Everything you learned for solving, run backwards.
Write the equation
Write a quadratic equation whose roots are 3 and
−5.
Sum = 3 + (−5) = −2 and product = 3(−5) = −15. The equation is x² − (sum)x + product, so x² − (−2)x − 15 = x² + 2x − 15. D forgets that the x-coefficient is minus the sum, so a negative sum turns positive.
1
A root of 3 means a factor of (x − 3).
The sign flips on the way into the bracket.
x = 3 → (x − 3)
move the 3 across
x = −5 → (x + 5)
a negative root gives a plus
(x − 3)(x + 5) = 0
factored form
x² + 2x − 15 = 0
expand: −3x + 5x = +2x
Not (x + 3)(x − 5)
That one has roots −3 and 5 — the same digits, the wrong equation. Read your
brackets back before expanding: each one should give the root you started with
when set to zero.
A root of 0 gives the factor (x − 0), which is just x. Roots 0 and 9: x(x − 9) = 0, so x² − 9x = 0, with no constant term.
2
Or go through the sum and the product.
x² − (sum)x + (product) = 0
sum = 3 + (−5) = −2
add the roots
product = 3 × (−5) = −15
multiply them
x² − (−2)x + (−15) = 0
into the pattern
x² + 2x − 15 = 0
same answer
The minus sign lives in the formula, so a sum of −2 becomes +2x. Multiplying the
brackets out is the safer route when the roots are simple; the sum-and-product
route wins when they are not.
3
There is more than one right answer.
x² + 2x − 15 = 0roots 3 and −5✓
2x² + 4x − 30 = 0every term doubledsame roots ✓
−x² − 2x + 15 = 0every term negatedsame roots ✓
Multiplying an equation that equals zero by any non-zero number changes nothing
about its solutions. Exam questions dodge the ambiguity by asking for the version
with integer coefficients and a leading coefficient of 1.
Fractional roots: clear them first
Roots ⅔ and −1 give (3x − 2)(x + 1) = 0, that is 3x² + x − 2 = 0. Writing
(x − ⅔)(x + 1) is correct but not in integer form, and the choices will be.
4
Irrational and imaginary roots arrive in pairs.
Which is what makes the sum and product come out tidy.
roots 2 + √3 and 2 − √3
conjugates
sum = 4
the radicals cancel
product = 2² − (√3)² = 1
difference of squares
x² − 4x + 1 = 0
integer coefficients again
The same trick handles 3 ± 2i: the sum is 6 and the product is 9 − 4i² = 13, so
the equation is x² − 6x + 13 = 0. If a question hands you one irrational or
imaginary root and asks for the equation, its partner is implied.
5
From a graph: start at the vertex.
Vertex form carries the vertex in plain sight — one more point fixes a.
y = a(x − h)² + k
vertex form
y = a(x − 2)² − 8
vertex (2, −8)
0 = a(4 − 2)² − 8
the curve passes through (4, 0)
4a = 8, a = 2
the stretch factor
y = 2(x − 2)² − 8
or y = 2x² − 8x expanded
The vertex alone is not enough
Infinitely many parabolas share a vertex — they differ in width. Without a
second point you cannot find a, and answering y = (x − 2)² − 8 assumes a = 1 for
no reason.
6
Roots plus one point: y = a(x − r)(x − s).
The roots fix the brackets; the extra point fixes a.
y = a(x − 1)(x − 4)
roots 1 and 4
8 = a(−1)(−4)
put in (0, 8)
8 = 4a, so a = 2
y = 2(x² − 5x + 4)
expand the brackets
y = 2x² − 10x + 8
If the extra fact is a maximum or minimum instead, use the vertex: it sits midway between the roots. Here x = 2.5 gives y = 2(1.5)(−1.5) = −4.5, the minimum — so a minimum of −4.5 would have fixed a = 2 just as well.
Right roots, wrong graph
y = x² − 5x + 4 has the same roots but crosses the y-axis at 4, not 8. When a picture gives a third point, a is rarely 1.
7
Variations: a repeated root, opposite roots, two fractions.
The pattern x² − (sum)x + (product) = 0 still does all the work.
One root, repeated: −4 is a double root
(x + 4)(x + 4) = 0
the same factor twice
x² + 8x + 16 = 0
sum −8, product 16
“Equal roots” or “a double root” still means two factors. Writing just x + 4 = 0 gives a line, not a quadratic.
Opposite roots: 32 and −32
(2x − 3)(2x + 3) = 0
x = 32 gives 2x − 3 = 0
4x² − 9 = 0
a difference of squares
Opposite roots add to 0, so there is no x term at all.
Two fractional roots: −12 and 34
sum = −24 + 34 = 14
common denominator
product = (−12)(34) = −38
multiply
x² − (14)x − 38 = 0
the pattern
8x² − 2x − 3 = 0
every term × 8, the LCD
Check by factoring: (2x + 1)(4x − 3) gives −12 and 34. Multiply by the LCD of the whole line, not just of one fraction.
Clearing only one fraction
Multiplying by 4 gives 4x² − x − 32 = 0, which still has a fraction. Times 8 clears both quarters and eighths at once.
8
New roots from old ones.
Find the old roots, change them, rebuild.
x² − 5x + 6 = (x − 2)(x − 3)
old roots 2 and 3
tripled: 6 and 9
sum 15, product 54
x² − 15x + 54 = 0
x² − (sum)x + product
reciprocals: 12 and 13
flip each root
(2x − 1)(3x − 1) = 0
integer factors
6x² − 5x + 1 = 0
the old coefficients, reversed
Reciprocal roots always reverse the coefficients: the roots of cx² + bx + a = 0 are the reciprocals of the roots of ax² + bx + c = 0.
On the exam
Expect the backwards direction: roots or a graph given, equation asked, usually “with integer coefficients”. Check a choice by its sum (−ba) and product (ca), and when a graph is shown, by its y-intercept as well.
✓
Check: solve your own equation.
Building it backwards is only right if solving it forwards returns the roots.
factorx² + 2x − 15 = (x − 3)(x + 5)3 and −5 ✓
second one2x² − 8x = 2x(x − 4)roots 0 and 4, as drawn ✓
vertex test−b2a = 84 = 2the axis is x = 2 ✓
Remember the flow
Roots → factors → expand · Graph → vertex form → one point for a
Then clear fractions so the coefficients are whole.
Now you try
Practice problem
Write the equation
Write the equation with roots −2 and 7.
Sum = −2 + 7 = 5 and product = −14, giving x² − 5x − 14 = 0. A flips the sign of the sum. Quick check: the product of the roots is negative, so the constant term must be negative too.
