Nature and relations of roots

This topic sits inside the Quadratic Equations sub-unit of Algebra, right after you learn to solve $ax^2+bx+c=0$ by the quadratic formula. Instead of stopping at "find the roots," it asks a sharper question: what structural facts about the roots can be read off the coefficients $a,b,c$ alone? The answer splits into two tools that work together — a single test number (the discriminant) that classifies what kind of roots exist, and two algebraic relations (sum and product) that let you compute with the roots symbolically, even before or without solving anything.

Classifying roots by the discriminant. For $ax^2+bx+c=0$ with real, rational $a,b,c$ ($a\neq0$), the roots come from

where: $D$ is the discriminant, and its sign (and, when positive, whether it is a perfect square) fixes the nature of the roots: $D>0$ and a perfect square gives real, rational, unequal roots; $D>0$ but not a perfect square gives real, irrational, unequal roots occurring as conjugate surd pairs $p\pm\sqrt q$; $D=0$ gives real, equal roots, each $-b/(2a)$; $D<0$ gives complex conjugate roots $\dfrac{-b}{2a}\pm i\dfrac{\sqrt{|D|}}{2a}$.

Sum and product of roots. From the quadratic formula directly,

where: $\alpha,\beta$ are the two roots. Any expression symmetric in $\alpha,\beta$ can be rewritten in terms of these two quantities alone, e.g. $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$ and $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$.

Framing a new equation. If a quadratic is to have roots $\alpha,\beta$, it must be

This works even when the desired roots are transforms of the original ones, such as $\alpha^3,\beta^3$ — compute $\alpha^3+\beta^3$ and $\alpha^3\beta^3=(\alpha\beta)^3$ from the identities above, then substitute.

Reciprocal roots and a common root. Roots $\alpha,1/\alpha$ occur exactly when $\alpha\cdot(1/\alpha)=1=c/a$, i.e. $a=c$ — a quick check that avoids solving the equation at all. If $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ share one root $\alpha$, eliminating $\alpha^2$ by cross-multiplication gives

and the two expressions for $\alpha$ being equal is the condition $(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)$ for a common root to exist. Note that this technique finds only the shared root — each equation may still have a second root of its own that the other equation does not have.

Why the discriminant governs rationality. When $a,b,c$ are rational and $D$ is a perfect square, $\sqrt D$ is itself rational, so $-b\pm\sqrt D$ and hence both roots are rational. When $D$ is positive but not a perfect square, $\sqrt D$ is an irrational surd, and the two roots differ only in the sign attached to that surd — the same pattern by which complex roots always arise as conjugate pairs $p\pm iq$ when $D<0$. In every case the "$\pm$" in the quadratic formula is the source of the pairing: whatever kind of number $\sqrt D$ turns out to be, both roots inherit it symmetrically.

For $3x^2-7x+2=0$, classify the roots, find $\alpha^2+\beta^2$, and then find $k$ such that $x^2+kx+2=0$ and $2x^2+3x+k=0$ share a common root.

Step 1. Here $a=3,b=-7,c=2$, so $D=(-7)^2-4(3)(2)=49-24=25$, a positive perfect square, so the roots are real, rational, and unequal.

Step 2. Sum and product: $\alpha+\beta=-b/a=7/3$, $\alpha\beta=c/a=2/3$. Then

Solving directly gives $\alpha=2,\beta=1/3$, and indeed $4+1/9=37/9$ ✓ — the shortcut and the direct check agree.

Step 3. For the common root, take $a_1=1,b_1=k,c_1=2$ and $a_2=2,b_2=3,c_2=k$:

Step 4. Equating the two expressions for $\alpha$: $k^2-6=4-k \Rightarrow k^2+k-10=0 \Rightarrow k=\dfrac{-1\pm\sqrt{41}}{2}$. For either value of $k$, the two given quadratics share exactly one common root, obtained by substituting $k$ back into $\alpha=(4-k)/(3-2k)$.

Competitive stretch

Frame the quadratic whose roots are $\alpha^3,\beta^3$ for the same equation $3x^2-7x+2=0$. Using $\alpha+\beta=7/3,\alpha\beta=2/3$: $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=(7/3)^3-3(2/3)(7/3)=343/27-126/27=217/27$, and $\alpha^3\beta^3=(\alpha\beta)^3=8/27$. The required equation is $27x^2-217x+8=0$ — matching the exact roots $\alpha^3=8,\beta^3=1/27$ found by solving $3x^2-7x+2=0$ directly. Notice the whole calculation used only $\alpha+\beta$ and $\alpha\beta$ — at no point did we need $\sqrt D$ or the individual roots, which is exactly the efficiency the sum-product relations are built for.

• Concluding "rational roots" from $D>0$ alone — $D>0$ only guarantees real roots; you must also check $D$ is a perfect square.
• Writing $\alpha+\beta=b/a$, dropping the negative sign — the correct relation is $\alpha+\beta=-b/a$.
• Using only $\alpha+\beta,\alpha\beta$ for expressions that are not symmetric, such as $\alpha^2-\beta^2$ or $\alpha-\beta$ — these need the discriminant too, not just the sum and product.
• Forgetting that irrational roots of a rational-coefficient quadratic always occur in conjugate surd pairs $p\pm\sqrt q$, never singly.
• Skipping the verification step after framing a new equation from transformed roots — plugging the original roots back into the sum-and-product identities (as in the worked example) quickly catches an arithmetic slip before it propagates into a wrong final equation.
• Applying the common-root cross-multiplication formula while forgetting it only isolates the shared root — each original equation can still carry a second, distinct root of its own.

Ask the tutor: If the common-root cross-multiplication trick feels mechanical, ask the AI tutor to walk through why eliminating $\alpha^2$ between the two equations forces a linear relation that only the shared root can satisfy.