0
u = −3 or u = 2
√x = −3 is impossible; √x = 2
a root is never negative
x = 4
check: 4 + 2 − 6 = 0 ✓
Answering "4 or 9" to the first equation gives the values of u, not x. Put x = 4 into x⁴ − 13x² + 36 and you get 256 − 208 + 36 = 84, not 0. The last step, u back to x, is part of the answer.
u can be a whole bracket: solve (x² − 2x)² − 11(x² − 2x) + 24 = 0.
u² − 11u + 24 = 0
let u = x² − 2x
(u − 3)(u − 8) = 0
u = 3 or u = 8
x² − 2x − 3 = 0, so x = 3 or x = −1
u = 3 is a quadratic in x
x² − 2x − 8 = 0, so x = 4 or x = −2
so is u = 8: four real solutions
Each value of u gives its own quadratic in x, and each of those can give two roots.
The same routine, with the endings students do not expect.
Every root rejected: x² − 1x − 1 = 2
x² − 1 = 2x − 2
x ≠ 1; × (x − 1)
x² − 2x + 1 = 0, a perfect square
one candidate, x = 1
no solution
x = 1 is the restriction ✗
When the only candidate breaks a denominator, the equation has no solution at all.
A fraction root: 3x = 5 + 2x
3x² = 5x + 2
x ≠ 0; × x
3x² − 5x − 2 = 0, so (3x + 1)(x − 2) = 0
standard form, then factor
x = 2 or x = −13
check: 5 + 2 ÷ (−13) = 5 − 6 = −1 = 3(−13) ✓
A leading number in front of x² usually means one root is a fraction.
A bracket on both sides: (x − 1)(x + 3) = 3x + 9
x² + 2x − 3 = 3x + 9
expand the left
x² − x − 12 = 0, so (x − 4)(x + 3) = 0
zero on one side
x = 4 or x = −3
check x = −3: (−4)(0) = 0 and −9 + 9 = 0 ✓
The right side is 3(x + 3). Dividing both sides by (x + 3) leaves x − 1 = 3 and loses x = −3.
Expect a rational equation, a radical equation, or a fourth-degree equation that is really a quadratic in x² or √x — often with one root that must be rejected, or a "how many real solutions" question. The fastest route is restrictions first, then clear, square or substitute, then solve the quadratic and go back to x. The 10-second check is non-negotiable here: put each candidate into the ORIGINAL equation, because that is where extraneous roots show themselves.
Solve for x
(x + 2)(x − 5) = 8
Expand: x² − 3x − 10 = 8, so x² − 3x − 18 = 0, which factors as (x − 6)(x + 3) = 0. So x = 6 or −3. Check: (6 + 2)(6 − 5) = 8 ✓ A sets each original bracket to zero, which only works when the product is zero — here it is 8, and those values give 0 instead.
Hint: expand first and move the 8 across. Only then does factoring tell you anything about the roots.
