No middle term, no formula needed — isolate the square and undo it.
Solve for x
3x² − 75 = 0
3x² = 75 gives x² = 25, so x = ±5. D keeps only the positive root — both signs square to 25, and both genuinely solve the equation. A forgets to take the root at all.
A quadratic with b = 0, or one already folded into a bracket.
| If it looks like | Best method |
|---|---|
| x² = k | extract the root — two steps |
| (x − h)² = k | extract the root |
| x² + bx + c = 0, factorable | factor |
| anything else | the quadratic formula |
The formula still works on 3x² − 75 = 0, with b = 0. It is just three times the writing for the same pair of answers.
Nothing may be attached to the x² when you take the root.
3x² − 75 = 0
as given
3x² = 75
add 75 to both sides
x² = 25
divide by 3 — the square is alone
x = ±5
root both sides
Going from 3x² = 75 straight to x√3 = √75 is legal but pointless, and most people drop the 3 instead. Clear the coefficient while it is still a plain division.
The radical sign asks for the positive root only. But solving x² = 25 asks which numbers square to 25, and there are two. The ± is put in by hand because the equation has two answers, not because the symbol produces them.
Root both sides, then unwrap what is left.
(x − 4)² = 49
the square is already alone
x − 4 = ±7
root both sides
x = 4 ± 7
add 4 — to both branches
x = 11 or x = −3
4 + 7 and 4 − 7
Same picture as before, slid along. The bracket says where the centre is; the root says how far out the answers sit.
A number in front of the bracket comes off first, as in Step 2. 3(x − 5)² − 12 = 15: add 12, 3(x − 5)² = 27; divide by 3, (x − 5)² = 9; then x − 5 = ±3, so x = 8 or x = 2.
In a story, the negative branch is usually thrown out. A square’s side grows by 5 cm and the new area is 81 cm²: (s + 5)² = 81, s + 5 = ±9, so s = 4 or s = −14. A side cannot be negative, so s = 4 and the original area is 16 cm².
(x − 4)² = 49 becomes x² − 8x − 33 = 0, and now you need the formula for something that was two lines away. Leave the bracket intact.
x² = 20
20 is not a perfect square
x = ±√20
still exact
x = ±2√5
√20 = √4 · √5
two irrational roots
simplify the radical, keep the ±
no real solution
nothing real squares to a negative
A positive number on the right gives two answers, zero gives exactly one (x = 0), and a negative gives none at all — unless the question allows imaginary numbers, where x = ±3i.
Same two moves every time — isolate the square, root with ± — but the unwrapping changes.
The x inside the bracket has a coefficient: (3x − 1)² = 16
3x − 1 = ±4
root both sides
3x = 5 or 3x = −3
add 1 to each branch
x = 53 or x = −1
divide each by 3
Split into two branches before dividing. The roots are no longer symmetric about a whole number: the centre is 13.
The right side is not a perfect square: (x + 3)² = 12
x + 3 = ±√12 = ±2√3
√12 = √4 · √3
x = −3 ± 2√3
subtract 3
x ≈ 0.46 or x ≈ −6.46
only if decimals are asked for
The answer stays in two parts, a whole number and a surd. They are unlike terms and do not combine.
Adding −3 and 2 as if both were “√3” gives −√3, which is wrong. A whole number and a surd are like x and 1: leave them side by side.
The square comes out negative: 5x² + 20 = 0
5x² = −20
subtract 20
x² = −4
divide by 5
no real solution
no real number squares to −4
Always isolate before deciding: the equation looked harmless until the square stood alone. Had the right side been 0, as in (x − 6)² = 0, there would be one root, x = 6.
(x − h)² = k has roots h ± √k: the centre h, then the distance √k each way.
x = 2 ± √k
the centre is 2
7 is 5 to the right of 2
so √k = 5 and k = 25
the other root is 5 to the left
2 − 5
x = −3
(−3 − 2)² = 25 ✓
Run it backwards when both roots are given: the centre h is their midpoint and √k is half their gap. Roots −1 and 11 give h = 5 and √k = 6, so the equation is (x − 5)² = 36.
(x + 4)² has its centre at −4, not 4 — the centre is the number that makes the bracket zero. Getting this sign wrong moves both roots.
Extraction items are quick ones: an isolated square, sometimes with a coefficient to divide out first, or one root given and the other asked. The 10-second check is symmetry — the two roots must sit the same distance either side of the bracket’s centre.
The negative one is where the mistakes hide, so test that one properly.
Solve for x
2(x + 1)² − 32 = 0
Divide by 2 first: (x + 1)² = 16, so x + 1 = ±4. That gives x = 3 or x = −5. B stops at ±4 — the ± step tells you about x + 1, and the 1 still has to come off.
