Formation of quadratic with given roots: Exercises

Before we start, the one tool we lean on everywhere. If a number $α$ is a root of a polynomial (plugging it in gives zero), then $(x - α)$ is a factor — this is the Factor theorem. Two roots $α, β$ give two factors, and the quadratic is their product.

Look at the picture below to see this before we write any formula. A parabola is a U-shaped curve. The places where it crosses the horizontal axis (the x-axis) are called the roots — there the height $P (x)$ is exactly $0$ . Each crossing at a number $α$ contributes a factor $(x - α)$ , because that factor is zero precisely at $x = α$ . The whole quadratic is the two factors multiplied, times a stretch dial we call $k$ .

Figure — Formation of quadratic with given roots

Now the shortcut from Vieta's formulas. The picture below shows why the sum and product of the roots pop out when you multiply the factors — the "middle" term collects both roots (their sum) and the "end" term multiplies both roots (their product).

Level 1 — Recognition

Goal: read off factors and the sum/product form without slips.

L1-Q1

Form a monic (leading coefficient $1$ ) quadratic with roots $4$ and $- 1$ .

Recall Solution — L1-Q1

A root of $4$ gives factor $(x - 4)$ . A root of $- 1$ gives factor $(x - (- 1)) = (x + 1)$ . $P (x) = (x - 4) (x + 1) = x^{2} + x - 4 x - 4 = x^{2} - 3 x - 4$ Answer: $x^{2} - 3 x - 4 = 0$ . Check with sum/product: sum $= 4 + (- 1) = 3$ , so the $x$ term is $- 3 x$ ✓; product $= 4 \cdot (- 1) = - 4$ ✓.

L1-Q2

The roots have sum $= - 2$ and product $= - 15$ . Write the monic quadratic.

Recall Solution — L1-Q2

Plug straight into $x^{2} - (sum) x + (product) = 0$ : $x^{2} - (- 2) x + (- 15) = x^{2} + 2 x - 15 = 0$ Answer: $x^{2} + 2 x - 15 = 0$ . (Its roots are $3$ and $- 5$ : sum $- 2$ , product $- 15$ ✓.)

L1-Q3

Which of these has roots $2$ and $5$ ? (a) $x^{2} + 7 x + 10$ (b) $x^{2} - 7 x + 10$ (c) $x^{2} - 7 x - 10$ .

Recall Solution — L1-Q3

Sum $= 7$ → the $x$ -coefficient must be $- 7$ . Product $= 10$ → constant $+ 10$ . That is exactly $x^{2} - 7 x + 10$ . Answer: (b).

Level 2 — Application

Goal: handle radicals, fractions, and the free constant $k$ .

L2-Q1

Roots are $3 + 5$ and $3 - 5$ . Form the monic quadratic.

Recall Solution — L2-Q1

These are conjugates, so use sum and product to dodge messy multiplication.

Sum $= (3 + 5) + (3 - 5) = 6$ (the $5$ cancels).
Product $= (3 + 5) (3 - 5) = 3^{2} - (5)^{2} = 9 - 5 = 4$ (difference of squares).

$x^{2} - 6 x + 4 = 0$ Answer: $x^{2} - 6 x + 4 = 0$ . Notice: irrational roots, but the equation is fully rational — that is the point of conjugate pairs.

L2-Q2

Roots are $\frac{1}{2}$ and $\frac{2}{3}$ . Form a quadratic with integer coefficients.

Recall Solution — L2-Q2

Monic first:

Sum $= \frac{1}{2} + \frac{2}{3} = \frac{3}{6} + \frac{4}{6} = \frac{7}{6}$
Product $= \frac{1}{2} \cdot \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$

$x^{2} - \frac{7}{6} x + \frac{1}{3} = 0$ Fractions are legal but ugly. Multiply through by the common denominator $6$ (allowed — it is just the free constant $k = 6$ , and it does not move the roots): $6 x^{2} - 7 x + 2 = 0$ Answer: $6 x^{2} - 7 x + 2 = 0$ . Quick check by factoring: $(2 x - 1) (3 x - 2) = 0$ gives $x = \frac{1}{2}, \frac{2}{3}$ ✓.

L2-Q3

Roots are $- 2$ and $3$ , and the parabola passes through the point $(1, 8)$ . Find the exact quadratic (fix $k$ ).

Recall Solution — L2-Q3

Start with the roots but keep $k$ free, because "passes through a point" is extra info that pins the vertical stretch: $P (x) = k (x + 2) (x - 3)$ Impose $P (1) = 8$ : $8 = k (1 + 2) (1 - 3) = k (3) (- 2) = - 6 k \Rightarrow k = - \frac{4}{3}$ So $P (x) = - \frac{4}{3} (x^{2} - x - 6) = - \frac{4}{3} x^{2} + \frac{4}{3} x + 8$ Clearing the fraction (multiply by $- 3$ ): $4 x^{2} - 4 x - 24 = 0$ , i.e. $x^{2} - x - 6 = 0$ if you only want the shape. Answer: $P (x) = - \frac{4}{3} (x + 2) (x - 3)$ , equivalently $- \frac{4}{3} x^{2} + \frac{4}{3} x + 8$ .

Level 3 — Analysis

Goal: work backwards from combined conditions, and handle complex roots.

L3-Q1

A quadratic $x^{2} + b x + c = 0$ has roots $α, β$ with $α + β = 4$ and $α^{2} + β^{2} = 20$ . Find $b$ and $c$ .

Recall Solution — L3-Q1

We know the sum but need the product to build the equation. Use the identity $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ Why this identity? It links a "square-sum" (given) to the ordinary sum and product (what Vieta needs). $20 = (4)^{2} - 2 α β = 16 - 2 α β \Rightarrow 2 α β = - 4 \Rightarrow α β = - 2$ Now $x^{2} - (sum) x + (product) = x^{2} - 4 x - 2$ , so $b = - 4, c = - 2$ . Answer: $b = - 4, c = - 2$ ; equation $x^{2} - 4 x - 2 = 0$ .

L3-Q2

Form a quadratic with real coefficients having root $2 - 3 i$ (where $i^{2} = - 1$ ).

The picture below shows why the second root is forced. On the complex plane — a flat map where the horizontal axis is the real part and the vertical axis is the imaginary part — the number $2 - 3 i$ sits at coordinates $(2, - 3)$ . For a quadratic with real coefficients the two roots must be mirror images across the real axis (a conjugate pair), so its partner $2 + 3 i$ sits at $(2, + 3)$ . When you add them the vertical parts cancel; when you multiply them the $i$ terms cancel too — that is why the equation stays fully real.

Recall Solution — L3-Q2

For a quadratic with real coefficients, complex roots must come as conjugate pairs (see Complex roots). So the other root is $2 + 3 i$ .

Sum $= (2 - 3 i) + (2 + 3 i) = 4$ (imaginary parts cancel).
Product $= (2 - 3 i) (2 + 3 i) = 2^{2} - (3 i)^{2} = 4 - 9 i^{2} = 4 + 9 = 13$ .

$x^{2} - 4 x + 13 = 0$ Answer: $x^{2} - 4 x + 13 = 0$ . (The quantity $b^{2} - 4 a c = (- 4)^{2} - 4 (1) (13) = 16 - 52 = - 36 < 0$ confirms the roots are complex, not real.)

L3-Q3

The roots of $x^{2} - 5 x + 6 = 0$ are $α, β$ . Form a new quadratic whose roots are $α + 1$ and $β + 1$ .

Recall Solution — L3-Q3

First read the old roots off Vieta: sum $α + β = 5$ , product $α β = 6$ . Now build sum/product of the new roots directly, without solving for $α, β$ :

New sum $= (α + 1) + (β + 1) = (α + β) + 2 = 5 + 2 = 7$ .
New product $= (α + 1) (β + 1) = α β + (α + β) + 1 = 6 + 5 + 1 = 12$ .

$x^{2} - 7 x + 12 = 0$ Answer: $x^{2} - 7 x + 12 = 0$ . (Old roots are $2, 3$ ; new roots $3, 4$ : sum $7$ , product $12$ ✓.)

Level 4 — Synthesis

Goal: chain several transformations, use graphs, mix conditions.

L4-Q1

The roots of $2 x^{2} - 3 x - 4 = 0$ are $α, β$ . Form a monic quadratic whose roots are $\frac{1}{α}$ and $\frac{1}{β}$ (the reciprocals).

Recall Solution — L4-Q1

From $2 x^{2} - 3 x - 4 = 0$ : divide the standard Vieta relations by the leading coefficient $2$ .

$α + β = - \frac{- 3}{2} = \frac{3}{2}$
$α β = \frac{- 4}{2} = - 2$

Reciprocal sum and product:

$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{3/2}{- 2} = - \frac{3}{4}$
$\frac{1}{α} \cdot \frac{1}{β} = \frac{1}{α β} = \frac{1}{- 2} = - \frac{1}{2}$

Monic quadratic: $x^{2} - (- \frac{3}{4}) x + (- \frac{1}{2}) = x^{2} + \frac{3}{4} x - \frac{1}{2} = 0$ Clearing fractions (multiply by $4$ ): $4 x^{2} + 3 x - 2 = 0$ . Answer: $x^{2} + \frac{3}{4} x - \frac{1}{2} = 0$ , or $4 x^{2} + 3 x - 2 = 0$ .

Why "reversing the coefficients" works (proof, not faith): if $α$ is a root of $2 x^{2} - 3 x - 4 = 0$ , substitute $x = \frac{1}{y}$ (so $y = \frac{1}{x}$ is a reciprocal root): $2 (\frac{1}{y})^{2} - 3 (\frac{1}{y}) - 4 = 0.$ Multiply every term by $y^{2}$ (allowed since $y \neq = 0$ because $α β = - 2 \neq = 0$ ): $2 - 3 y - 4 y^{2} = 0 ⟹ 4 y^{2} + 3 y - 2 = 0.$ The coefficients $2, - 3, - 4$ came out in reverse order as $- 4, - 3, 2$ , i.e. $4 y^{2} + 3 y - 2$ after multiplying by $- 1$ — exactly the shortcut, now derived.

L4-Q2

A parabola cuts the x-axis at $- 3$ and $1$ , and passes through $(- 1, - 8)$ . Find its equation, then find the vertex — using the graph below to guide each step.

Recall Solution — L4-Q2

Read the roots off the figure. The pink dots mark where the curve meets the axis: $x = - 3$ and $x = 1$ . Those give factors $(x + 3)$ and $(x - 1)$ , so $P (x) = k (x + 3) (x - 1) .$ Use the blue point on the figure to fix the stretch $k$ . The curve is drawn passing through the blue square at $(- 1, - 8)$ ; the height there is what pins $k$ (roots alone cannot): $- 8 = k (- 1 + 3) (- 1 - 1) = k (2) (- 2) = - 4 k \Rightarrow k = 2.$ So $P (x) = 2 (x + 3) (x - 1) = 2 (x^{2} + 2 x - 3) = 2 x^{2} + 4 x - 6$ . Locate the vertex using the figure's symmetry. The dashed yellow line is the axis of symmetry, sitting halfway between the two crossings at $x = \frac{- 3 + 1}{2} = - 1$ . Its height is $P (- 1) = 2 (- 1)^{2} + 4 (- 1) - 6 = 2 - 4 - 6 = - 8$ . So the lowest point of the U — the vertex — is $(- 1, - 8)$ , which happens to be the same blue point that fixed $k$ . Answer: $P (x) = 2 x^{2} + 4 x - 6 = 0$ , vertex $(- 1, - 8)$ .

Level 5 — Mastery

Goal: full problem-solving under multiple simultaneous constraints and edge cases.

L5-Q1

A monic quadratic has roots $α, β$ . You are told $α - β = 2$ and $α β = 8$ with $α > β$ . Form the quadratic, then find $α$ and $β$ .

Recall Solution — L5-Q1

We have the product but a difference, not a sum. Bridge with the identity $(α + β)^{2} = (α - β)^{2} + 4 α β$ Why? Expanding both sides: $(α + β)^{2} - (α - β)^{2} = 4 α β$ . It converts a known difference and product into the sum that Vieta wants. $(α + β)^{2} = 2^{2} + 4 (8) = 4 + 32 = 36 \Rightarrow α + β = \pm 6$ Two cases (cover both!):

Sum $= 6$ : $x^{2} - 6 x + 8 = 0 \Rightarrow (x - 4) (x - 2) = 0$ , roots $4, 2$ . Then $α = 4, β = 2$ , and $α - β = 2$ ✓.
Sum $= - 6$ : $x^{2} + 6 x + 8 = 0 \Rightarrow (x + 2) (x + 4) = 0$ , roots $- 2, - 4$ . Here $α - β = (- 2) - (- 4) = 2$ ✓ as well.

Answer: two valid quadratics — $x^{2} - 6 x + 8 = 0$ (roots $4, 2$ ) and $x^{2} + 6 x + 8 = 0$ (roots $- 2, - 4$ ). Both satisfy the difference and product; the problem does not force positivity, so both count.

L5-Q2

Find all real $m$ for which the quadratic with roots $α, β$ satisfying $α + β = m$ and $α β = m$ has equal roots (a repeated root). State the quadratic(s).

Recall Solution — L5-Q2

The quadratic is $x^{2} - m x + m = 0$ . Equal roots happen exactly when the discriminant is zero. The discriminant is the quantity under the square root in the Quadratic formula; for $a x^{2} + b x + c$ we write it as $Δ = b^{2} - 4 a c$ (the Greek letter $Δ$ , "delta", is just a name for this number). When $Δ = 0$ the parabola only touches the axis at a single point, so the two roots merge into one. $Δ = (- m)^{2} - 4 (1) (m) = m^{2} - 4 m = 0$ $m (m - 4) = 0 \Rightarrow m = 0 or m = 4$ Cover both, including the degenerate one:

$m = 0$ : equation $x^{2} = 0$ , repeated root $x = 0$ . Sum $0$ , product $0$ ✓ (edge case: both roots collapse to zero).
$m = 4$ : equation $x^{2} - 4 x + 4 = (x - 2)^{2} = 0$ , repeated root $x = 2$ . Sum $4$ , product $4$ ✓.

Answer: $m = 0$ giving $x^{2} = 0$ , and $m = 4$ giving $x^{2} - 4 x + 4 = 0$ .

L5-Q3

A quadratic $x^{2} + p x + q = 0$ has roots $α, β$ . A second quadratic has roots $α^{2}$ and $β^{2}$ . Express the second quadratic's coefficients in terms of $p$ and $q$ , and test it on $x^{2} - 3 x + 2 = 0$ .

Recall Solution — L5-Q3

From the first: sum $α + β = - p$ , product $α β = q$ . New roots $α^{2}, β^{2}$ :

New sum $= α^{2} + β^{2} = (α + β)^{2} - 2 α β = (- p)^{2} - 2 q = p^{2} - 2 q$ .
New product $= α^{2} β^{2} = (α β)^{2} = q^{2}$ .

Second quadratic (monic): $x^{2} - (p^{2} - 2 q) x + q^{2} = 0$ Test on $x^{2} - 3 x + 2 = 0$ : here $p = - 3, q = 2$ , roots $1$ and $2$ .

Predicted: $x^{2} - ((- 3)^{2} - 2 \cdot 2) x + 2^{2} = x^{2} - (9 - 4) x + 4 = x^{2} - 5 x + 4 = 0$ .
Direct: roots $1^{2} = 1$ and $2^{2} = 4$ ; sum $5$ , product $4$ → $x^{2} - 5 x + 4 = 0$ ✓.

Answer: $x^{2} - (p^{2} - 2 q) x + q^{2} = 0$ ; for $x^{2} - 3 x + 2$ it gives $x^{2} - 5 x + 4 = 0$ .

Wrap-up

Recall One-line summary of every level

L1: read off factors and sum/product ::: factor $= (x - root)$ ; $x^{2} - (sum) x + (product)$ . L2: radicals, fractions, and fixing $k$ ::: use sum/product for conjugates; a given point pins $k$ . L3: build from combined conditions ::: use identities like $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ ; complex roots come in conjugate pairs. L4: chain transformations, use graphs ::: reciprocal roots use $\frac{α + β}{α β}$ and $\frac{1}{α β}$ (guard against zero product). L5: multiple constraints, edge cases ::: differences need $(α + β)^{2} = (α - β)^{2} + 4 α β$ with $\pm$ ; equal roots mean discriminant $Δ = 0$ .

Back to the parent: Formation of quadratic with given roots. Related tools you leaned on: Vieta's formulas, Factor theorem, Quadratic formula, Complex roots, Completing the square, Graph transformations.