Sign and transformation of roots

Two skills sit on top of the quadratic formula once you already know how to find α,β\alpha,\beta: reading off the sign of ax2+bx+cax^2+bx+c for every real xx without plotting the curve, and building a new quadratic whose roots are some transformation of α,β\alpha,\beta — shifted, scaled, negated, or inverted — without ever solving for α,β\alpha,\beta individually. Both tricks lean on the same two numbers, α+β=−b/a\alpha+\beta=-b/a and αβ=c/a\alpha\beta=c/a. The sign rule turns the position of xx relative to the roots into a same-as-aa or opposite-to-aa verdict. The transformation rule turns a rule like "add 2 to each root" into a new sum and a new product, which is all a quadratic ever needs. Together they power the wavy-curve method for inequalities and quick construction of related equations — both staples of the Quadratic Equations unit.

The core insight

The sign of ax2+bx+cax^2+bx+c flips only at its real roots and always matches aa outside them, while a transformed pair of roots gives a new quadratic through the transformed sum and product alone, with no need to solve for the original roots.

The core idea

Sign of the expression. For real coefficients with roots α≤β\alpha\le\beta, write ax2+bx+c=a(x−α)(x−β)ax^2+bx+c=a(x-\alpha)(x-\beta) whenever the roots are real. Outside the roots, (x−α)(x-\alpha) and (x−β)(x-\beta) carry the same sign, so their product is positive and the expression matches the sign of aa. Between the roots, the two factors disagree in sign, the product is negative, and the expression takes the opposite sign to aa.

sign(ax2+bx+c)={sign(a),x<α or x>β−sign(a),α<x<βsign(a),all x, if D<0\text{sign}\big(ax^2+bx+c\big)= \begin{cases} \text{sign}(a), & x<\alpha \text{ or } x>\beta\\[2pt] -\text{sign}(a), & \alpha<x<\beta\\[2pt] \text{sign}(a), & \text{all } x, \text{ if } D<0 \end{cases}

where: D=b2−4acD=b^2-4ac is the discriminant, α,β\alpha,\beta are the real roots when D≥0D\ge0 (equal if D=0D=0, in which case the expression never goes negative), and "sign(a)(a)" means positive if a>0a>0, negative if a<0a<0. Three cases fall out of this factorisation. When D>0D>0 the roots are distinct and the number line splits into three intervals, alternating sign around α\alpha and β\beta. When D=0D=0 the two roots coincide at a single point, so (x−α)2≥0(x-\alpha)^2\ge0 always, and the expression is a perfect square times aa — it touches zero once but never crosses to the other sign. When D<0D<0 the quadratic has no real roots at all, so a(x−α)(x−β)a(x-\alpha)(x-\beta) is never zero for real xx, and by continuity it cannot switch sign anywhere; it is entirely governed by aa. This same three-way split is exactly what feeds the wavy-curve method for solving inequalities like ax2+bx+c>0ax^2+bx+c>0.

Transformation of roots. A quadratic ax2+bx+c=0ax^2+bx+c=0 with roots α,β\alpha,\beta satisfies α+β=−b/a\alpha+\beta=-b/a and αβ=c/a\alpha\beta=c/a. Because any quadratic is fixed entirely by its sum and product of roots — x2−(sum)x+(product)=0x^2-(\text{sum})x+(\text{product})=0 reproduces exactly those roots — a transformed pair of roots needs only a transformed sum and product, never the individual surd values of α,β\alpha,\beta.

x2−(α′+β′) x+α′β′=0x^2-(\alpha'+\beta')\,x+\alpha'\beta'=0

where: α′,β′\alpha',\beta' is the new pair of roots (e.g. α+h,β+h\alpha+h,\beta+h for a shift by hh, or kα,kβk\alpha,k\beta for scaling by kk), and α′+β′\alpha'+\beta', α′β′\alpha'\beta' are computed directly from α+β\alpha+\beta and αβ\alpha\beta using the transformation rule. The point of working this way is speed and safety. Solving ax2+bx+c=0ax^2+bx+c=0 for α,β\alpha,\beta individually often produces surds, and tracking a surd through a further transformation (adding, scaling, inverting) invites arithmetic slips. The sum-and-product route sidesteps that entirely — it never asks what α\alpha and β\beta are, only how their combination transforms, so the working stays rational even when the roots themselves are irrational or complex.

Figure
Number line showing where a upward parabola is positive and negative relative to its two roots
Sign of ax squared plus bx plus c, roots alpha and beta, a positive alpha beta positive (same as a) negative (opposite a) positive (same as a)
Derivation ANew equation from transformed roots

The goal is to show, using only algebra a Class 11 student already owns, that the equation whose roots are α+h,β+h\alpha+h,\beta+h can be written down purely from the original sum and product of roots — without ever computing α\alpha and β\beta as individual numbers, which matters most when they are irrational.

Step 1. Start from ax2+bx+c=0ax^2+bx+c=0 with roots α,β\alpha,\beta, so α+β=−ba\alpha+\beta=-\dfrac{b}{a} and αβ=ca\alpha\beta=\dfrac{c}{a}. These two values are all the information the original equation carries about its roots.

Step 2. Define the transformed roots α′=α+h, β′=β+h\alpha'=\alpha+h,\ \beta'=\beta+h for a chosen shift hh. Their sum is α′+β′=(α+h)+(β+h)=(α+β)+2h\alpha'+\beta'=(\alpha+h)+(\beta+h)=(\alpha+\beta)+2h, obtained by simply expanding the brackets and collecting the α+β\alpha+\beta term.

Step 3. Their product is α′β′=(α+h)(β+h)=αβ+hα+hβ+h2=αβ+h(α+β)+h2\alpha'\beta'=(\alpha+h)(\beta+h)=\alpha\beta+h\alpha+h\beta+h^2=\alpha\beta+h(\alpha+\beta)+h^2, again by direct expansion, being careful to keep the cross term h(α+β)h(\alpha+\beta).

Step 4. Substitute the known α+β\alpha+\beta and αβ\alpha\beta into x2−(α′+β′)x+α′β′=0x^2-(\alpha'+\beta')x+\alpha'\beta'=0 to get the required equation. Since α′,β′\alpha',\beta' were never computed individually, this method works equally well when α,β\alpha,\beta are irrational or complex conjugates — the sum-and-product route never needs to "see" the surds.

Worked example

If α,β\alpha,\beta are the roots of 2x2−7x+3=02x^2-7x+3=0, find the quadratic equation whose roots are α+2,β+2\alpha+2,\beta+2.

Step 1. Read off α+β=72\alpha+\beta=\dfrac{7}{2} and αβ=32\alpha\beta=\dfrac{3}{2} from the coefficients.

Step 2. New sum: (α+2)+(β+2)=(α+β)+4=72+4=152(\alpha+2)+(\beta+2)=(\alpha+\beta)+4=\dfrac72+4=\dfrac{15}{2}.

Step 3. New product: (α+2)(β+2)=αβ+2(α+β)+4=32+7+4=252(\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4=\dfrac32+7+4=\dfrac{25}{2}.

Step 4. Required equation: x2−152x+252=0x^2-\dfrac{15}{2}x+\dfrac{25}{2}=0, or clearing fractions, 2x2−15x+25=02x^2-15x+25=0. Checking against the exact roots α,β=3,12\alpha,\beta=3,\dfrac12 (from α,β=7±54\alpha,\beta=\dfrac{7\pm5}{4}): α+2,β+2=5,52\alpha+2,\beta+2=5,\dfrac52, with sum 7.5=15/27.5=15/2 and product 12.5=25/212.5=25/2, confirming the answer. The required equation is 2x2−15x+25=02x^2-15x+25=0.

Competitive stretch

For the same equation 2x2−7x+3=02x^2-7x+3=0, the equation with reciprocal roots 1/α,1/β1/\alpha,1/\beta has new sum α+βαβ=7/23/2=73\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{7/2}{3/2}=\dfrac73 and new product 1αβ=23\dfrac{1}{\alpha\beta}=\dfrac23, giving x2−73x+23=0x^2-\dfrac73x+\dfrac23=0, or 3x2−7x+2=03x^2-7x+2=0 — notice the coefficients of the original equation simply reverse order, which is always true for a reciprocal-roots transformation, since replacing xx by 1/x1/x in the original equation and clearing denominators achieves exactly this reversal. A quick sign check on the same original equation is also worth doing alongside the transformation: 2x2−7x+3=02x^2-7x+3=0 has D=49−24=25>0D=49-24=25>0 and a=2>0a=2>0, so with roots 12,3\dfrac12,3, the expression is positive outside [12,3]\left[\dfrac12,3\right] and negative inside it — the same three-interval pattern from the core idea, now on the very quadratic whose roots were just transformed.

Common traps

  • Reading "between the roots" as positive — it is the opposite of aa's sign there, and matches aa's sign only outside the roots.
  • Applying the sign rule's "split by roots" logic when D<0D<0; with no real roots the expression never changes sign at all, so it equals sign(a)(a) everywhere.
  • Dropping the cross term h(α+β)h(\alpha+\beta) when expanding (α+h)(β+h)(\alpha+h)(\beta+h) for a shift, giving a wrong new product.
  • Mixing up transformation rules — negation (−α,−β-\alpha,-\beta) leaves the product unchanged but flips the sum's sign, while a shift changes both; treating them the same gives a wrong equation even with correct arithmetic.

Ask the tutor: try asking the tutor to walk through the sign of −x2+4x−3-x^2+4x-3 (negative leading coefficient) so you can see the same-as-aa/opposite-to-aa rule work with a<0a<0.

Why does the sign of the expression flip exactly at the roots and nowhere else?

Because ax2+bx+c=a(x−α)(x−β)ax^2+bx+c=a(x-\alpha)(x-\beta) is a continuous function whose only zeros (for real roots) are x=αx=\alpha and x=βx=\beta. A continuous function that never touches zero on some stretch of the number line cannot switch from positive to negative there — it would have to pass through zero to do so, by the intermediate value idea you already use for locating roots. So the sign is forced to stay constant on each of the three intervals cut out by α\alpha and β\beta, and testing one convenient point per interval (or applying the same-as-aa/opposite-to-aa rule directly) fixes the sign for every xx in that interval, not just the one you tested.

Do I always have to expand brackets to find the new sum and product, or is there a shortcut?

For the four standard transformations — shift by hh, scale by kk, negate, or invert — memorising the results is faster than re-deriving them each time: shift gives sum (α+β)+2h(\alpha+\beta)+2h and product αβ+h(α+β)+h2\alpha\beta+h(\alpha+\beta)+h^2; scale gives k(α+β)k(\alpha+\beta) and k2αβk^2\alpha\beta; negation gives −(α+β)-(\alpha+\beta) and unchanged αβ\alpha\beta; reciprocal gives (α+β)/(αβ)(\alpha+\beta)/(\alpha\beta) and 1/(αβ)1/(\alpha\beta). For anything unfamiliar — say the roots are α2,β2\alpha^2,\beta^2, or α+β\alpha+\beta isn't linear in the shift — do not force-fit one of the four standard rules; instead expand directly from the definition of the new roots, exactly as the derivation above does for a shift, and you will never go wrong even on an unseen transformation.

My final equation has different coefficients from the answer key but the roots check out — did I make a mistake?

Probably not. Once you clear fractions, x2−(sum)x+(product)=0x^2-(\text{sum})x+(\text{product})=0 can be scaled by any nonzero constant and still have the same roots — 2x2−15x+25=02x^2-15x+25=0 and 4x2−30x+50=04x^2-30x+50=0 are the same equation in disguise, since multiplying an equation through by a constant does not change which values of xx satisfy it. Compare your equation to the expected one by checking if one is a constant multiple of the other (divide corresponding coefficients and see if the ratio is the same throughout), rather than expecting identical coefficients — many textbooks quote the "simplest integer form," which is one particular choice of that constant, not the only correct one.

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