The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:

$$Base=10cm and Altitiude=5cm$$

$$Base=14cm and Altitude=10cm$$

$$Base=12cm and Altitude=10cm$$

Theory of Equations - Detailed Explanations with Solved Examples

Every equation of the 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).

In algebra, the study of algebraic equations, which are equations defined by a polynomial, is called the theory of equations. A polynomial is an expression consisting of one or more terms. The main difficulty of the theory of equations was to know when an algebraic equation has an algebraic solution. In this article, we will learn about the theory of equations and examples of solving equations.

The following are some important concepts covered under the theory of equations.

Linear equations
Simultaneous linear equations
Finding the integer solutions of an equation or of a system of equations
Systems of polynomial equations

Important Points to Remember

The important concepts in the theory of equations are given below:

The general form of a quadratic equation in x is given by ax² + bx + c = 0
The roots are given by x = (-b±√(b² – 4ac))/2a
If α and β are the roots of the equation ax² + bx + c = 0, a ≠ 0, then the sum of roots, α + β = -b/a.

Product of roots, αβ = c/a

If the sum and product of roots are known, then the quadratic equation is given by x² – (sum of roots)x + product of roots = 0

For a quadratic equation, b² – 4ac is known as the discriminant denoted by D.
If D = 0, the equation will have two equal real roots.
If D > 0, then the equation will have two distinct real roots.
If D < 0, then the equation has no real roots.
The graph of a quadratic equation is a parabola. The parabola will open upwards if a >0, and open downwards if a < 0.
If a > 0, when x = -b/2a, f(x) attains its minimum value.
If a < 0, when x = -b/2a, f(x) attains its maximum value.

Relationship between Roots and Coefficients

If α₁, α₂, α₃, α₄, α₅, α₆,…., α_n are the roots of the quadratic equation:

$\begin{array}{l}a_{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\end{array} $

Then, the sum of roots:

$\begin{array}{l}\sum \alpha_{1} = \alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} . . . . . . \alpha_{n}=\mathbf{-\;\frac{a_{1}}{a_{o}}}\end{array} $

The sum of the product of roots taken two at a time:

$\begin{array}{l}\sum \alpha_{1} \alpha_{2} = \mathbf{\frac{a_{2}}{a_{o}}}\end{array} $

The sum of the product of roots taken three at a time:

$\begin{array}{l}\sum \alpha_{1} . \alpha_{2} .\alpha_{3}=\mathbf{-\;\frac{a_{3}}{a_{o}}}\end{array} $

Product of Roots:

$\begin{array}{l}\alpha_{1} . \alpha_{2} . \alpha_{3} . \alpha_{4} . \alpha_{5}. . . . . . \alpha_{n} = (- 1)^{n} \times \frac{a_{n}}{a_{o}}\end{array} $

Therefore, for a cubic equation

$\begin{array}{l}ax^{3} + bx^{2} + cx + d = 0\end{array} $

$\begin{array}{l}\alpha{1} + \alpha_{2} + \alpha_{3}=\mathbf{-\;\frac{a_{1}}{a_{o}}}=\mathbf{-\;\frac{b}{a}}\end{array} $

$\begin{array}{l}\alpha_{1}.\alpha_{2} + \alpha_{2}. \alpha_{3}+ \alpha_{3}.\alpha_{1}=\mathbf{\frac{a_{2}}{a_{o}}}=\mathbf{\frac{c}{a}}\end{array} $

$\begin{array}{l}\alpha_{1}.\alpha_{2} . \alpha_{3} = (- 1)^{n} \times \frac{a_{3}}{a_{o}}=-\frac{d}{a}\end{array} $

Theory of Equation Solved Problems

Example 1: If one root of a cubic equation is double of another, then find all the roots of equation x³ + 36 = 7x².

Solution:

Given, f (x) = x³ – 7x² + 0.x + 36

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

Therefore, α + β + γ = -b/a = 7 . . . . . . (1)

Also, α β + γ β + αγ = c/a = 0 . . . . . . . . . . (2)

And, αβγ = -d/a = – 36 . . . . . . . . (3)

Since, α = 2β [Given],

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

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

β (2β + 3γ) = 0

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

And, 2β² 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 = 7x² 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:

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 + 5y – y² – yx + 5x – xy – x² = 8

Or, y² – (5 – x)y – 5x + 8 + x² = 0

Now, b² – 4ac (D) ≥ 0 [Since, y is real]

i.e., (5 – x)² – 4 (x² – 5x + 8) ≥ 0,

Or, x² + 25 – 10x – 4x² + 20x – 32 ≥ 0,

Or, – 3x² + 10x – 7 ≥ 0,

Or, 3x² – 10x + 7 ≤ 0 . . . . . . . . . . . (2)

Now, the roots of the quadratic equation 3x² – 10x + 7 = 0

$\begin{array}{l}x = \mathbf{\frac{+\;10\;\pm \;\sqrt{100\;-\;84}}{6}\;=\;\frac{10\;\pm \;4}{6}\;=\;\frac{7}{3},\;1}\end{array} $

Therefore, from Equation (2),

$\begin{array}{l}\mathbf{\left ( x\;-\;1 \right )\;\left ( x\;-\;\frac{7}{3} \right )}\le 0\end{array} $

$\begin{array}{l}\mathbf{x\;\in \;\left [ 1, \;\frac{7}{3} \right ]}\end{array} $

Example 3: If α and β are the roots of the quadratic equation 2x² + 6x + k = 0, find the maximum value of [(α/β) + (β/α)] if k < 0.

Solution:

From the given quadratic equation 2x² + 6x + k = 0,

α + β = – 3 and αβ = k/2

Since, k < 0, D = (b² – 4ac) = 36 – 4k > 0.

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

Now,

$\begin{array}{l}\mathbf{\left [\;\frac{\alpha }{\beta }\;+\;\frac{\beta }{\alpha }\; \right ]}=\mathbf{\left [\;\frac{\alpha^{2} \;+\;\beta ^{2}}{\alpha \;\beta }\right ]}\end{array} $

$\begin{array}{l}=\mathbf{\frac{\left ( \alpha \;+\; \beta \right )^{2}\;-\;2\;\alpha \;\beta}{\alpha \;\beta }\;=\; \frac{\left ( \alpha \;+\; \beta \right )^{2}}{\alpha\; \beta }\;-\;2}\end{array} $

$\begin{array}{l}=\mathbf{\frac{18}{k}\;-\;2}\end{array} $

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

Hence, the maximum value of [(α/β) + (β/α)]= – 2.

Example 4: In a polynomial equation

$\begin{array}{l}px^{4} + qx^{3} + rx^{3} + sx^{2} + tx + u = 0\end{array} $

, 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

$\begin{array}{l}f(x) = px^{4} + qx^{3} + rx^{3} + sx^{2} + tx + u \end{array} $

The product of all the roots = u/p

And, the sum of the product of roots taken two at a time = r/p

Now, according to the given condition,

$\begin{array}{l}\mathbf{\frac{r}{p}}=3\;\mathbf{\frac{u}{p}}\end{array} $

Therefore, r = 3u.

Example 5: Solve the equation

$\begin{array}{l}log_{4} (2x^{2} + x + 1) – log_{2} (2x – 1) = 1\end{array} $

Solution:

Given,

$\begin{array}{l}f (x) = log_{4} (2x^{2} + x + 1) – log_{2} (2x – 1) – 1\end{array} $

i.e.,

$\begin{array}{l}\mathbf{\frac{log_{e}\;(2\;x^{2}\;+\;x\;+\;1)}{log_{e}\;4}\;-\;\frac{log_{e}\;(2x\;-\;1)}{log_{e}\;2}\;=\;1}\end{array} $

Or,

$\begin{array}{l}\mathbf{\frac{log_{e}\;(2\;x^{2}\;+\;x\;+\;1)}{log_{e}\;4}\;-\;\frac{log_{e}\;(2x\;-\;1)^{2}}{log_{e}\;4}\;=\;1}\end{array} $

Or,

$\begin{array}{l}\mathbf{log_{e}\;\frac{(2\;x^{2}\;+\;x\;+\;1)}{(2x\;-\;1)^{2}}\;=\;log_{e}\;4}\end{array} $

Or, 2x² + x + 1 = 16x² + 4 – 16x

Or, 14x² – 17x + 3 = 0

i.e., (x – 1) (14x – 3) = 0

So, x = 3/14 does not satisfy the given equation.

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

Example 6: Find all the roots of equation f (x) = x⁴ – 21x² – 2x³ + 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 = -b/a = 2

Therefore, a = 1/2

And, the product of roots = (a – 3d) (a – d) (a + d) (a + 3d)

= (a² – 9d²) (a² – d²)

= e/a = 40

Since, a = 1/2

$\begin{array}{l}\mathbf{\left ( \frac{1}{4}\;-\;9\;d^{2} \right )\;\left ( \frac{1}{4}\;-\;d^{2} \right )\;=\;40}\end{array} $

Or,

$\begin{array}{l}144d^{4} – 40d^{2} – 639 = 0\end{array} $

Let, d² = y

Therefore,

$\begin{array}{l}144y^{2} – 40y^{2} – 639 = 0\end{array} $

Hence, y = 9/4 or y = -7/36

Neglecting y = -7/36

Therefore, d² = y = 9/4

i.e., d = -3/2 and d = 3/2

Therefore, the roots of equation f (x):

$\begin{array}{l}(a – 3d) = \mathbf{\left ( \frac{1}{2}\;-\;3\times \frac{3}{2}\right )\;=\;-\;4}\end{array} $

$\begin{array}{l}(a + 3d) =\mathbf{\left ( \frac{1}{2}\;+\;3\times \frac{3}{2}\right )\;=\;5}\end{array} $

$\begin{array}{l}(a + d) =\mathbf{\left ( \frac{1}{2}\;+\; \frac{3}{2}\right )\;=\;2}\end{array} $

$\begin{array}{l}(a – d) = \mathbf{\left ( \frac{1}{2}\;-\; \frac{3}{2}\right )\;=\;-\;1}\end{array} $

Example 7: Solve the equation 3x³ – x + 88 = 4x² if one of the roots of the given cubic polynomial equation is 2 – i√7.

Solution:

Let,

$\begin{array}{l}f(x) = 3x^{3} – x + 88 – 4x^{2} \end{array} $

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

Therefore, α = x = 2 – i√7

And, β = x = 2 + i√7 [Since imaginary roots occur in conjugate pairs]

Therefore,

(x – 2 + i√7)(x – 2 – i√7) = (x – 2)² + 7

= x² – 4x + 11

= g (x)

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

Therefore, 3x + 8 = 0

Hence, γ = -8/3

Therefore, the roots of the given equation f (x) are 2 – i√7, 2 + i√7, γ = -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 8: Find how many total positive real roots, negative real roots, and imaginary roots the equation

$\begin{array}{l}3x^{7} – x^{5} + 2x^{3} – 7 = 0\end{array} $

will have.

Solution:

Let, f (x) =

$\begin{array}{l}3x^{7} – x^{5} + 2x^{3} – 7 \end{array} $

Now, sign of f (x) =

$\begin{array}{l}\mathbf{+\;ve\;\rightarrow \;-\;ve\;\rightarrow \;+\;ve\;\rightarrow \;-\;ve}\end{array} $

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

Now, sign of f (- x) =

$\begin{array}{l}-3x^{7} + x^{5} – 2x^{3} – 7 = 0\end{array} $

$\begin{array}{l}\mathbf{-\;ve\;\rightarrow \;+\;ve\;\rightarrow \;-\;ve\;\rightarrow \;-\;ve}\end{array} $

Since the sign of f (- x) changes twice, 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.

Quadratic Equations