# Theory of Equations Study Material For IIT JEE

Quadratic Equations IIT JEE Study material has been prepared by our experts to help the students in preparing all the important concepts included in Quadratic Equation IIT JEE Syllabus in a much better and effective ways. These revision notes facilitate the students to get an overview of all the important concepts like Theory of polynomial equations, Higher degree polynomial equations and more along with the detailed summary, important formulas, and IIT JEE quadratic equation solved examples. These are valuable revision notes with neatly illustrated solved examples for the best revision of IIT JEE and other entrance examinations like VITEEE, BITSAT, SRMEEE, etc.

## The various topics covered in Quadratic Equations JEE Revision Notes include:

Theory of Equations

1: Every Equation of nth degree has total ‘n’ real or imaginary roots.

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

Solution:

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

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

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

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

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

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

Therefore, 3x + 8 = 0

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

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

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 $3x^{7} – x^{5} + 2x^{3} – 7 = 0$ will have.

Solution:

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

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

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) = $-3x^{7} + x^{5} – 2x^{3} – 7 = 0$ = $\mathbf{-\;ve\;\rightarrow \;+\;ve\;\rightarrow \;-\;ve\;\rightarrow \;-\;ve}$

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 $α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, α_{6}, . . . . . . . . . . . . , α_{n}$ are the roots of the quadratic Equation:

$a_{0} ­x^{n} + a_{1} ­x^{n-1} + a_{2} ­x^{n – 2} + a_{3} ­x^{n – 3} + a_{4} ­x^{n – 4 }+ . . . . . . . . . . . . . . . . + a_{n-1} x + a_{n} = 0$

Then, the Sum of Roots:

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

Sum of Product of roots taken two at a time:

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

Sum of Product of roots taken three at a time:

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

Product of Roots:

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

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

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

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

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

## Quadratic Equation IIT JEE Problems

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

Solution:

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

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

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

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

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

Since, α = 2β [Given]

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

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

β (2β + 3γ) = 0

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

And, 2β$^{2}$ 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 $x^{3} + 36 = 7x_{2}$ 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, $xy + 5y – y^{2} – yx + 5x – xy – x^{2} = 8$

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

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

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

Or, $x^{2} + 25 – 10x – 4x^{2} + 20x – 32 ≥ 0$,

Or,$- 3x^{2} + 10x – 7 ≥ 0$,

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

Now, the roots of Quadratic Equation $3x^{2} – 10x + 7 = 0$

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

Therefore, From Equation (2),

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

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

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

Solution:

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

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

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

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

Now, $\mathbf{\left [\;\frac{\alpha }{\beta }\;+\;\frac{\beta }{\alpha }\; \right ]}$ = $\mathbf{\left [\;\frac{\alpha^{2} \;+\;\beta ^{2}}{\alpha \;\beta }\right ]}$

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

= $\mathbf{\frac{18}{k}\;-\;2}$

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

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

Example: In a polynomial equation $px^{4} + qx^{3} + rx^{3} + sx^{2} + tx + u = 0$, 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) = $px^{4} + qx^{3} + rx^{3} + sx^{2} + tx + u$

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

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

Now, according to the given condition:

$\mathbf{\frac{r}{p}}$ = $3\;\mathbf{\frac{u}{p}}$

Therefore, r = 3u

Example: Solve the equation $log_{4} (2x^{2} + x + 1) – log_{2} (2x – 1) = 1$.

Solution:

Given, f (x) = $log_{4} (2x^{2} + x + 1) – log_{2} (2x – 1) – 1$

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

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

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

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

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

i.e. (x – 1) (14x – 3) = 0

Since, x = $\mathbf{\frac{3}{14}}$ 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) = $x^{4} – 21 x^{2} – 2 x^{3} + 22x + 40$, if its roots are in A.P.

Solution:

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

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

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

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

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

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

Or, $144d^{4} – 40d^{2} – 639 = 0$,

Let, $d^{2} = y$

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

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

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

Therefore, $d^{2} = y$ = $\mathbf{\frac{9}{4}}$

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

Therefore, the roots of equation f (x):

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

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

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

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

#### Practise This Question

Which of the following elements represent highly electro positive as well as highly electro negative element in its period?