Theory of Equation – Relation between Roots and Coefficients of an Equation


Theory of Equation / Monday, October 21st, 2019

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, …, αn be the n roots of the equation a0xn + a1xn-1 + a2xn-2 + … + an = 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 Sr is the sum of all the products of α1, α2, α3, …, αn 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.

 Example 01

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.

 Example 02

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.

 Example 03

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

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

 Example 04

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

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

 Example 05

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.

 Example 06

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,

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

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

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

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

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

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

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

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

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

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

 

 

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