One number tells you the roots — no need to solve the equation at all.
Find the discriminant
2x2 − 4x + 5 = 0
b² − 4ac = (−4)² − 4(2)(5) = 16 − 40 = −24. Negative means no real roots. Note (−4)² is +16, so the sign of the answer comes entirely from the −4ac part — B drops that minus.
The equation must already equal zero. This one does.
2x2 − 4x + 5 = 0
The minus in front of 4x belongs to b. Writing b = 4 changes the answer and is the single most common slip here.
If it is not = 0 yet, move everything to one side first.
D = b2 − 4ac
It is the piece under the square root sign.
D = (−4)2 − 4(2)(5)
(−4)² not −4²
4 · a · c = 4(2)(5)
Brackets keep the minus attached to the 4.
(−4)² = +16
a minus times a minus
4(2)(5) = 40
4 · 2 = 8, then 8 · 5
D = 16 − 40
put the pieces back
D = −24
16 is smaller than 40
A negative discriminant is perfectly normal.
| If D is | Roots | What it looks like |
|---|---|---|
| positive, a perfect square | 2 rational roots | cuts the x-axis twice; the quadratic factors |
| positive, not a perfect square | 2 irrational roots | cuts twice, at surd values |
| zero | 1 real root | just touches the axis |
| negative | no real roots | never reaches the axis |
D = −24 is less than zero
The two roots are imaginary, and they come as a pair.
Real roots are x-intercepts. This curve has none.
y = 3
stays above zero
a is positive so the curve opens upward, and the whole thing sits above the axis.
Turn the wording into a condition on D, then solve for k.
| The question says | Condition on D |
|---|---|
| two distinct real roots | D > 0 |
| equal roots, exactly one real root, touches the x-axis | D = 0 |
| no real roots, never meets the x-axis | D < 0 |
| real roots (equal allowed) | D ≥ 0 |
| rational roots (whole-number a, b, c) | D is a perfect square |
D = k² − 4(1)(25) = k² − 100
a = 1, b = k, c = 25
k² − 100 < 0
no real roots
k² < 100
−10 < k < 10
k is trapped between −10 and 10
k = −20 gives k² = 400, which is not below 100. A squared unknown that must stay small gives a sandwich, bounded on both sides.
For the least whole number that works, solve first and then round the right way: x² + 6x + k = 0 has no real roots when 36 − 4k < 0, so k > 9 and the least integer is 10. (k = 9 gives D = 0, one real root.)
“Touches the x-axis” means D = 0, and the point it touches is the repeated root. For y = x² + 10x + c: 100 − 4c = 0, so c = 25 and y = (x + 5)², which touches the axis at (−5, 0).
Three ways the same b² − 4ac gets disguised.
Not in standard form: describe the roots of 2x² = 3 − x
2x² + x − 3 = 0
everything to one side
D = 1² − 4(2)(−3) = 1 + 24 = 25
a = 2, b = 1, c = −3
two distinct rational roots
25 is a perfect square
Read a, b, c only after the equation equals zero; reading b = −1 and c = 3 from the original gives D = 1 − 24 = −23 and the wrong verdict. The roots are 1 and −32.
k in both b and c: x² + 2kx + (k + 6) = 0 has equal roots
D = (2k)² − 4(1)(k + 6) = 0
equal roots means D = 0
4k² − 4k − 24 = 0
(2k)² = 4k², not 2k²
k² − k − 6 = 0, so (k − 3)(k + 2) = 0
divide by 4, factor
k = 3 or k = −2
both values work
The condition on D is itself a quadratic in k, so two values of k is normal. k = 3 gives (x + 3)²; k = −2 gives (x − 2)².
A negative a: −x² + 6x + k = 0 has real roots
D = 36 − 4(−1)(k) = 36 + 4k
a = −1 turns −4ac into +4k
36 + 4k ≥ 0
“real roots” allows equal ones
k ≥ −9
k = −9 gives the one root x = 3
“Real roots” is ≥, not >. The boundary value belongs to the answer.
Writing −4(1)(k) instead of −4(−1)(k) gives 36 − 4k ≥ 0 and k ≤ 9 — every sign wrong. Bracket a and c with their signs before multiplying.
The discriminant condition, plus one rule: the x² term must survive.
D = 36 − 4(k)(3) = 36 − 12k
a = k
36 − 12k > 0
two distinct roots
k < 3
and k ≠ 0
k = 0 leaves −6x + 3 = 0
With k = 0 the equation is the line −6x + 3 = 0, which has one root (x = ½), not two. Always test the value that kills the x² term on its own: it may leave a line with one root, or a false statement with none.
Discriminant items almost always run backwards: a condition on the roots is given and k is asked. Translate the words with the table, solve for k, then test the boundary values and any k that makes a = 0 — that is where the trap choices sit.
x = [ 4 ± √−24 ] ÷ 4
The part under the root is our D.
Find k
For what value of k does
x2 + kx + 9 = 0
have exactly ONE real root?
Exactly one real root means the discriminant is zero: k² − 4(1)(9) = 0, so k² = 36 and k = ±6. C keeps only the positive value — a square root always allows both signs, and both really do give a repeated root.
Hint: one real root means D is exactly zero. Set b² − 4ac = 0 and solve for k — expect two answers.
