Visual walkthrough — Formation of quadratic with given roots

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Step 1 — What a picture of a quadratic looks like

WHAT. A quadratic is a rule that turns a number $x$ into another number $y$. When you plot every pair $(x,y)$, the dots trace a smooth U-shape called a parabola.

WHY start here. Before touching any symbol, we need to agree on what we are drawing. Everything later is just "reading" this curve.

PICTURE. The picture has two axes. The horizontal line is the $x$-axis — the "input" line, where we read which number we fed in. The vertical line is the $y$-axis — the "height" line, where we read how tall the curve is at that input. The two orange dots are the only two places where the curve dips down to height $y=0$ and touches the horizontal axis.

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Step 2 — Turn one crossing into a "distance from" measurement

WHAT. Pick any input $x$ on the horizontal axis. The quantity $(x-\alpha )$ measures how far $x$ is to the right of the first root (it is negative if $x$ is to the left).

$\underset{\text{where we are}}{\underbrace{(x}}-\underset{\text{first crossing}}{\underbrace{\alpha )}}$

WHY this expression and not something fancier. We want a gadget that becomes exactly zero the moment $x$ lands on the root, and is non-zero everywhere else. Subtraction does precisely that: $x-\alpha =0$ only when $x=\alpha$. This is the whole trick, wearing no disguise.

PICTURE. The magenta bracket shows the gap $(x-\alpha )$ shrinking to nothing as the moving point $x$ slides onto the orange root.

Recall Why not

$(x+\alpha )$? Because $(x+\alpha )$ hits zero at $x=-\alpha$, the wrong place. To zero out at $\alpha$ you must subtract $\alpha$. This is the single most common sign slip. ::: Subtract the root: factor is $(x-\alpha )$.

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Step 3 — Two roots, two "distance" gadgets, multiplied

WHAT. Do the same for the second root: $(x-\beta )$ is the distance from $x$ to $\beta$. Now multiply the two distances together:

$P(x)=\underset{\text{zero at }\alpha }{\underbrace{(x-\alpha )}}\cdot \underset{\text{zero at }\beta }{\underbrace{(x-\beta )}}$

WHY multiply? A product is zero the instant any one of its parts is zero. So $P(x)=0$ exactly when $x=\alpha$ or $x=\beta$ — and at no other input. That is precisely the behaviour a quadratic with those two roots must have. We didn't guess the formula; we forced it.

PICTURE. Two coloured factor-curves (each a straight line crossing zero at its own root) get multiplied into the U-shaped $P(x)$ that touches zero at both orange dots.

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Step 4 — Expand the product into standard shape

WHAT. Multiply out $(x-\alpha )(x-\beta )$ using the ordinary "each term times each term" rule. First write out all four products:

$$(x-\alpha )(x-\beta )=\underset{x^2}{\underbrace{x\cdot x}}\underset{-\beta x}{\underbrace{+x\cdot (-\beta )}}\underset{-\alpha x}{\underbrace{+(-\alpha )\cdot x}}\underset{+\alpha \beta }{\underbrace{+(-\alpha )(-\beta )}}$$

Now combine the two middle terms. Both $-\alpha x$ and $-\beta x$ carry a factor of $x$, so we pull the shared $x$ out (this is the distributive law used in reverse):

$-\alpha x-\beta x=(-\alpha -\beta )x=-(\alpha +\beta )x$

Here the two negative signs were factored into one leading minus, leaving $(\alpha +\beta )$ inside. Putting the pieces back together:

$P(x)=x^2-(\alpha +\beta )x+\alpha \beta$

Term by term:

• $x^2$ — the leading piece; it makes the shape a parabola rather than a line.
• $-(\alpha +\beta )x$ — the middle coefficient carries the sum of the roots, wearing a minus sign.
• $\alpha \beta$ — the constant carries the product of the roots, no sign flip.

WHY expand. The factor form is beautiful, but exams and further algebra want the tidy $x^2+bx+c$ shape. Expanding just re-dresses the same curve.

PICTURE. A rectangle-area diagram: each sub-rectangle is one of the four products $x\cdot x$, $x\cdot (-\beta )$, $(-\alpha )\cdot x$, $(-\alpha )(-\beta )$ — and the two orange strips are the ones we merge into $-(\alpha +\beta )x$.

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Step 5 — The freedom knob $k$ (why one answer becomes infinitely many)

WHAT. Roots only fix where the curve crosses the axis, not how steep it is, nor even which way it opens. So we may multiply the whole thing by any non-zero number $k$:

$P(x)=k(x-\alpha )(x-\beta ),k\ne 0$

WHY $k$ matters. Squash or stretch the parabola vertically and the two crossing points stay put, because at a root the factor part is already $0$ and $k\cdot 0=0$ no matter what $k$ is.

• $k>0$ — parabola opens upward (a valley). Bigger $k$ = narrower; smaller $k$ = wider.
• $k<0$ — the sign of every height flips, so the whole curve turns upside down and opens downward (a hill). The roots do not move, because $k\cdot 0=0$ even when $k$ is negative.

Setting $k=1$ just picks the simplest upward member.

PICTURE. Four parabolas through the same two orange roots: narrow-up ($k=2$), standard-up ($k=1$), wide-up ($k=0.5$), and a flipped downward one ($k=-1$). All share the crossings; only their width and their opening direction change.

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Step 6 — Fixing $k$ with one extra point

WHAT. Give the curve one more condition — say it must pass through $(1,8)$ with roots $-2$ and $3$. Then: $8=k(1+2)(1-3)=k(3)(-2)=-6k\Rightarrow k=-\frac{4}{3}$

Notice $k$ came out negative, so this particular parabola opens downward — exactly the flipped case from Step 5.

WHY. Two roots leave a one-number gap (the height/steepness/opening-direction). One extra known point supplies exactly that one missing number, locking $k$.

PICTURE. The two roots pin the shape; the purple point $(1,8)$ tilts and stretches it until it passes through — selecting a unique (here, downward-opening) parabola.

Now put this curve into standard form. We do it in two separate, legal moves:

Move 1 — expand the factored form (just multiplying out, the curve is unchanged): $P(x)=-\frac{4}{3}(x+2)(x-3)=-\frac{4}{3}{x}^2+\frac{4}{3}x+8$

Move 2 — set the equation to zero, then clear fractions. To solve or state the equation we ask where the height is zero, i.e. $P(x)=0$: $-\frac{4}{3}{x}^2+\frac{4}{3}x+8=0$ Multiplying both sides of an equation by the same non-zero number $(-3)$ keeps the equality true and does not move the roots (a zero on the right stays zero). This just removes the fractions: $4x^2-4x-24=0$

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Step 7 — The degenerate cases (never leave a scenario unshown)

WHAT & WHY & PICTURE. Three special situations that beginners hit and panic over:

(a) Both roots equal, $\alpha =\beta$. Then $P(x)=k(x-\alpha {)}^2$. The two crossings merge into one touch — the parabola kisses the axis and bounces back. Product $={\alpha }^2$, sum $=2\alpha$.

(b) A root at zero, $\alpha =0$. The factor $(x-0)=x$, so the constant term $\alpha \beta$ vanishes: $P(x)=x^2-\beta x$. The curve passes through the origin.

(c) Conjugate/complex roots. If the roots are $2\pm \sqrt{3}$ (or $p\pm qi$), the $\sqrt{3}$ or $i$ parts cancel in both sum and product: $\text{sum}=(2+\sqrt{3})+(2-\sqrt{3})=4,\text{product}=(2+\sqrt{3})(2-\sqrt{3})=4-3=1$ giving the clean real equation $x^2-4x+1=0$. See Complex roots.

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The one-picture summary

Everything in one frame: two roots → two subtraction-factors → multiply → expand → optional $k$ knob (positive = upward, negative = downward) → standard equation.

Recall Feynman retelling (explain it to a 12-year-old)

Two treasures are buried on a number line at spots $\alpha$ and $\beta$. Stand at any spot $x$. Measure your distance to the first treasure: that's $(x-\alpha )$. Measure your distance to the second: $(x-\beta )$. Multiply the two distances. The answer is zero only when you're standing right on a treasure — because a product is zero only if one part is zero. So $(x-\alpha )(x-\beta )$ is a "treasure detector" that beeps zero at exactly the right spots. Multiply it out — combining the two middle pieces by pulling out their shared $x$ — and you get $x^2-(\text{sum})x+(\text{product})$: the sum of the treasure positions sits in the middle (with a minus), the product sits at the end. You may amplify the whole detector by any loudness $k$ without moving the treasures; if $k$ is positive the bowl opens up, if $k$ is negative it flips upside-down like a hill, but the treasures stay put because $k$ times zero is still zero. One extra clue (a known point) tells you the exact loudness and even its sign. And if the two treasures land on the same spot, sit at zero, or come as a mirror pair — the same recipe still works, it just merges, drops a term, or cancels its square roots.