Roots of polynomial equations
Introduced| English | Meaning |
|---|---|
| root/ruːt/ | Values of a variable that make an equation or polynomial expression equal to zero. |
| coefficient/ˌkəʊɪˈfɪʃənt/ | A numerical or constant factor multiplying a term in an expression. |
| quadratic/kwɒˈdrætɪk/ | Describing a polynomial or relationship whose highest relevant variable power is two. |
| cubic/ˈkjuːbɪk/ | Having degree three, as in a polynomial whose highest power of the variable is three. |
| quartic equation/ˈkwɔːtɪk ɪˈkweɪʒn/ | A polynomial equation in one variable whose highest power is four, with a non-zero coefficient of that power. |
| symmetric functions/sɪˈmetrɪk ˈfʌŋkʃnz/ | Expressions unchanged when the roots exchange names, such as sums of roots, sums of their pairwise products or their total product. |
| substitution/ˌsʌbstɪˈtjuːʃn/ | Replacing a variable or expression with another value or expression to simplify or transform a problem. |
Read information without solving every root 根
- Before trusting a program's calculated roots, check their sum and product against its stored polynomial coefficients 系数. These relations also give useful sums when finding the individual roots would be difficult.
- A root makes the polynomial zero. Count repeated roots with their multiplicities, and include complex roots: a real graph shows only the real ones. We will use quadratic 二次, cubic 三次 and fourth-degree equations.
Expand the factors to see the signs
- If $a\ne0$, write a quadratic as $a(x-\alpha)(x-\beta)$. Expanding and comparing coefficients gives $\alpha+\beta=-b/a$ and $\alpha\beta=c/a$ for $ax^2+bx+c=0$.
- For $ax^3+bx^2+cx+d=0$, expansion gives the sum, sum of pairs and product below. The alternating signs come from the minus signs in each factor; they do not mean the roots themselves must be negative.
For $2x^2-7x+3=0$, the sum is $7/2$ and product $3/2$. The roots $3$ and $1/2$ check both results. For $(x-1)^2(x-3)=0$, count the roots as $1,1,3$, giving sum 5, not 4.
For x² − 5x + 6 = 0, what is the sum of the roots α + β = −b/a?
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α + β = −b/a = −(−5)/1 = 5.
For x² − 5x + 6 = 0, what is the product αβ = c/a?
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αβ = c/a = 6/1 = 6.
For x³ − 6x² + 11x − 6 = 0, the sum of roots Σα = −b/a. Find it.
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Σα = −(−6)/1 = 6. (Roots are 1, 2, 3: sum = 6.)
For ax³ + bx² + cx + d = 0, the product of roots αβγ = d/a.
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The product of three roots is αβγ = −d/a (negative), not d/a.
Match each root relationship to its formula (leading coefficient a).
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Quadratic: α+β = −b/a, αβ = c/a. Cubic: Σα = −b/a, αβγ = −d/a (sign alternates with degree).
A quartic equation 四次方程 has four roots
- For $ax^4+bx^3+cx^2+dx+e=0$, let $e_1$ be the sum of the four roots, $e_2$ the sum of all six different pairs, $e_3$ the sum of all four different triples, and $e_4$ their product.
- For roots $\alpha,\beta,\gamma,\delta$, the pairs are $\alpha\beta,\alpha\gamma,\alpha\delta,\beta\gamma,\beta\delta,\gamma\delta$. Use each pair once. These symmetric functions 对称函数 are unchanged when the roots exchange names.
The roots of $2x^4-6x^3+3x^2+5x-7=0$ are α,β,γ,δ. Find the sum of the roots, the sum of pairs, the sum of triples, the product, the sum of their squares and the sum of their reciprocals.
$e_1=3,e_2=3/2,e_3=-5/2,e_4=-7/2$. The sum of squares is $3^2-2(3/2)=6$. No root is zero because the constant term is non-zero. The reciprocal sum is $e_3/e_4=5/7$.
The polynomial $(x-2)^2(x+1)$ is expanded as $x^3-3x^2+4$. Find the sum of roots, sum of pairs and sum of squares. Explain how the repeated root enters.
The root list is 2,2,−1. Its sum is 3, pair sum $4-2-2=0$, and sum of squares $4+4+1=9$. The root 2 contributes twice because its factor is squared; using only distinct values gives the wrong coefficient relations.
Find sums of powers
- Write $S_k=\alpha^k+\beta^k+\cdots$, with one contribution per root. For degree 2, 3 or 4, $S_2=e_1^2-2e_2$. For degree 3 or 4, $S_3=e_1^3-3e_1e_2+3e_3$.
- For higher powers, use the original equation at each root and add. If a quartic is $x^4+bx^3+cx^2+dx+e=0$, then $S_4+bS_3+cS_2+dS_1+4e=0$. The final term is $4e$, because there are four roots.
For $x^4-2x^3+3x^2-4x+5=0$, $(e_1,e_2,e_3,e_4)=(2,3,4,5)$. Thus $S_2=-2$, $S_3=2$ and $S_4=2(2)-3(-2)+4(2)-4(5)=-2$. A negative sum of squares is possible when the roots are complex.
For x² − 5x + 6 = 0, α² + β² = (α+β)² − 2αβ. Find it.
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(α+β)² − 2αβ = 25 − 12 = 13.
For the four roots of $x^4+2x^2-3x+1=0$, find S₁,S₂,S₃,S₄ without solving the equation. Explain why a negative S₂ is possible.
$e_1=0,e_2=2,e_3=3,e_4=1$. Thus $S_1=0,S_2=-4,S_3=9$. Summing each root equation gives $S_4+2S_2-3S_1+4=0$, so $S_4=4$. The roots need not all be real, so their squared values need not all be nonnegative real numbers.
Recover unknown coefficients
- Match the given sums to the coefficient relations first. For a cubic $x^3+bx^2+cx+d=0$, $b=-S_1$ and $c=(S_1^2-S_2)/2$.
- If $d$ is unknown, sum the original equation to get $S_3+bS_2+cS_1+3d=0$. Multiply the root equation by that root and sum again: $S_4+bS_3+cS_2+dS_1=0$. These two equations can determine $d$ from a given $S_4$.
Use the number of roots in the constant contribution. In the multiplied equation the constant $d$ becomes $d\alpha,d\beta,d\gamma$, so the sum is $dS_1$, not $3d$ again.
A cubic $x^3+bx^2+cx+d=0$ has root power sums S₁=3, S₂=5 and S₄=29. Find b,c,d, deriving the equations you use for the higher sums.
$b=-3$ and $5=3^2-2c$ give c=2. Summing the root equation gives $S_3-3(5)+2(3)+3d=0$, so $S_3=9-3d$. Multiplying by a root before summing gives $29-3S_3+2(5)+3d=0$. Substitute: $12+12d=0$, hence d=−1.
Shift and scale roots with a substitution 代换
- If new roots are $y=mx+k$ with $m\ne0$, rearrange to $x=(y-k)/m$. Substitute this expression into the original polynomial, clear denominators and collect powers of $y$.
- Check the new root sum $mS_1+nk$, where $n$ is the degree. Multiplying a whole equation by a non-zero constant does not change its roots.
For $P(x)=x^3-3x+1$, the new roots $2\alpha+1,2\beta+1,2\gamma+1$ use $x=(y-1)/2$. Multiply $P((y-1)/2)=0$ by 8:
The new sum is 3, agreeing with $2(0)+3(1)$.
Real roots and coefficients
y = ax³ + bx² + cx + d
Only real roots appear as x-intercepts. The coefficient relations include all roots, with multiplicity, including complex ones. Keep the leading coefficient non-zero when using cubic relations.
With α + β = 5 and αβ = 6, what is the product (α+1)(β+1)?
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(α+1)(β+1) = αβ + α + β + 1 = 6 + 5 + 1 = 12.
The four roots of $P(x)=x^4-x^3+2x^2+x-3$ are α,β,γ,δ. Find an equation with roots 2α−1,2β−1,2γ−1,2δ−1. Verify its root sum.
Use x=(y+1)/2 and multiply P by 16. The result is $(y+1)^4-2(y+1)^3+8(y+1)^2+8(y+1)-48=0$, hence $y^4+2y^3+8y^2+22y-33=0$. The original root sum is 1, so the new sum is 2(1)−4=−2, agreeing with the coefficient of y³.
Reciprocal roots and excluded values
- For new roots $y=1/x$, use $x=1/y$ and multiply by $y^n$. The original constant term must be non-zero: zero has no reciprocal. A cubic $ax^3+bx^2+cx+d=0$ becomes $dy^3+cy^2+by+a=0$.
- For $y=1/(x-k)$, use $x=k+1/y$. Check $P(k)\ne0$ before claiming all roots have images. Clearing denominators is an algebraic step; it does not make a forbidden original root valid.
$2x^3-3x^2+4x-5=0$ has reciprocal-root equation $5y^3-4y^2+3y-2=0$. Its new sum is $4/5$, also $e_2/e_3$. Check $y=0$ is not a root of the transformed equation.
Find an equation with roots 1/α,1/β,1/γ when α,β,γ solve $3x^3-2x^2+5x-4=0$. Check the new root sum and product.
Put x=1/y and multiply by y³: $3-2y+5y^2-4y^3=0$, or $4y^3-5y^2+2y-3=0$. Its root sum is 5/4 and product 3/4. The original constant −4 ensures no original root is zero, and the new constant −3 excludes y=0.
Squared roots without guessing signs
- Separate even and odd powers: $P(x)=E(x^2)+xO(x^2)$. For $y=x^2$, squaring $E(y)=-xO(y)$ gives $E(y)^2-yO(y)^2=0$.
- This uses both possible signs of $x$: one of $\sqrt y$ or $-\sqrt y$ is an original root. Keep multiplicities when two different roots have the same square; the transformed equation may have a repeated root.
For $x^3-2x^2-x+2=0$, $E(y)=2-2y$ and $O(y)=y-1$. Therefore
The original roots $1,-1,2$ become $1,1,4$. Do not remove the repeated factor just because there are only two different squared values.
Obtain an equation whose roots are the squares of the roots of $x^3+2x^2-3x-4=0$, without solving for those roots.
With y=x², the equation is $x(y-3)=4-2y$. Squaring gives $y(y-3)^2=(4-2y)^2$. Collecting terms yields $y^3-10y^2+25y-16=0$. Check its root sum 10 against original $e_1^2-2e_2=(-2)^2-2(-3)=10$.
Obtain an equation whose roots are the squares of all four roots of $x^4-3x^2+x+2=0$. State the even and odd parts you use.
$E(y)=y^2-3y+2$ and $O(y)=1$. Thus $(y^2-3y+2)^2-y=0$, or $y^4-6y^3+13y^2-13y+4=0$. The new sum 6 equals the original sum of squares $0^2-2(-3)$.
Reduce a higher power before substituting
- If $x^3+3x-2=0$, every root satisfies $x^3=2-3x$. A proposed new root $y=x^3+1$ is therefore $y=3-3x$, a simple linear transformation.
- Use $x=(3-y)/3$ in the original equation. After multiplying by 27 and collecting terms, $y^3-9y^2+54y-54=0$. Coefficient relations now give sums of the transformed roots without finding them individually.
The roots of $x^3+2x-3=0$ are α,β,γ. Find an equation with roots α³+1,β³+1,γ³+1. Then find the sum of the squares and sum of the cubes of these new roots.
The original root relation gives y=x³+1=4−2x, hence x=(4−y)/2. Substitution gives $(4-y)^3+8(4-y)-24=0$, or $y^3-12y^2+56y-72=0$. New sums are $T_1=12,T_2=12^2-2(56)=32$. Summing the new root equation gives $T_3=12(32)-56(12)+3(72)=-72$.
Check the equation and its roots
- State what each sum includes, divide by the leading coefficient, and check alternating signs. Complex and repeated roots remain part of the algebra even when a real graph does not show them.
- Define the new variable, rearrange back to the old variable, clear denominators and check the new degree and root sum. A condition such as a non-zero denominator must survive the calculation.
For x squared minus 7x plus 10 = 0, what is the sum of the roots, what is the sum of their squares, and why is the sum not b over a?
For ax²+bx+c=0, the sum is −b/a. Here a=1 and b=−7, giving 7.
For x²−7x+10=0, what is the sum of its two roots?
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Minus b over a is 7. The sum of the squares is the separate result 29. Other options — 5 (the lesson sum for x squared minus 5x plus 6); 29 (the sum of the squares).
For the roots of $P(x)=x^3-2x^2-x+2$, find an equation for the transformed roots y=1/(x−3). Explain why replacing 3 by 1 would not give images of all original roots.
Use x=3+1/y and multiply by y³: $(3y+1)^3-2y(3y+1)^2-y^2(3y+1)+2y^3=0$. This is $8y^3+14y^2+7y+1=0$. Since P(3)=8≠0, all three images exist. But P(1)=0, so y=1/(x−1) is undefined for an original root. A lower-degree equation obtained after clearing denominators would describe only the remaining images.