Theory of Equations

Every Equation of nth degree has a total ‘n’ real or imaginary roots. If α is the root of Equation f (x) = 0, then the polynomial f (x) is exactly divisible by (x – α) i.e. (x – α) is the factor of the given polynomial f (x).

Example: Solve the Equation 3x3x+88=4x23x^{3} – x + 88 = 4x^{2} if one of the root of the given cubic polynomial equation is 2    7  i\mathbf{2\;-\;\sqrt{7}\;i}.

Solution:

Let, f (x) = 3x3x+884x23x^{3} – x + 88 – 4x^{2}

The given cubic function will have three roots i.e. α, β, and γ.

Therefore, α = x = 2    7  i\mathbf{2\;-\;\sqrt{7}\;i}

And, β = x = 2  +  7  i\mathbf{2\;+\;\sqrt{7}\;i} [Since, imaginary roots occur in conjugate pairs]

Therefore, (x    2  +  7  i)  (x    2    7  i)\mathbf{(x\;-\;2\;+\;\sqrt{7}\;i)\;(x\;-\;2\;-\;\sqrt{7}\;i)} = (x2)2+7=x24x+11(x – 2)^{2} + 7 = x^{2} – 4x + 11 = g (x)

On dividing f (x) by g (x) we will get 3x + 8 as quotient.

Therefore, 3x + 8 = 0

Hence, γ  =    83\mathbf{\gamma \;=\;-\;\frac{8}{3}}

Therefore, the roots of given Equation f (x) are 2    7  i\mathbf{2\;-\;\sqrt{7}\;i}, 2  +  7  i\mathbf{2\;+\;\sqrt{7}\;i}, γ  =    83\mathbf{\gamma \;=\;-\;\frac{8}{3}}

⇒ If any given equation f (x) of nth degree has maximum ‘p’ positive real roots and ‘q’ negative real roots, then the given equation has at least n – (p + q) imaginary roots.

Example: Find how many total positive real roots, negative real roots and imaginary roots does the equation 3x7x5+2x37=03x^{7} – x^{5} + 2x^{3} – 7 = 0 will have.

Solution:

Let, f (x) = 3x7x5+2x373x^{7} – x^{5} + 2x^{3} – 7

Now, Sign of f (x) = +  ve      ve    +  ve      ve\mathbf{+\;ve\;\rightarrow \;-\;ve\;\rightarrow \;+\;ve\;\rightarrow \;-\;ve}

Since, the sign of f (x) changes thrice. Therefore, f (x) can’t have more than 3 positive real roots. [p = 3]

Now, Sign of f (- x) = 3x7+x52x37=0-3x^{7} + x^{5} – 2x^{3} – 7 = 0 =   ve    +  ve      ve      ve\mathbf{-\;ve\;\rightarrow \;+\;ve\;\rightarrow \;-\;ve\;\rightarrow \;-\;ve}

Since, the sign of f (- x) changes twice. Therefore, f (x) can’t have more than 2 negative real roots. [q = 2]

For Imaginary Roots: n – (p + q) = 7 + (3 + 2) = 2.

Therefore, f (x) will have at least two imaginary roots.

Relationship between Roots and Coefficients

If α1,α2,α3,α4,α5,α6,............,αnα_{1}, α_{2}, α_{3}, α_{4}, α_{5}, α_{6}, . . . . . . . . . . . . , α_{n} are the roots of the quadratic Equation:

a0­xn+a1­xn1+a2­xn2+a3­xn3+a4­xn4+................+an1x+an=0a_{0} ­x^{n} + a_{1} ­x^{n-1} + a_{2} ­x^{n – 2} + a_{3} ­x^{n – 3} + a_{4} ­x^{n – 4 }+ . . . . . . . . . . . . . . . . + a_{n-1} x + a_{n} = 0

Then, the Sum of Roots:

α1=α1+α2+α3+α4......αn\sum \alpha_{1} = \alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} . . . . . . \alpha_{n} =   a1ao\mathbf{-\;\frac{a_{1}}{a_{o}}}

Sum of Product of roots taken two at a time:

α1α2\sum \alpha_{1} \alpha_{2} = a2ao\mathbf{\frac{a_{2}}{a_{o}}}

Sum of Product of roots taken three at a time:

α1.α2.α3\sum \alpha_{1} . \alpha_{2} .\alpha_{3} =   a3ao\mathbf{-\;\frac{a_{3}}{a_{o}}}

Product of Roots:

α1.α2.α3.α4.α5......αn=(1)n×anao\alpha_{1} . \alpha_{2} . \alpha_{3} . \alpha_{4} . \alpha_{5}. . . . . . \alpha_{n} = (- 1)^{n} \times \frac{a_{n}}{a_{o}}

Therefore, For a Cubic Equation ax3+bx2+cx+d=0ax^{3} + bx^{2} + cx + d = 0

α1+α2+α3\alpha{1} + \alpha_{2} + \alpha_{3} =   a1ao\mathbf{-\;\frac{a_{1}}{a_{o}}} =   ba\mathbf{-\;\frac{b}{a}}

α1.α2+α2.α3+α3.α1\alpha_{1}.\alpha_{2} + \alpha_{2}. \alpha_{3}+ \alpha_{3}.\alpha_{1} = a2ao\mathbf{\frac{a_{2}}{a_{o}}} = ca\mathbf{\frac{c}{a}}

α1.α2.α3=(1)n×a3ao\alpha_{1}.\alpha_{2} . \alpha_{3} = (- 1)^{n} \times \frac{a_{3}}{a_{o}} = da-\frac{d}{a}

Theory of Equation Problems

Example 1: If one of the root of cubic equation is double of another, then find all the roots of Equation x3+36=7x2x^{3} + 36 = 7x_{2}.

Solution:

Given, f (x) = x37x2+0x+36x^{3} – 7x_{2} + 0x + 36.

Let α, β, and γ be the roots of the given cubic function f (x).

Therefore, α + β + γ =   ba\mathbf{-\;\frac{b}{a}} = 7 . . . . . . (1)

Also, α β + γ β + αγ = ca\mathbf{\frac{c}{a}} = 0 . . . . . . . . . . (2)

And, αβγ =   da\mathbf{-\;\frac{d}{a}} = – 36 . . . . . . . . (3)

Since, α = 2β [Given]

Therefore, 3β + γ = 7 [From Equation (1)] . . . . . . . . . (4)

Also, 2β2^{2} + 3βγ = 0 [From Equation (2)]

β (2β + 3γ) = 0

Since, β ≠ 0, Therefore, 2β + 3γ = 0 . . . . . . . . . . . . . (5)

And, 2β2^{2} γ = – 36 [From Equation (3)] . . . . . . . . . . . . . . (6)

On Solving Equation (4) and Equation (5) we get,

β = 3 and γ = – 2

Therefore, the roots of equation x3+36=7x2x^{3} + 36 = 7x_{2} are 3, 6, and – 2.

Example 2: If x, y, and z are real variables satisfying the equations x + y + z = 5 and xy + yz+ zx = 8. Determine the range of x.

Solution:

Since, x + y + z = 5 (Given)

Therefore, z = 5 – (y + x) . . . . . . . . . . . . (1)

On Substituting the values of Equation (1) in xy + yz+ zx = 8 we get,

xy + z (y+ z) = 8,

i.e. xy + (y + z) (5 – y- x) = 8,

Or, xy+5yy2yx+5xxyx2=8xy + 5y – y^{2} – yx + 5x – xy – x^{2} = 8

Or, y2(5x)y5x+8+x2=0y^{2} – (5 – x)y – 5x + 8 + x^{2} = 0

Now, b24ac(D)0b^{2} – 4ac (D) ≥ 0 [Since, y is real]

i.e. (5x)24(x25x+8)0(5 – x)^{2} – 4 (x^{2} – 5x + 8) ≥ 0,

Or, x2+2510x4x2+20x320x^{2} + 25 – 10x – 4x^{2} + 20x – 32 ≥ 0,

Or,3x2+10x70- 3x^{2} + 10x – 7 ≥ 0,

Or, 3x210x+703x^{2} – 10x + 7 ≤ 0 . . . . . . . . . . . (2)

Now, the roots of quadratic equation 3x210x+7=03x^{2} – 10x + 7 = 0

x = +  10  ±  100    846  =  10  ±  46  =  73,  1\mathbf{\frac{+\;10\;\pm \;\sqrt{100\;-\;84}}{6}\;=\;\frac{10\;\pm \;4}{6}\;=\;\frac{7}{3},\;1}

Therefore, From Equation (2),

(x    1)  (x    73)\mathbf{\left ( x\;-\;1 \right )\;\left ( x\;-\;\frac{7}{3} \right )} ≤ 0

i.e. x    [1,  73]\mathbf{x\;\in \;\left [ 1, \;\frac{7}{3} \right ]}

Example: If α and β are the roots of the Quadratic Equation 2x2+6x+k=02x^{2} + 6x + k = 0. Find the Maximum value of [  αβ  +  βα  ]\mathbf{\left [\;\frac{\alpha }{\beta }\;+\;\frac{\beta }{\alpha }\; \right ]} if k < 0.

Solution:

From the Given Quadratic Equation 2x2+6x+k=02x^{2} + 6x + k = 0,

α + β = – 3 and αβ = k2\mathbf{\frac{k}{2}}

Since, k < 0, therefore, D = (b24acb^{2} – 4ac) = 36 – 4k > 0.

Hence, the roots of quadratic equation α and β are real.

Now, [  αβ  +  βα  ]\mathbf{\left [\;\frac{\alpha }{\beta }\;+\;\frac{\beta }{\alpha }\; \right ]} = [  α2  +  β2α  β]\mathbf{\left [\;\frac{\alpha^{2} \;+\;\beta ^{2}}{\alpha \;\beta }\right ]}

= (α  +  β)2    2  α  βα  β  =  (α  +  β)2α  β    2\mathbf{\frac{\left ( \alpha \;+\; \beta \right )^{2}\;-\;2\;\alpha \;\beta}{\alpha \;\beta }\;=\; \frac{\left ( \alpha \;+\; \beta \right )^{2}}{\alpha\; \beta }\;-\;2}

= 18k    2\mathbf{\frac{18}{k}\;-\;2}

Since, k < 0, therefore, the maximum possible value of the above expression is -2.

Hence the maximum value of [  αβ  +  βα  ]\mathbf{\left [\;\frac{\alpha }{\beta }\;+\;\frac{\beta }{\alpha }\; \right ]} = – 2.

Example: In a polynomial equation px4+qx3+rx3+sx2+tx+u=0px^{4} + qx^{3} + rx^{3} + sx^{2} + tx + u = 0, the product of all the roots taken at a time is one third of the sum of the product of roots taken two at a time. Find the relationship between ‘r’ and ‘u’.

Solution:

Let f (x) = px4+qx3+rx3+sx2+tx+upx^{4} + qx^{3} + rx^{3} + sx^{2} + tx + u

The product of all the roots = up\mathbf{\frac{u}{p}}

And, the sum of the product of roots taken two at a time = rp\mathbf{\frac{r}{p}}

Now, according to the given condition:

rp\mathbf{\frac{r}{p}} = 3  up3\;\mathbf{\frac{u}{p}}

Therefore, r = 3u

Example: Solve the equation log4(2x2+x+1)log2(2x1)=1log_{4} (2x^{2} + x + 1) – log_{2} (2x – 1) = 1.

Solution:

Given, f (x) = log4(2x2+x+1)log2(2x1)1log_{4} (2x^{2} + x + 1) – log_{2} (2x – 1) – 1

i.e. loge  (2  x2  +  x  +  1)loge  4    loge  (2x    1)loge  2  =  1\mathbf{\frac{log_{e}\;(2\;x^{2}\;+\;x\;+\;1)}{log_{e}\;4}\;-\;\frac{log_{e}\;(2x\;-\;1)}{log_{e}\;2}\;=\;1}

Or, loge  (2  x2  +  x  +  1)loge  4    loge  (2x    1)2loge  4  =  1\mathbf{\frac{log_{e}\;(2\;x^{2}\;+\;x\;+\;1)}{log_{e}\;4}\;-\;\frac{log_{e}\;(2x\;-\;1)^{2}}{log_{e}\;4}\;=\;1}

Or, loge  (2  x2  +  x  +  1)(2x    1)2  =  loge  4\mathbf{log_{e}\;\frac{(2\;x^{2}\;+\;x\;+\;1)}{(2x\;-\;1)^{2}}\;=\;log_{e}\;4}

Or, 2  x2  +  x  +  1  =  16  x2  +  4    16x\mathbf{2\;x^{2}\;+\;x\;+\;1 \;=\; 16\;x^{2}\;+\;4\;-\;16x}

Or, 14  x2    17  x  +  3\mathbf{14\;x^{2}\;-\;17\;x\;+\;3} = 0

i.e. (x – 1) (14x – 3) = 0

Since, x = 314\mathbf{\frac{3}{14}} does not satisfy the given equation.

Therefore, x = 1 is the solution of the given equation f (x).

Example: Find all the roots of equation f (x) = x421x22x3+22x+40x^{4} – 21 x^{2} – 2 x^{3} + 22x + 40, if its roots are in A.P.

Solution:

Let a – 3d, a – d, a + d, and a + 3d be the roots of the given equation.

Now, The Sum of Roots = 4a =   ba\mathbf{-\;\frac{b}{a}} = 2

Therefore, a = 12\mathbf{\frac{1}{2}}

And, the product of roots = (a3d)(ad)(a+d)(a+3d)=(a29d2)(a2d2)(a – 3d) (a – d) (a + d) (a + 3d) = (a^{2} – 9d^{2}) (a^{2} – d^{2}) = ea\mathbf{\frac{e}{a}} = 40

Since, a = 12\mathbf{\frac{1}{2}}

Therefore, (14    9  d2)  (14    d2)  =  40\mathbf{\left ( \frac{1}{4}\;-\;9\;d^{2} \right )\;\left ( \frac{1}{4}\;-\;d^{2} \right )\;=\;40}

Or, 144d440d2639=0144d^{4} – 40d^{2} – 639 = 0,

Let, d2=yd^{2} = y

Therefore, 144y240y2639=0144y^{2} – 40y^{2} – 639 = 0

Hence, y = 94\mathbf{\frac{9}{4}} or y =   736\mathbf{-\;\frac{7}{36}}

Neglecting y =   736\mathbf{-\;\frac{7}{36}}

Therefore, d2=yd^{2} = y = 94\mathbf{\frac{9}{4}}

i.e. d =   32\mathbf{-\;\frac{3}{2}} and d = 32\mathbf{\frac{3}{2}}

Therefore, the roots of equation f (x):

= (a – 3d) = (12    3×32)  =    4\mathbf{\left ( \frac{1}{2}\;-\;3\times \frac{3}{2}\right )\;=\;-\;4}

= (a + 3d) = (12  +  3×32)  =  5\mathbf{\left ( \frac{1}{2}\;+\;3\times \frac{3}{2}\right )\;=\;5}

= (a + d) = (12  +  32)  =  2\mathbf{\left ( \frac{1}{2}\;+\; \frac{3}{2}\right )\;=\;2}

= (a – d) = (12    32)  =    1\mathbf{\left ( \frac{1}{2}\;-\; \frac{3}{2}\right )\;=\;-\;1}

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