The method that works when factoring will not.
Solve for x
2x² + 3x − 5 = 0
a = 2, b = 3, c = −5, so b² − 4ac = 9 + 40 = 49 and √49 = 7. That gives x = (−3 ± 7) ÷ 4, so x = 1 or x = −52. A divides by 2 instead of 2a = 4: (−3 ± 7) ÷ 2 gives 2 and −5. D uses +3 instead of −b: (3 ± 7) ÷ 4 gives 52 and −1, both signs flipped. B makes both slips at once.
The equation must equal zero before you read anything off.
2x² + 3x − 5 = 0
a
2
b
3
c
−5
c is −5, not 5. Nearly every wrong answer in this topic starts with a sign dropped at this step, before any arithmetic has happened.
Not equal to zero yet? Expand and move every term to one side first. 2x(x + 4) = 3 − x becomes 2x² + 8x = 3 − x, then 2x² + 9x − 3 = 0, so a = 2, b = 9 and c = −3.
x = −b ± √b² − 4ac 2a
flip the sign of b
this is why there are two roots
the whole top sits over it
The bar runs under everything on top, including the −b. Dividing only the square root by 2a is a common slip.
That piece is the discriminant, and it decides everything.
b² − 4ac
the discriminant
3² − 4(2)(−5)
drop the values in
9 + 40
−4 · 2 · −5 = +40
= 49
a perfect square
√49 = 7, so the answers will be tidy.
−4ac with c negative gives a plus. Writing 9 − 40 = −31 here would suggest no real roots, which is wrong.
x = −3 ± 74
−3 + 74 = 44
x = 1
−3 − 74 = −104
x = −2.5
2x² + 3x − 5 = (2x + 5)(x − 1)
setting each bracket to zero gives the same two answers
| If the discriminant is | Then | Factoring |
|---|---|---|
| a perfect square | two tidy roots | would have worked |
| positive, not a square | two roots with surds | will not work |
| zero | one repeated root | a perfect square |
| negative | no real roots | will not work |
The formula never fails, so when a quadratic resists factoring for more than a minute, stop hunting and use it.
Most formula answers are not whole numbers, and the choices show them simplified.
D = 64 − 52 = 12
b² − 4ac
x = 8 ± √122
the formula
√12 = √4 · √3 = 2√3
pull out the square factor
x = 8 ± 2√32
every term on top is even
x = 4 ± √3
divide each term by 2
8 ± 2√32 is 4 ± √3. Writing 8 ± √3 halves only the surd, and 4 ± 2√3 halves only the 8 — both are planted among the choices.
In a word problem a length keeps only the positive root. If a width w satisfies w(w + 4) = 11, then w² + 4w − 11 = 0, D = 16 + 44 = 60 = 4 · 15, and w = −4 ± 2√152 = −2 ± √15. The width is √15 − 2, about 1.87; −2 − √15 is negative and is thrown out.
The formula is the same each time; what changes is how the last line comes out.
Negative a, and 2a does not cancel fully: −2x² + 4x + 1 = 0
a = −2, b = 4, c = 1
a carries the minus
D = 16 − 4(−2)(1) = 16 + 8 = 24
√24 = 2√6
x = −4 ± 2√6−4
2a = −4
x = 2 ± √62
divide every term by −2
The negative 2a flips every sign on top, and ± just swaps its two roots. Only a 2 cancels, so a fraction stays.
Halving only the 2 leaves √6 undivided. Either keep the fraction or write 1 ± √62; the two roots are about 2.22 and −0.22.
Discriminant zero: 4x² − 12x + 9 = 0
D = 144 − 4(4)(9) = 144 − 144 = 0
√0 = 0
x = 12 ± 08
both branches are the same
x = 32 only
it is (2x − 3)² = 0
The ± adds and subtracts nothing, so there is one repeated root, not two.
Discriminant negative: x² − 2x + 5 = 0
D = 4 − 4(1)(5) = 4 − 20 = −16
a negative under the root
no real roots
stop here
This is why we work out D first: a negative D ends the problem in one line. Only if complex roots are asked do we go on, to x = 2 ± 4i2 = 1 ± 2i.
Multiply every term by the LCD, and it becomes an ordinary quadratic.
x2 = 1 + 1x
the LCD is 2x
x² = 2x + 2
every term × 2x
x² − 2x − 2 = 0
zero first
D = 4 + 8 = 12
a = 1, b = −2, c = −2
x = 2 ± 2√32 = 1 ± √3
simplify as in Step 6
Last, make sure no root makes a denominator zero. Here only x = 0 is banned, and neither 1 + √3 nor 1 − √3 is 0, so both stand.
Forgetting to multiply the 1 by 2x turns the equation into x² = 1 + 2, a different problem with different roots.
Formula items usually test the last line: surds simplified with 2a cancelled through every term, or an equation that must first be cleared of fractions and set to 0. Quick check: the roots of ax² + bx + c = 0 add to −ba — here 1 + √3 and 1 − √3 add to 2, matching x² − 2x − 2 = 0.
Solve for x
3x² − 5x − 2 = 0
a = 3, b = −5, c = −2, so −b = +5 and −4ac = +24, giving b² − 4ac = 25 + 24 = 49. Then x = (5 ± 7) ÷ 6, so x = 2 or x = −13. C flips both signs — the trap when b and c are already negative.
Hint: both b and c are negative, so watch the two sign traps — −b becomes +5, and −4ac becomes +24. The discriminant is a perfect square again.
