Formation of quadratic with given roots: Worked examples

This page is a stress test. The parent note showed you the machine:

Below we hit every kind of input this machine can be fed. First the map, then the worked cases.

The scenario matrix

Every problem you will ever see is one of these cells. Column "Covered by" points to the example that nails it.

#	Case class	What makes it tricky	Covered by
A	Both roots positive integers	baseline, warm-up	Ex 1
B	Mixed signs (one $+$ , one $-$ )	sign of the factor $(x - α)$	Ex 2
C	Both roots negative	double sign flip	Ex 2
D	A repeated root ( $α = β$ )	parabola touches, doesn't cross	Ex 3
E	One root is zero	the constant term vanishes	Ex 4
F	Irrational conjugate roots $p \pm q$	radicals must cancel	Ex 5
G	Complex conjugate roots $p \pm q i$	real coefficients from imaginary roots	Ex 6
H	Given sum & product only (roots unknown)	use Vieta directly	Ex 7
I	Extra point fixes $k$ (scaling)	roots alone are not enough	Ex 8
J	Real-world word problem	translate words → sum/product	Ex 9
K	Exam twist: new roots built from old	transform, don't re-solve	Ex 10
L	Negative stretch ( $k < 0$ , opens downward)	leading coefficient flips sign	Ex 11

Tools we will invoke and why: the Factor theorem (turns "is a root" into "is a factor"), Vieta's formulas (lets us skip the roots entirely when only sum & product are known), the Quadratic formula (to check by going backwards), Complex roots (Case G), and Graph transformations (Cases K and L, where we move or flip the curve).

Figure — Formation of quadratic with given roots

What the figure shows (in words, in case it does not load): three parabolas over the same grey x-axis. The blue curve dips below the axis and comes back up, cutting it at two separate points ( $x = 4$ and $x = 6$ ) — this is the ordinary "two real roots / crossing" case. The orange curve just dips down to kiss the axis at a single point ( $x = 5$ ) and rises again without passing through — this is the "double root / touching" case. The green curve floats entirely above the axis, its lowest point still positive — it never meets the axis, the "complex roots" case. So: crossing = two real roots, touching = repeated root, never-touching = complex roots.

This figure is your reference chart for the "how many times does it meet the axis?" question. Cases D (touching) and G (never meeting) are the two unusual ways captured by the orange and green curves. Cases A, B, C and E are all ordinary crossings — two real roots, the blue-curve behaviour (Case E's crossing simply happens to sit at the origin). Keep the picture in mind: we will point back to it in Ex 3 and Ex 6.

Case A — both roots positive integers

Worked example Ex 1 — Cell A

Statement: Form the monic quadratic with roots $α = 4$ and $β = 6$ . ("Monic" means leading coefficient $= 1$ , i.e. we take $k = 1$ .)

Forecast: Before reading on — will the middle coefficient be $+ 10$ or $- 10$ ? Guess now. (Predicting the sign first sharpens your feel for the minus in Vieta's template.)

Step 1. Write the factor form: $P (x) = (x - 4) (x - 6)$ . Why this step? Factor theorem: a root of $4$ forces the factor $(x - 4)$ , because plugging $x = 4$ makes it zero.

Step 2. Expand: $(x - 4) (x - 6) = x^{2} - 6 x - 4 x + 24 = x^{2} - 10 x + 24$ . Why this step? Standard form $a x^{2} + b x + c$ is the format examiners expect.

Answer: $x^{2} - 10 x + 24 = 0$ . (The $- 10$ — did you forecast the minus?) This is an ordinary two-real-root crossing — the blue curve type in the top figure.

Verify: Sum of roots $4 + 6 = 10 = - \frac{b}{a} = - \frac{- 10}{1}$ ✓. Product $4 \cdot 6 = 24 = \frac{c}{a} = 24$ ✓.

Cases B & C — mixed signs and both negative

Worked example Ex 2 — Cells B and C

Statement: (B) roots $5$ and $- 2$ . (C) roots $- 3$ and $- 7$ . Form both monic quadratics.

Forecast: Which one has a positive constant term? Guess before Step 3. (Guessing forces you to reason about products of signs before touching algebra.)

Step 1 (B). Factors: root $5 \to (x - 5)$ ; root $- 2 \to (x - (- 2)) = (x + 2)$ . Why this step? The rule is "subtract the root". A negative root flips into a plus inside the bracket. This is the #1 sign trap.

Step 2 (B). $(x - 5) (x + 2) = x^{2} + 2 x - 5 x - 10 = x^{2} - 3 x - 10$ . Why this step? Expanding to read off coefficients.

Answer (B): $x^{2} - 3 x - 10 = 0$ . Constant is negative — makes sense, opposite-sign roots multiply to a negative number.

Step 1 (C). Factors: root $- 3 \to (x + 3)$ ; root $- 7 \to (x + 7)$ . Why this step? Both roots negative, so both brackets flip to plus.

Step 2 (C). $(x + 3) (x + 7) = x^{2} + 10 x + 21$ . Why this step? Same expansion.

Answer (C): $x^{2} + 10 x + 21 = 0$ . Let us read off the two coefficients carefully, from two independent viewpoints that must agree:

Constant term $c = + 21$ . By direct expansion, $(+ 3) (+ 7) = + 21$ . (Vieta agrees: product of roots $= (- 3) (- 7) = + 21$ .) Positive, because two negatives multiply to a positive.
Middle coefficient $b = + 10$ . By expansion: the cross terms give $+ 3 x + 7 x = + 10 x$ , so $b = + 10$ directly. By Vieta (separate check): the sum of roots is $- 3 + (- 7) = - 10$ , and Vieta's rule is $b = - (sum) = - (- 10) = + 10$ . Both routes give $+ 10$ — that agreement is the point.

Verify (B): sum $5 + (- 2) = 3 = - (- 3)$ ✓, product $5 \cdot (- 2) = - 10$ ✓. Verify (C): sum $- 3 + (- 7) = - 10 = - (10)$ ✓, product $(- 3) (- 7) = 21$ ✓.

Case D — repeated (double) root

Worked example Ex 3 — Cell D

Statement: Form the monic quadratic whose only root is $α = β = 5$ .

Forecast: Will the graph cross the x-axis or just kiss it? Guess. (Your gut answer here is the intuition we then confirm with the discriminant.)

Step 1. Both roots are $5$ , so both factors are $(x - 5)$ : $P (x) = (x - 5) (x - 5) = (x - 5)^{2} .$ Why this step? A "double root" literally means the factor $(x - 5)$ appears twice. There is no second, different crossing.

Step 2. Expand: $(x - 5)^{2} = x^{2} - 10 x + 25$ . Why this step? Standard form.

Answer: $x^{2} - 10 x + 25 = 0$ . The graph touches the axis at $x = 5$ (a tangent), it does not pass through — this is exactly the orange (touching) curve in the top reference figure. Look back at it now: the vertex sits on the axis.

Verify: The Quadratic formula gives discriminant $b^{2} - 4 a c = (- 10)^{2} - 4 (1) (25) = 100 - 100 = 0$ . A zero discriminant is exactly the fingerprint of a repeated root ✓. Sum $5 + 5 = 10 = - (- 10)$ ✓, product $5 \cdot 5 = 25$ ✓.

Case E — one root is zero

Worked example Ex 4 — Cell E

Statement: Form the monic quadratic with roots $0$ and $- 8$ .

Forecast: What will the constant term $c$ be? Think about the product of the roots. (Forecasting the constant before computing tests whether you have linked $c$ to the product.)

Step 1. Factors: root $0 \to (x - 0) = x$ ; root $- 8 \to (x + 8)$ . Why this step? Zero is a perfectly ordinary root; its factor is just $x$ .

Step 2. $P (x) = x (x + 8) = x^{2} + 8 x$ . Why this step? Multiply out.

Answer: $x^{2} + 8 x = 0$ . Notice no constant term: $c = 0$ . This is still an ordinary two-real-root crossing (blue-curve type) — one of its crossings just happens to land exactly at the origin. Why? Product of roots $= 0 \cdot (- 8) = 0 = c$ . Whenever a root is $0$ , the constant vanishes and the parabola passes through the origin.

Verify: Sum $0 + (- 8) = - 8 = - (8)$ ✓, product $0 \cdot (- 8) = 0 = c$ ✓.

Case F — irrational conjugate roots

Worked example Ex 5 — Cell F

Statement: Roots are $3 + 2$ and $3 - 2$ . Form the monic quadratic.

Forecast: The roots contain $2$ . Will $2$ survive in the final equation? Guess yes/no. (This forecast previews the whole reason conjugate pairs exist — worth committing to an answer.)

Step 1. Use sum & product instead of grinding the brackets. Sum $= (3 + 2) + (3 - 2) = 6$ — the $2$ terms cancel. Why this step? Conjugates $p + q$ and $p - q$ are built to cancel their irrational halves on adding. This is why they come in pairs in a real-coefficient equation.

Step 2. Product $= (3 + 2) (3 - 2) = 3^{2} - (2)^{2} = 9 - 2 = 7$ . Why this step? Difference-of-squares $(a - b) (a + b) = a^{2} - b^{2}$ turns the radical into an ordinary square.

Step 3. Assemble with $x^{2} - (sum) x + (product)$ : $x^{2} - 6 x + 7 = 0.$ Why this step? Vieta's template; note the minus in front of the sum.

Answer: $x^{2} - 6 x + 7 = 0$ — no radical survives. (Answer to forecast: no.)

Verify: Discriminant $= (- 6)^{2} - 4 (1) (7) = 36 - 28 = 8 > 0$ , and $8 /2 = 2$ , giving roots $\frac{6 \pm 8}{2} = 3 \pm 2$ ✓.

Case G — complex conjugate roots

Worked example Ex 6 — Cell G

Statement: Roots are $2 + 3 i$ and $2 - 3 i$ , where $i = - 1$ . Form the real quadratic.

Forecast: The roots are imaginary. Can the final equation still have real coefficients? Guess. (Whatever you guess, the surprise-or-confirmation is exactly what fixes this idea in memory.)

Step 1. Sum $= (2 + 3 i) + (2 - 3 i) = 4$ — the imaginary parts cancel. Why this step? Just like radicals in Case F, conjugate pairs $p \pm q i$ cancel their imaginary halves on adding. See Complex roots.

Step 2. Product $= (2 + 3 i) (2 - 3 i) = 2^{2} - (3 i)^{2} = 4 - 9 i^{2} = 4 - 9 (- 1) = 4 + 9 = 13$ . Why this step? Difference of squares again, plus the key fact $i^{2} = - 1$ , which turns the imaginary part into a positive real number.

Step 3. Assemble: $x^{2} - 4 x + 13 = 0$ . Why this step? Vieta template with sum $4$ , product $13$ .

Answer: $x^{2} - 4 x + 13 = 0$ — all coefficients real (answer to forecast: yes). This parabola never touches the x-axis — exactly the green (never-touching) curve in the top reference figure. Look back: its lowest point floats above the axis.

Verify: Discriminant $= (- 4)^{2} - 4 (1) (13) = 16 - 52 = - 36 < 0$ , confirming no real roots; $\frac{4 \pm - 36}{2} = \frac{4 \pm 6 i}{2} = 2 \pm 3 i$ ✓.

Case H — sum & product given, roots hidden

Worked example Ex 7 — Cell H

Statement: A quadratic has roots that sum to $- \frac{1}{2}$ and multiply to $- 3$ . Form the monic equation, then find the roots.

Forecast: Will the equation have integer coefficients? Guess. (Committing to yes/no makes the fraction-clearing step feel earned rather than arbitrary.)

Step 1. Plug straight into $x^{2} - (sum) x + (product)$ : $x^{2} - (- \frac{1}{2}) x + (- 3) = x^{2} + \frac{1}{2} x - 3 = 0.$ Why this step? Vieta's formulas let us build the quadratic without ever knowing the individual roots — that is the whole point of Case H.

Step 2. Clear the fraction by multiplying through by $2$ (allowed — it is the $k = 2$ scaling, roots unchanged): $2 x^{2} + x - 6 = 0.$ Why this step? Integer coefficients are tidier and reveal factorability.

Step 3. Factor: $2 x^{2} + x - 6 = (2 x - 3) (x + 2)$ , giving roots $x = \frac{3}{2}$ and $x = - 2$ . Why this step? To check the hidden roots are consistent.

Answer: $2 x^{2} + x - 6 = 0$ (or monic $x^{2} + \frac{1}{2} x - 3 = 0$ ).

Verify: $\frac{3}{2} + (- 2) = - \frac{1}{2}$ ✓ (matches given sum), $\frac{3}{2} \cdot (- 2) = - 3$ ✓ (matches given product).

Case I — an extra point fixes the stretch $k$

Worked example Ex 8 — Cell I

Statement: A parabola has roots $- 1$ and $4$ and passes through the point $(2, - 9)$ . Find its equation.

Forecast: Is the monic version $(x + 1) (x - 4)$ enough? Why or why not? (Guessing "not enough, we need $k$ " primes you to see what the extra point is for.)

Step 1. Roots give the shape but not the stretch, so keep $k$ : $P (x) = k (x + 1) (x - 4) .$ Why this step? Infinitely many parabolas share these two roots (Mistake 2 in the parent). The point $(2, - 9)$ is the extra fact that pins down $k$ .

Step 2. Impose $P (2) = - 9$ : $- 9 = k (2 + 1) (2 - 4) = k (3) (- 2) = - 6 k ⟹ k = \frac{- 9}{- 6} = \frac{3}{2} .$ Why this step? One equation, one unknown $k$ — exactly solvable.

Step 3. Substitute back and expand: $P (x) = \frac{3}{2} (x + 1) (x - 4) = \frac{3}{2} (x^{2} - 3 x - 4) = \frac{3}{2} x^{2} - \frac{9}{2} x - 6.$ Why this step? Deliver in standard form.

Answer: $\frac{3}{2} x^{2} - \frac{9}{2} x - 6 = 0$ , or clearing fractions (multiply by $2$ ), $3 x^{2} - 9 x - 12 = 0$ .

Verify: $P (2) = \frac{3}{2} (3) (- 2) = \frac{3}{2} \cdot (- 6) = - 9$ ✓ (passes through the point). Roots of $3 x^{2} - 9 x - 12$ : divide by 3 → $x^{2} - 3 x - 4 = (x + 1) (x - 4)$ , roots $- 1, 4$ ✓.

Case J — real-world word problem

Worked example Ex 9 — Cell J

Statement: A rectangular garden has perimeter $26$ m and area $40 m^{2}$ . The two side lengths are the roots of a quadratic. Find that quadratic and the dimensions.

Forecast: From "perimeter $26$ " — what is the sum of the two sides? Work it out before Step 1. (Making this prediction first is exactly the word-to-Vieta translation the problem is testing.)

Step 1. Let the sides be $α$ and $β$ . Perimeter $= 2 (α + β) = 26 \Rightarrow α + β = 13$ . Why this step? Translating the word "perimeter" into a sum of roots — the exact ingredient Vieta needs.

Step 2. Area $= α β = 40$ . Why this step? Translating "area" into a product of roots.

Step 3. Build the quadratic $x^{2} - (sum) x + (product)$ : $x^{2} - 13 x + 40 = 0.$ Why this step? Same template — the word problem is just Case H in disguise.

Step 4. Factor to read off the physical dimensions: $(x - 5) (x - 8) = 0 \Rightarrow x = 5$ or $x = 8$ . Why this step? The two roots are literally the two side lengths of the garden.

Answer: Quadratic $x^{2} - 13 x + 40 = 0$ ; the garden is $5 m \times 8 m$ .

Verify (with units): perimeter $= 2 (5 + 8) = 2 (13) = 26$ m ✓; area $= 5 \times 8 = 40 m^{2}$ ✓. Both roots positive — sensible for lengths ✓.

Case K — exam twist: new roots from old

Worked example Ex 10 — Cell K

Statement: The equation $x^{2} - 5 x + 6 = 0$ has roots $α, β$ . Without solving for $α, β$ first, form the quadratic whose roots are $α + 1$ and $β + 1$ .

Forecast: New roots are each shifted up by $1$ . Should the graph move left or right? Guess. (Predicting the shift direction connects the algebra to a picture before you compute.)

Step 1. Read off old Vieta values: $α + β = 5$ , $α β = 6$ . Why this step? Vieta's formulas extract sum and product straight from the coefficients — no root-finding needed.

Step 2. New sum $= (α + 1) + (β + 1) = (α + β) + 2 = 5 + 2 = 7$ . Why this step? Shifting both roots by $+ 1$ adds $2$ to their sum.

Step 3. New product $= (α + 1) (β + 1) = α β + (α + β) + 1 = 6 + 5 + 1 = 12$ . Why this step? Expand the product and substitute the known sum & product — pure algebra, still no roots.

Step 4. Assemble: $x^{2} - 7 x + 12 = 0$ . Why this step? Vieta template with new sum $7$ , new product $12$ .

Answer: $x^{2} - 7 x + 12 = 0$ . (Its roots are $3$ and $4$ — the old roots $2, 3$ each pushed up by $1$ , so the graph shifts right; a "replace $x$ with $x - 1$ " shift.)

Verify: Old roots: $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ , so $α, β = 2, 3$ . New roots $2 + 1 = 3$ , $3 + 1 = 4$ ; and $x^{2} - 7 x + 12 = (x - 3) (x - 4)$ ✓. Sum $3 + 4 = 7$ ✓, product $3 \cdot 4 = 12$ ✓.

Case L — negative stretch, parabola opens downward

Worked example Ex 11 — Cell L

Statement: A parabola has roots $- 1$ and $3$ , and its maximum value is $8$ (so it opens downward). Find its equation.

Forecast: For a curve that opens downward, will the leading coefficient $k$ be positive or negative? Guess. (Committing here links the sign of $k$ to the direction the parabola opens before any algebra.)

Step 1. Start from the factor form, keeping $k$ unknown: $P (x) = k (x + 1) (x - 3) .$ Why this step? Roots fix the two crossings; the "maximum value" is the extra fact that will pin down $k$ (just like Case I, but now the pinning info is a peak height, not a passing point).

Step 2. A downward parabola peaks at its vertex, halfway between the roots: $x = \frac{- 1 + 3}{2} = 1$ . Why this step? By symmetry the highest point sits exactly between the two roots. That is where the maximum of $8$ occurs.

Step 3. Impose $P (1) = 8$ : $8 = k (1 + 1) (1 - 3) = k (2) (- 2) = - 4 k ⟹ k = \frac{8}{- 4} = - 2.$ Why this step? One equation, one unknown $k$ . The negative result $k = - 2$ is precisely what makes the parabola open downward (answer to forecast: negative).

Step 4. Substitute back and expand: $P (x) = - 2 (x + 1) (x - 3) = - 2 (x^{2} - 2 x - 3) = - 2 x^{2} + 4 x + 6.$ Why this step? Standard form, now with a negative leading coefficient $a = - 2$ .

Answer: $- 2 x^{2} + 4 x + 6 = 0$ (equivalently, dividing by $- 2$ , the roots are unchanged: $x^{2} - 2 x - 3 = 0$ ).

Verify: Vertex value $P (1) = - 2 (2) (- 2) = 8$ ✓ (matches the stated maximum). Roots of $- 2 x^{2} + 4 x + 6$ : divide by $- 2$ → $x^{2} - 2 x - 3 = (x + 1) (x - 3)$ , roots $- 1, 3$ ✓. Leading coefficient $- 2 < 0$ ✓ confirms it opens downward.

Wrap-up

Recall Which cell is this problem? (self-test)

"Roots $1 \pm 5$ " is which case? ::: Case F — irrational conjugates; use sum $= 2$ , product $= 1 - 5 = - 4$ , giving $x^{2} - 2 x - 4 = 0$ . "Roots $0$ and $9$ " is which case? ::: Case E — one zero root; constant term is $0$ , equation $x^{2} - 9 x = 0$ . "Roots sum $8$ , product $16$ " — what special feature? ::: Case D-flavoured — discriminant $64 - 64 = 0$ , so it is a repeated root ( $x = 4$ twice), equation $x^{2} - 8 x + 16 = 0$ . "Roots $2, 5$ but opens downward with peak height $10$ " — which case? ::: Case L — negative $k$ ; vertex at $x = 3.5$ , solve $k (3.5 - 2) (3.5 - 5) = 10$ for a negative $k$ .