Relation between Roots and Coefficients of an Equation

To do the sums in theory of equation we have to first understand the Relation between Roots and Coefficients of an Equation, and here they are

Let α₁, α₂, α₃, …, α_n be the n roots of the equation a₀xⁿ + a₁x^n-1 + a₂x^n-2 + … + a_n = 0, then we have the identity

\[{{\text{a}}_{0}}{{\text{x}}^{\text{n}}}+\text{ }{{\text{a}}_{\text{1}}}{{\text{x}}^{\text{n}-\text{1}}}+\text{ }{{\text{a}}_{\text{2}}}{{\text{x}}^{\text{n}-\text{2}}}+\text{ }\ldots \text{ }+\text{ }{{\text{a}}_{\text{n}}}={{a}_{0}}(x-{{\alpha }_{1}})(x-{{\alpha }_{2}})…(x-{{\alpha }_{n}})\]

\[={{a}_{0}}\left\{ {{x}^{n}}-{{S}_{1}}{{x}^{n-1}}+{{S}_{2}}{{x}^{n-2}}+…+{{(-1)}^{n}}{{S}_{n}} \right\}\]

Where S_r is the sum of all the products of α₁, α₂, α₃, …, α_n has taken r at a time.

Equating the coefficients of like powers of x on the two sides of the above identity, we have

\[{{S}_{1}}=\sum{{{\alpha }_{1}}=-\frac{{{a}_{1}}}{{{a}_{0}}}}\]

\[{{S}_{2}}=\sum{{{\alpha }_{1}}{{\alpha }_{2}}=\frac{{{a}_{2}}}{{{a}_{0}}}}\]

\[{{S}_{3}}=\sum{{{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}=-\frac{{{a}_{3}}}{{{a}_{0}}}}\]

…..   ……….   ……….   …..

…..   ……….   ……….   …..

\[{{S}_{n}}={{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}…{{\alpha }_{n}}={{(-1)}^{n}}\frac{{{a}_{n}}}{{{a}_{0}}}\]

The Relation between Roots and Coefficients of a Quadratic Equation

If α, β be the roots of the equation a₀x² + a₁x + a₂ = 0, then

\[\alpha +\beta =-\frac{{{a}_{1}}}{{{a}_{0}}}\]

\[\alpha \beta =\frac{{{a}_{2}}}{{{a}_{0}}}\]

The Relation between Roots and Coefficients of a Cubic Equation

If α, β, γ be the roots of the equation a₀x³ + a₁x² + a₂x + a₃= 0, then

\[\alpha +\beta +\gamma =-\frac{{{a}_{1}}}{{{a}_{0}}}\]

\[\alpha \beta +\beta \gamma +\gamma \alpha =\frac{{{a}_{2}}}{{{a}_{0}}}\]

\[\alpha \beta \gamma =-\frac{{{a}_{3}}}{{{a}_{0}}}\]

The Relation between Roots and Coefficients of a Biquadratic Equation

If α, β, γ, δ be the roots of the equation a₀x⁴ + a₁x³ + a₂x² + a₃x + a₄= 0, then

\[\alpha +\beta +\gamma +\delta =-\frac{{{a}_{1}}}{{{a}_{0}}}\]

\[\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac{{{a}_{2}}}{{{a}_{0}}}\]

\[\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =-\frac{{{a}_{3}}}{{{a}_{0}}}\]

\[\alpha \beta \gamma \delta =\frac{{{a}_{4}}}{{{a}_{0}}}\]

Observation

We notice that the number of relations between the roots and the coefficients of an equation is equal to the number of roots of the equation. But these relations are not sufficient to solve the given equation. Some other relation between the roots is required to solve the equation.

Solve the equation 4x³ + 16x² – 9x – 36 = 0, when the sum of the roots is zero.

Solution:

Since the sum of two roots of the given cubic equation is zero, let us take the roots as α, -α, β.

Then the sum of roots,

\[\alpha +(-\alpha )+\beta =-\frac{16}{4}\]

\[\therefore \beta =-4\]

The product of the three roots

\[\alpha (-\alpha )\beta =\frac{36}{4}\]

\[\therefore {{\alpha }^{2}}=\frac{9}{4},~~as~~\beta =-4\]

\[\therefore \alpha =\pm \frac{3}{2}\]

Hence the roots are 3/2, -3/2, -4.

Solve the equation x³ – 3x² – 6x + 8 = 0, if the roots are in A.P.

Solution:

Since the roots of the given cubic equation are in A.P., we take the roots as: α – β, α, α + β

Sum of the roots,

\[\left( \alpha -\beta  \right)+\alpha +\left( \alpha +\beta  \right)=3\]

\[i.e.,\alpha =1\]

Product of the roots,

\[\left( \alpha -\beta  \right)\alpha \left( \alpha +\beta  \right)=-8\]

\[\Rightarrow \alpha \left( {{\alpha }^{2}}-{{\beta }^{2}} \right)=-8\]

\[\Rightarrow 1\left( 1-{{\beta }^{2}} \right)=-8,~~as~~\alpha =1\]

\[\Rightarrow {{\beta }^{2}}=9\]

\[\therefore \beta =\pm 3\]

Therefore the roots are -2, 1, 4.

Solve the equation 8x³ – 52x² + 78x – 27 = 0, when the roots are in geometric progression.

Solution:

We assume the roots of the given cubic equation as (α/β), α, αβ, which are obviously in G.P.

Product of the roots,

\[\frac{\alpha }{\beta }.\alpha .\alpha \beta =\frac{27}{8}\]

\[\Rightarrow {{\alpha }^{3}}=\frac{27}{8}\]

\[\therefore \alpha =\frac{3}{2}\]

Again, sum of the roots,

\[\frac{\alpha }{\beta }+\alpha +\alpha \beta =\frac{52}{8}\]

\[\Rightarrow \alpha \left( 1+\beta +{{\beta }^{2}} \right)=\frac{13}{2}\beta \]

\[\Rightarrow \frac{3}{2}\left( 1+\beta +{{\beta }^{2}} \right)=\frac{13}{2}\beta ,~~as~~\alpha =\frac{3}{2}\]

\[\Rightarrow 3{{\beta }^{2}}-10\beta +3=0\]

\[\Rightarrow \left( 3\beta -1 \right)\left( \beta -3 \right)=0\]

\[\therefore \beta =\frac{1}{3},3\]

Hence the roots are

\[\frac{1}{2},\frac{3}{2},\frac{9}{2}\]

Solve the equation 6x³ – 11x² – 3x + 2 = 0, given that the roots are in Harmonic Progression.

Solution:

If α, β, γ be the roots of the given equation, then

\[\alpha +\beta +\gamma =\frac{11}{6}……….(i)\]

\[\alpha \beta +\beta \gamma +\gamma \alpha =-\frac{1}{2}……….(ii)\]

\[\alpha \beta \gamma =-\frac{1}{3}……….(iii)\]

Since α, β, γ are H.P., (1/α), (1/β), (1/γ) are in A.P.

\[i.e.,\frac{1}{\beta }-\frac{1}{\alpha }=\frac{1}{\gamma }-\frac{1}{\beta }\]

\[\Rightarrow \beta =\frac{2\gamma \alpha }{\gamma +\alpha }……….(iv)\]

\[\Rightarrow \alpha \beta +\beta \gamma =2\gamma \alpha \]

\[\therefore \alpha \beta +\beta \gamma +\gamma \alpha =3\gamma \alpha =-\frac{1}{2}~~~\left[ from~~~(ii) \right]\]

\[i.e.,\gamma \alpha =-\frac{1}{6}\]

Again from (iii) we have,

\[\alpha \beta \gamma =-\frac{1}{3}\]

\[\therefore \beta =2~~~as~~~\gamma \alpha =-\frac{1}{6}\]

Thus 2 is a root of the given equation.

Now, we have

\[\text{6}{{\text{x}}^{\text{3}}}\text{ 11}{{\text{x}}^{\text{2}}}\text{ 3x }+\text{ 2 }=\text{ }0\]

\[\Rightarrow \left( x-2 \right)\left( 6{{x}^{2}}+x-1 \right)=0\]

\[\Rightarrow \left( x-2 \right)\left( 3x-1 \right)\left( 2x+1 \right)=0\]

\[\therefore x=2,\frac{1}{3},-\frac{1}{2}\]

Hence the roots are

\[-\frac{1}{2},\frac{1}{3},2\]

If the roots of the equation x⁴ + ax³ + bx² + cx + d = 0 are in G.P., then show that c² = ad.

Solution:

Let the roots of the equation be (α/r³), (α/r), (αr), and (αr³), which are in G.P.

Then,

\[\frac{\alpha }{{{r}^{3}}}+\frac{\alpha }{r}+\alpha r+\alpha {{r}^{3}}=-a\]

\[\Rightarrow \alpha \left( \frac{1}{{{r}^{3}}}+\frac{1}{r}+r+{{r}^{3}} \right)=-a………(i)\]

\[\frac{{{\alpha }^{2}}}{{{r}^{4}}}+\frac{{{\alpha }^{2}}}{{{r}^{2}}}+{{\alpha }^{2}}+{{\alpha }^{2}}+{{\alpha }^{2}}{{r}^{2}}+{{\alpha }^{2}}{{r}^{4}}=b………(ii)\]

\[\frac{{{\alpha }^{3}}}{{{r}^{3}}}+\frac{{{\alpha }^{3}}}{r}+{{\alpha }^{3}}r+{{\alpha }^{3}}{{r}^{3}}=-c\]

\[\Rightarrow {{\alpha }^{3}}\left( \frac{1}{{{r}^{3}}}+\frac{1}{r}+r+{{r}^{3}} \right)=-c……….(iii)\]

\[\frac{\alpha }{{{r}^{3}}}.\frac{\alpha }{r}.\alpha r.\alpha {{r}^{3}}=d\]

\[\Rightarrow {{\alpha }^{4}}=d……….(iv)\]

By dividing (iii) by (i) we have,

\[{{\alpha }^{2}}=\frac{c}{a}\]

\[\Rightarrow {{\alpha }^{4}}=\frac{{{c}^{2}}}{{{a}^{2}}}=d~~\left[ by~~(iv) \right]\]

Hence c² = ad is the required condition.

If one of the roots of the equation x³ + px² + qx + r = 0 is equal to the sum of the other two roots, then prove that p³ + 8r = 4pq.

Solution:

Let the roots of the equation be α, β, α + β.

Then,

\[\alpha +\beta +\left( \alpha +\beta  \right)=-p\]

\[\Rightarrow \alpha +\beta =-\frac{p}{2}……….(i)\]

\[\alpha \beta +\beta \left( \alpha +\beta  \right)+\left( \alpha +\beta  \right)\alpha =q\]

\[\Rightarrow \alpha \beta +{{\left( \alpha +\beta  \right)}^{2}}=q………(ii)\]

\[\Rightarrow \alpha \beta +\frac{{{p}^{2}}}{4}=q~~~\left[ by~~~(i) \right]\]

\[\Rightarrow \alpha \beta =q-\frac{{{p}^{2}}}{4}~\]

\[\alpha \beta \left( \alpha +\beta  \right)=-r\]

\[\Rightarrow \left( q-\frac{{{p}^{2}}}{4}~ \right)\left( -\frac{p}{2}. \right)=-r\]

\[\Rightarrow 4pq-{{p}^{3}}=8r\]

\[\therefore {{p}^{3}}+8r=4pq\]

 ;