Question bank — Vieta's formulas — sum and product of roots

False. You must divide by $a=2$ first: $\alpha +\beta =-\frac{8}{2}=-4$. Forgetting to divide by the leading coefficient gives the answer for the different equation $x^2+8x+6=0$.

False. The formula comes purely from matching coefficients after expanding $a(x-\alpha )(x-\beta )$, which is an algebraic identity that holds even when $\alpha ,\beta$ are complex. It never uses realness.

False (for real coefficients). Complex roots of a real quadratic come as conjugates $p\pm qi$; their sum $2p$ and product $p^2+q^2$ are both real, so $-b/a$ and $c/a$ stay real.

False. There are two equal roots, both $3$ (a repeated root). Vieta's counts multiplicity: $\alpha +\beta =3+3=6$ and $\alpha \beta =9$, matching $-b/a=6$, $c/a=9$.

False. $5(a{x}^2+bx+c)=0$ has the same roots, and $-\frac{5b}{5a}=-\frac{b}{a}$, $\frac{5c}{5a}=\frac{c}{a}$ — both ratios are unchanged. Vieta's depends only on the ratios of coefficients.

False. A positive product only means the roots have the same sign. They could both be negative (e.g. roots $-2,-3$ give product $6>0$). You also need $\alpha +\beta >0$ to conclude both are positive.

True. Sum is $-b/a=-1$ regardless of the Discriminant being negative. The complex roots are conjugates whose real parts add to $-1$.

True. A negative product forces opposite signs (one positive, one negative), so the roots lie on either side of $0$.

True. Sum and product uniquely determine the monic quadratic $x^2-(\alpha +\beta )x+\alpha \beta$, hence the same root pair — this is exactly how Systems of Equations in sum/product are solved.

The negative sign is missing. Sum is $-\frac{b}{a}=-\frac{-12}{3}=+4$. The factor form $(x-\alpha )(x-\beta )$ produces $-(\alpha +\beta )$ as the $x$-coefficient, which is where the flip lives.

The sum term should be subtracted: $x^2-(\text{sum})x+(\text{product})$. Sum $=3$, so it's $x^2-3x-10=0$. The sign trap is the same $-(\alpha +\beta )$ that appears in the sum formula.

Missing the cross term. $(\alpha +\beta {)}^2={\alpha }^2+2\alpha \beta +{\beta }^2$, so ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$. You must subtract $2\alpha \beta$; the square of a sum is not the sum of squares.

The product term is wrong. New product is ${\alpha }^2{\beta }^2=(\alpha \beta {)}^2=3^2=9$, not $19$. The correct equation is $x^2-19x+9=0$; $19$ is the sum of squared roots, not the product.

The sum is perfectly defined: $-b/a=-0/1=0$. A missing $x$-term means the roots are equal and opposite ($2$ and $-2$), which sum to zero — a genuine value, not a missing one.

You cannot split it that way. The correct combine is $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-b/a}{c/a}=-\frac{b}{c}$. Symmetric expressions must be rewritten in terms of $\alpha +\beta$ and $\alpha \beta$ first.

Because expanding $a(x-\alpha )(x-\beta )$ shows the coefficients are built from the sum and product; reading the coefficients backwards recovers them directly, bypassing the Quadratic Formula.

The clean identity $x^2-(\alpha +\beta )x+\alpha \beta$ is monic (leading coefficient $1$). Dividing $a{x}^2+bx+c$ by $a$ produces $x^2+\frac{b}{a}x+\frac{c}{a}$, whose coefficients are exactly $-(\text{sum})$ and product.

We factor with $(x-\alpha )$, so expansion gives $-\alpha x-\beta x=-(\alpha +\beta )x$ (one minus per root, summed) but $(-\alpha )(-\beta )=+\alpha \beta$ (two minuses cancel). The parity of minus signs differs.

Because $\alpha +\beta$ and $\alpha \beta$ are the elementary symmetric polynomials for two variables, and a fundamental theorem says every symmetric polynomial is a combination of these — so Vieta's gives you the raw materials for all of them.

Because their sum $-b/a$ and product $c/a$ are real. If one root were $p+qi$, the other must make both a real sum and real product, forcing it to be $p-qi$.

The discriminant only decides whether roots are real, equal, or complex. Vieta's is a statement about coefficients versus roots (with multiplicity/complex allowed), so it holds in every discriminant case.

Each factor contributes a minus root, so the constant term picks up $(-1{)}^n$ where $n$ is the degree. For $n=2$: $(-1{)}^2=+$; for $n=3$: $(-1{)}^3=-$. The sign alternates with degree.

Here $b=0$, $c=-4$: sum $=-0/1=0$, product $=-4$. Roots are $2$ and $-2$ — equal magnitude, opposite sign, consistent with zero sum and negative product.

At least one root is zero, which happens exactly when $c=0$. Then $a{x}^2+bx=0$ factors as $x(ax+b)=0$, giving roots $0$ and $-b/a$.

No. If $a=0$ the equation $bx+c=0$ is linear, not quadratic — there is only one root and no product of two roots to speak of. Vieta's for degree two requires $a\ne 0$.

With $\alpha =\beta =r$: sum $=2r=-b/a$ and product $=r^2=c/a$. The repeated root is $r=-b/(2a)$, matching the vertex from Completing the Square.

Sum $=-1$ and product $=1$, both real, even though each individual root $-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$ is complex. Conjugation cancels the imaginary parts.

Sum $=0$, product $=0$, so the monic equation is $x^2-0\cdot x+0=x^2=0$. Both $b$ and $c$ vanish, and $x=0$ is a double root.

The roots are $\pm k$ for some nonzero $k$ (negatives of each other), so $\alpha \beta =-k^2<0$. The equation is $x^2+\alpha \beta =x^2-k^2=0$ with no $x$-term.