# Quadratic Equations IIT JEE Study Material

Quadratic Equations IIT JEE Revision Notes summarise all the key concepts included in IIT JEE Syllabus in an easy to understand format. These IIT JEE revision notes provide an extra edge to students by developing self-confidence in them before their final JEE Mains and JEE Advanced examinations. Quadratic Equations constitute an important part of Algebra. Quadratic Equation lays the foundation for various other important topics included in JEE Mains and Advanced Syllabus, hence, it is essential for students to master this unit before moving on to the other concepts included in IIT JEE. Quadratic Equations are comparatively easier in comparison to the other concepts, therefore, with proper study material and practice, the students can easily fetch good marks from this unit. The various topics covered in Quadratic Equations IIT JEE Revision Notes include:

The 2nd order polynomial Equations are known as quadratic equations. The General form of a quadratic Equation is $\mathbf{ax^{2}\;+\; bx\;+\;c}$. Where, $\mathbf{a\;\neq \;0}$ and a, b, and c are real numbers. The values of x satisfying the quadratic equation are the roots of the quadratic equation $\mathbf{\left ( \alpha , \beta \right )}$. The quadratic equation will always have two roots. The nature of roots may be either real or imaginary.

The solution or roots of a quadratic Equation are given by the quadratic formula:

$\mathbf{\left ( \alpha ,\;\beta \right )\;=\;\frac{-b\;\pm \;\sqrt{b^{2}\;-\;4ac}}{2\;ac}}$

### The Discriminant Of A Quadratic Equation

The term $\mathbf{b^{2}\;-\;4ac}$ in a quadratic Formula is known as the Discriminant of a Quadratic Equation (D). The discriminant reveals the nature of roots a quadratic equation will have.

Nature of Roots of a Quadratic Equation

If the values of Discriminant (D) = 0 i.e. $\mathbf{b^{2}\;-\;4ac} \;= \;0$ The Quadratic Equation will have equal roots i.e. $\mathbf{\alpha \;=\; \beta \;=\; \frac{-b}{2a}}$
If the values of Discriminant (D) < 0 i.e. $\mathbf{b^{2}\;-\;4ac} = 0$ The Quadratic Equation will have imaginary roots i.e.$\mathbf{ \alpha \;=\;\left ( p\;+\;iq\right )}$ and $\mathbf{\beta \;=\;\left ( p\;-\;iq\right )}$. Where “iq” is the imaginary part of a complex number
If the values of Discriminant (D) > 0 i.e. $\mathbf{b^{2}\;-\;4ac} \;> \;0$ The Quadratic Equation will have real roots
If the values of Discriminant (D) > 0 and D is a perfect Square The Quadratic Equation will have Rational roots
If the values of Discriminant (D) > 0 and D is not a perfect Square The Quadratic Equation will have Irrational roots i.e. $\mathbf{ \alpha \;=\;\left ( p\;+\;\sqrt{q}\right )}$ and $\mathbf{ \beta \;=\;\left ( p\;-\;\sqrt{q}\right )}$
If the values of Discriminant (D) > 0, D is a perfect Square, a = 1 and b and c are integers The quadratic Equations will have integral roots

Example 1: Find the values of k such that the quadratic Equation $\mathbf {\frac{p}{x\;+\;r}\;+\;\frac{q}{x\;-\;r}\;=\;\frac{k}{2\;x}}$ has two equal roots.

Solution:

The Quadratic Equation $\mathbf {\frac{p}{x\;+\;r}\;+\;\frac{q}{x\;-\;r}\;=\;\frac{k}{2\;x}}$ can be rewritten as:

$\left [2p + 2q – k \right ] x^{2} – 2r \left [p – q \right ] x + r^{2} k = 0$

For Equal Roots, the Discriminant (D) = 0, i.e.$b^{2} – 4ac = 0$

Here, a = [ 2p + 2q – k ], b = – 2r [ p – q ] and c = $r^{2} k$

i.e. $[ – 2r (p – q) ]^{2} – 4 [ ( 2p + 2q – k ) (r^{2} k) ] = 0$

$r^{2} (p – q)^{2} – r^{2}k (2p + 2q – k) = 0$

Since r ≠ 0, therefore, $(p – q)^{2} – k (2p + 2q – k) = 0$

i.e. $k^{2} – 2(p + q) k + (p – q)^{2}$

Therefore, k = $\mathbf {\frac{2\;\left ( p\;+\;q \right )\;\pm \;\sqrt{4\;\left ( p\;+\;q \right )^{2}\;-\; 4\;\left ( p\;-\;q \right )^{2}}}{2}\;=\;-\;(p\;+\;q)\;\pm \;\sqrt{4pq}}$

i.e. k = $\mathbf {\;(p\;+\;q)\;\pm \;2\;\sqrt{p\;q}\;=\;\left ( \sqrt{p}\;\pm \;\sqrt{q} \right )^{2}}$

Example 2: Find the values of k for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots.

Solution:

The Given Quadratic Equation can be rewritten as, $x^{2} – (10 + k) x + 1 + 10k = 0$.

$D = b^{2} – 4ac = 100 + k^{2} + 20k – 40k = k^{2} – 20k + 96 = ( k – 10 )^{2} – 4$

The quadratic Equations will have integral roots, If the values of Discriminant (D) > 0, D is a perfect Square, a = 1 and b and c are integers.

i.e. $(k – 10)^{2} – D = 4$

Since, Discriminant (D) is a perfect square. Hence, the difference of two perfect square in R.H.S will be 4 only when D = 0 and $( k – 10 )^{2} = 4$.

That Implies, k – 10 = ± 2. Therefore, the values of k = 8 and 12.

### Quadratic Equation IIT JEE Important Tips

1: If one root of a quadratic equation is a + ib then, a – ib will be the other root of the quadratic equation, where $\mathbf{i \;= \;\sqrt{-1}}$.

2: Similarly, If one root of a quadratic equation is $\mathbf{a\;+ \;\sqrt{b}}$ then, $\mathbf{a\;- \;\sqrt{b}}$ will be the other root of the quadratic equation. [where, $\mathbf{\sqrt{b}}$ is an irrational number]

3: A quadratic equation becomes an identity (a, b, c = 0) if the equation is satisfied by more than two numbers i.e. having more than two roots or solutions either real or complex.

4: The Quadratic Expression$g(x, y) = ax^{2} +2hxy + by^{2} +2 gx+ 2fy + c$ can be resolved into two linear factors if:

$af^{2} + bg^{2} + ch^{2} – abc – 2 fgh = 0$ or$\mathbf{\begin{vmatrix} a & h & g\\ h & b & f\\ g & f & c \end{vmatrix}}$

### Relationships Between The Coefficient And The Roots Of A Quadratic equation

If $\mathbf{\alpha }$ and $\mathbf{\beta }$ are roots of a Quadratic Equation $\mathbf{ax^{2}\;+\;bx\;+\;c}$ then,

$\mathbf{\alpha\;+\; \beta \;=\;\frac{-b}{a}}$, $\mathbf{\alpha\; \beta \;=\;\frac{c}{a}}$ and, $\mathbf{\left | \alpha \;+\;\beta \right |\;=\;\frac{\sqrt{D}}{\left | a \right |}}$

If $\mathbf{\alpha }$ and $\mathbf{\beta }$ are roots of a Quadratic Equation $\mathbf{ax^{2}\;+\;bx\;+\;c}$ then, the quadratic Equation can be written as:

$\mathbf{\left ( x\;-\;\alpha \right )\; \left ( x\;-\;\beta \right ) \;=\;0}$

Or, $\mathbf{x^{2}\;-\;\left ( \alpha \;+\;\beta \right )x\;+\;\alpha \;\beta \;= \;0}$

Or, $\mathbf{x^{2}\;+\;\left (\frac{b}{a} \right )x\;+ \frac{c}{a} \;= \;0}$

Example 1: If the coefficient of x in the quadratic equation $x^{2} + bx + c = 0$ was taken as 17 in place of 13, its roots were found to be – 2 and – 15. Find the roots of the original quadratic equation.

Solution:

Since there is no change in the coefficient of x2 and c, therefore, the product of zeros will remain same in both the Quadratic Equations.

Therefore, the Product of Zeros (c) = – 2 × – 15 = 30,

Since, the original value of b is 13.

Therefore, The Sum of Zeros = $\mathbf{\frac{-b}{a}}$ = -13.

Hence, the original quadratic equation is:

$x^{2}$ – (Sum of Zeros)x + (Product of Zeros) = 0 i.e. $x^{2} + 13x + 30 = 0$

i.e. (x + 10) (x + 3) = 0

Therefore, the roots of original quadratic equations are -3 and -10.

Example 2: If $\mathbf{\alpha }$and $\mathbf{\beta }$ are roots of a quadratic Equation $px^{2} + qx + r = 0$, then find the quadratic Equation whose roots are, $\mathbf{\alpha^{2}\;+\;\beta ^{2} }$ and $\mathbf{\frac{1}{\alpha^{2}}\;+\; \frac{1}{\beta ^{2}}}$.

Solution:

From the quadratic Equation $px^{2} + qx + r = 0$,

$\mathbf{\alpha }$ + $\mathbf{\beta }$ = $\mathbf{\frac{-q}{p}}$ . . . . . . . . . . . . . (1)

$\mathbf{\alpha \;\beta }$ = $\mathbf{\frac{r}{p}}$ . . . . . . . . . . . . . (2)

Now, The Product of roots $\mathbf{\alpha^{2}\;+\;\beta ^{2} }$ and $\mathbf{\frac{1}{\alpha^{2}}\;+\; \frac{1}{\beta ^{2}}}$:

= $\mathbf{\left ( \alpha^{2}\;+\;\beta ^{2} \right )\;\left ( \frac{1}{\alpha^{2}}\;+\; \frac{1}{\beta ^{2}} \right )\;=\; \frac{\left ( \alpha^{2}\;+\;\beta ^{2} \right )^{2} }{\alpha ^{2}\;\beta ^{2}}}$

i.e. $\mathbf{\frac{\left ( \alpha^{2}\;+\;\beta ^{2} \right )^{2} }{\alpha ^{2}\;\beta ^{2}}\;=\;\frac{\left ( \alpha \;+\;\beta \right )^{2}\;-\;2\;\left ( \alpha \;\beta \right )}{\left ( \alpha \;\beta \right )^{2}}\;=\; \frac{\left ( \frac {p^{2}}{q^{2}} \right )\;-\;2\;\left ( \frac{r}{q} \right )}{\left ( \frac{r^{2}}{q^{2}} \right )}}$ [Using Equation (1) and (2)]

Now, The Sum of roots $\mathbf{\alpha^{2}\;+\;\beta ^{2} }$ and $\mathbf{\frac{1}{\alpha^{2}}\;+\; \frac{1}{\beta ^{2}}}$:

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

i.e. $\mathbf{\left ( \alpha +\beta \right )^{2}\;-\;2\;\alpha \;\beta \;+\; \frac{ \left ( \alpha +\beta \right )^{2}\;-\;2\;\alpha \;\beta }{\left ( \alpha \;\beta \right )^{2}}\;=\;\frac{q^{2}}{p^{2}}\;-\frac{2r}{p}\;+\;\frac{\frac{q^{2}}{p^{2}}\;-\frac{2r}{p}}{\frac{r^{2}}{p^{2}}}}$

= $\mathbf{\left ( q^{2}\;-\;2pr \right )\;\left ( \frac{p^{2}\;+\;r^{2}}{r^{2}\;p^{2}} \right )}$ [Using Equation (1) and (2)]

Hence, the required quadratic Equation is: $x^{2}$ – (Sum of Zeros)x + (Product of Zeros) = 0

i.e. $\mathbf{x^{2}\;-\;\left [\left ( q^{2}\;-\;2pr \right )\;\left ( \frac{p^{2}\;+\;r^{2}}{r^{2}\;p^{2}} \right ) \right ]x\;+\;\frac{\left ( \frac {p^{2}}{q^{2}} \right )\;-\;2\;\left ( \frac{r}{q} \right )}{\left ( \frac{r^{2}}{q^{2}} \right )} \;= \;0 }$

Therefore, $\mathbf{\left ( prx \right )^{2}-\left [ \left ( q^{2}\;-\;2pr \right )\;\left ( p^{2}\;+\;r^{2}\right ) \right ]+\left ( p^{2}-2rq \right )\;=\;0 }$ is the required Quadratic Equation.

### Common Roots Of A Quadratic Equation

Let $\mathbf{\beta }$ be the common root (solution) of quadratic Equations $a_{1}x^{2} + b_{1}x + c_{1}$ and $a_{2}x^{2} + b_{2}x + c_{2}$.

This implies that $\mathbf{a_{1}\beta ^{2}\;+\;b_{1}\beta \;+\;c_{1}\;=\;0}$and $\mathbf{a_{2}\beta ^{2}\;+\;b_{2}\beta \;+\;c_{2}\;=\;0}$

Now, Solving for $\mathbf{\beta ^{2}}$ and $\mathbf{\beta }$ we will get:

$\mathbf{\begin{vmatrix} \beta ^{2} & \beta & 1\\ a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2} \end{vmatrix} \;=\; \frac{\beta ^{2}}{b_{1}c_{2}\;-\;b_{2}c_{1}}\;=\;\frac{-\beta }{a_{1}\;c_{2}\;-\;a_{2}\;c_{1}}\;=\;\frac{1}{a_{1}\;b_{2}\;-\;a_{2}\;b_{1}}}$

Therefore, $\mathbf{\beta ^{2}\;=\;\frac{b_{1}\;c_{2}\;-\;b_{2}\;c_{1}}{a_{1}\;b_{2}\;-\;a_{2}\;b_{1}}}$. . . . . . . . . . . . . . . . (1)

And, $\mathbf{\beta \;=\;\frac{a_{2}\;c_{1}\;-\;a_{1}\;c_{2}}{a_{1}\;b_{2}\;-\;a_{2}\;b_{1}}}$. . . . . . . . . . . . . . . . (2)

On Squaring Equation (2) and equating it with Equation (1) we get:

$(a_{1} b_{2} – a_{2} b_{1}) (b_{1} c_{2} – b_{2} c_{1}) = (a_{2} c_{1} – a_{1} c_{2})^{2}$

Hence, it is the required condition for quadratic equations having one common root.

If both the roots of quadratic equations $a_{1} x^{2} + b_{1} x + c_{1}$ and $a_{2} x^{2} + b_{2} x + c_{2}$ are common then:

$\mathbf{\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}}$

Example: For what value of k, both the quadratic equations $6x^{2} – 17x + 12 = 0$ and $3x^{2} – 2x + k = 0$ will have a common root.

Solution:

If one of the root of quadratic Equations $a_{1} x^{2} + b_{1} x + c_{1}$ and $a_{2} x^{2} + b_{2} x + c_{2}$ is common then: $(a_{1} b_{2} – a_{2} b_{1}) (b_{1} c_{2} – b_{2} c_{1}) = (a_{2} c_{1} – a_{1} c_{2})^{2}$ . . . . . . . . . . . . . (1)

Form the given quadratic Equations, $a_{1}$ = 6, $b_{1}$ = -17, $c_{1}$, $a_{2}$ = 3, $b_{2}$ = -2 and $c_{2}$ = k

On substituting these values in Equation (1), we will get:

$[(6 \times -2) – (3 \times -17)] [ – 17k – (-2 \times 12)] = (3 \times 12 – 6k)^{2}$

$-663k + 936 = 1296 + 36k^{2} – 432k$

$36k^{2} + 231k + 360 = 0$

$12k^{2} + 125k + 120 = 0$

(4k + 15) (3k + 8) = 0

Therefore, the values of k are $\mathbf {\frac{-15}{4}, \;\frac{-8}{3}}$

### Descartes Rule Of Sign

This is a very famous rule that helps in getting an idea about the roots of a polynomial equation. The rule states that the number of positive real roots of $P_{n}(x) = 0$ cannot be more than the number of sign changes. Similarly, the number of negative roots cannot be more than the number of sign changes in $P_{n}(-x)$.

Example 1: How many real roots does the equation $2x^{5} + 2x^{4} – 11x^{3} + 9x^{2} – 4x + 2 = 0$ will have?

Since the given equation has 4 sign changes so it can have a maximum of 4 positive real roots. Now for f(-x), the equation has only one sign change i.e. f (-x) =$-x^{5} + 2x^{4} + x^{3} + x^{2} + x + 2 = 0$. Hence, the equation will have only one negative real root.

Example 2: How many real roots does the quadratic equation $x^{2} + 3 |x| + 2$ will have?

Solution:

Since the given quadratic equation has no sign change for both f(x) and f(-x) Therefore, the quadratic equation will have no real root.

### Quadratic Equation IIT JEE Practice Problems

Q.1: If $\mathbf{\alpha }$ and $\mathbf{\beta }$ are roots of the quadratic Equation $x^{2} – x + 1 = 0$, then Find the value of $\mathbf{\alpha ^{2009}}$ + $\mathbf{\beta ^{2009}}$.

Solution:

The roots of the quadratic Equation $x^{2} – x + 1 = 0$ are given by:

$\mathbf{\left ( \alpha ,\;\beta \right )\;=\;\frac{1\;\pm \;\sqrt{1\;-\;4}}{2} =\frac{1\;\pm \;\sqrt{3}}{2}}$

Therefore, $\mathbf{\alpha }$ = $\mathbf{ \frac{1}{2}\;+\;\frac{\sqrt{3}\;i}{2}}$ and, $\mathbf{\alpha }$ = $\mathbf{ \frac{1}{2}\;-\;\frac{\sqrt{3}\;i}{2}}$

i.e. $\mathbf{\alpha }$ = $\mathbf{ cos \;\frac{\pi }{3}\;+\;i\;sin\;\frac{\pi }{3}}$and, $\mathbf{\alpha }$ = $\mathbf{ cos \;\frac{\pi }{3}\;-\;i\;sin\;\frac{\pi }{3}}$

Since, $\mathbf{z^{p}=cos\;\left ( p\;\theta \right )\;+\;i\;sin\;\left ( p\;\theta \right ) }$ [De Moivre’s Theorem]

Therefore, $\mathbf{\alpha ^{2009}}$ = $\mathbf{z^{p}=cos\;\left ( 2009\;\theta \right )\;+\;i\;sin\;\left ( 2009\;\theta \right ) }$

And, $\mathbf{\beta ^{2009}}$ = $\mathbf{z^{p}=cos\;\left ( 2009\;\theta \right )\;-\;i\;sin\;\left ( 2009\;\theta \right ) }$

i.e. $\mathbf{\alpha ^{2009}}$ + $\mathbf{\beta ^{2009}}$ = $\mathbf{ 2\;cos \;2009\;\frac{\pi }{3}}$

= $\mathbf{ 2\;cos\;\left [ 668\;\pi \;+\; \pi \;+\;\frac{2\;\pi }{3} \right ]\;=\;2\;cos\;\left [ \pi \;+\frac{2\;\pi }{3} \right ]}$

= $\mathbf{ -2\;cos\;\frac{2\;\pi }{3}\;=\;-2\;\left ( \frac{-1}{2} \right )\;=\;1}$

Therefore, the value of $\mathbf{\alpha ^{2009}}$ + $\mathbf{\beta ^{2009}}$ = 1

Q.2: If $\mathbf{\alpha }$ and $\mathbf{\beta }$ are roots of a quadratic Equation $x^{2} + bx – c = 0$ and $\mathbf{\gamma }$ and $\mathbf{\delta }$ are roots of a quadratic equation $x^{2} + bx + s = 0$, where q and r ≠ 0, then find the value of $\mathbf{\frac{\left ( \alpha \;-\;\gamma \right )\;\left ( \alpha \;-\;\delta \right )}{\left ( \beta \;-\;\gamma \right )\;\left ( \beta \;-\;\delta \right )}}$.

Solution:

From the given quadratic equations $\mathbf{\alpha \;+\;\beta }$ = – b and $\mathbf{\gamma \;+\; \delta }$ = – b . . . . . . . . (1)

Therefore, $\mathbf{\alpha \;+\;\beta = \gamma \;+\; \delta }$ . . . . . . . . . . (2)

And, $\mathbf{\alpha \;\beta }$ = – c and $\mathbf{\gamma \;\delta }$ = s . . . . . . . . . . . . . (3)

Now,

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

= $\mathbf{\alpha ^{2}\;-\;\left ( \alpha \;+\;\beta \right )\;\alpha \;+ \;\gamma \delta }$ [From Equation (2)]

= $\mathbf{-\;\alpha \beta \;+ \;\gamma \delta }= c + s$ [From Equation (3)]

Therefore, $\mathbf{\left ( \alpha \;-\;\gamma \right )\;\left ( \alpha \;-\;\delta \right )}$ = c + s

Similarly, $\mathbf{\left ( \beta \;-\;\gamma \right )\;\left ( \beta \;-\;\delta \right )}$ = c + s

Therefore, the value of $\mathbf{\frac{\left ( \alpha \;-\;\gamma \right )\;\left ( \alpha \;-\;\delta \right )}{\left ( \beta \;-\;\gamma \right )\;\left ( \beta \;-\;\delta \right )}}$ = 1.

Q.3: If $\mathbf{\alpha }$ and $\mathbf{\beta }$ are roots of a quadratic equation $px^{2} + qx + r = 0$ and $\mathbf{\gamma }$ and $\mathbf{\delta }$ are roots of a quadratic equation $gx^{2} + hx + i = 0$, then find the equation whose roots are $\mathbf{\left ( \alpha \;\gamma \;+\;\beta \;\delta \right )}$ and $\mathbf{\left ( \alpha \;\delta \;+\;\beta \;\gamma \right )}$.

Solution:

From the above quadratic Equations $\mathbf{\alpha \;+\;\beta }$ = $\mathbf{\frac{-q}{p}}$ and $\mathbf{\alpha \;\beta }$ = $\mathbf{\frac{-r}{p}}$ . . . . . . . . . . . . (1)

And, $\mathbf{\gamma \;+\;\delta }$ =$\mathbf{\frac{-h}{g}}$ and $\mathbf{\gamma \;\delta }$ = $\mathbf{\frac{i}{g}}$ . . . . . . . . (2)

Now, The Product and Sum of Roots $\mathbf{\left ( \alpha \;\gamma \;+\;\beta \;\delta \right )}$ and $\mathbf{\left ( \alpha \;\delta \;+\;\beta \;\gamma \right )}$:

Product = $\mathbf{\left ( \alpha \;\gamma \;+\;\beta \;\delta \right ) \; \left ( \alpha \;\delta \;+\;\beta \;\gamma \right )}$

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

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

= $\mathbf{\gamma \;\delta \;\left [ \left ( \alpha \;+\;\beta \right )^{2}\;-2\;\left ( \alpha \;\beta \right ) \right ]\; +\; \alpha \;\beta \left [ \left ( \gamma \;+\;\delta \right )^{2} \;-\;2\;\left ( \gamma \;\delta \right ) \right ] }$

Now, on substituting the values of Equation (1) and (2), we will get:

= $\mathbf{\frac{i}{g} \;\left ( \frac{-q}{p}\right )^{2}\;-2\;\left ( \frac{r}{p}\right )\; +\; \frac{r}{p} \;\left ( \frac{-h}{g}\right )^{2} \;-\;2\;\left (\frac{i}{g} \right )}$

=$\mathbf{\frac{i\;g\;q^{2}\;-\;2\;g^{2}\;p\;r\;+\;r\;p\;h^{2}\;-\;2\;i\;g\;p^{2}}{g^{2}\;p^{2}} }$

= $\mathbf{\frac{i}{g} \;\left [ \left ( \frac{-q}{p}\right )^{2}\;-2\;\left ( \frac{r}{p}\right ) \right ]\; +\; \frac{r}{p} \;\left [ \left ( \frac{-h}{g}\right )^{2} \;-\;2\;\left (\frac{i}{g} \right ) \right ]}$

=$\mathbf{\frac{q^{2}\;i}{g\;p^{2}}\;-\;\frac{2\;r\;i}{g\;p}\;+\;\frac{h^{2}\;r}{p\;g^{2}}\;-\;\frac{2\;i\;r}{p\;g}}$

= $\mathbf{ \frac{q^{2}\;i\;g\;-\;4\;r\;i\;g\;p\;+\;h^{2}\;r\;p}{g^{2}\;p^{2}}}$

Sum = $\mathbf{\left ( \alpha \;\gamma \;+\;\beta \;\delta \right ) \; +\;\left ( \alpha \;\delta \;+\;\beta \;\gamma \right )}$

= $\mathbf{\alpha \;\gamma \;+\;\beta \;\delta \;+\; \alpha \;\delta \;+\;\beta \;\gamma \;=\;\alpha \left ( \gamma \;+\;\delta \right )\;+\;\beta \left ( \gamma \;+\;\delta \right )}$

= $\mathbf{\left ( \alpha \;+\;\beta \right )\;\left ( \gamma \;+\;\delta \right )\;=\;\frac{-\;q}{p}\;\cdot \;\frac{-\;h}{g}\;=\;\frac{q\;h}{p\;g}}$ [From Equation (1) and (2)]

The quadratic Equation is given by: $x^{2}$ – (Sum of Roots)x + (Product of Roots) = 0

i.e. $\mathbf {x^{2}\;-\;\left ( \frac{q\;h}{p\;g} \right )x\;+\;\frac{q^{2}\;i\;g\;-\;4\;r\;i\;g\;p+h^{2}\;r\;p}{g^{2}\;p^{2}}\;=\;0}$

Therefore, $\mathbf {\;g^{2}\;p^{2}x^{2}\;-\;\left ( p\;g\;q\;h \right )x\;+\;q^{2}\;i\;g\;-\;4\;r\;i\;g\;p\;+\;h^{2}\;r\;p\;=\;0}$ is the required quadratic Equation.

Q.4: Find the values of k such that the Quadratic Equations $x^{2} – 11x + k$ and $x^{2} – 14x + 2k$ have a common factor.

Solution:

Let $\mathbf {\left ( x\;-\;\alpha \right )}$ be the common factor of quadratic Equations $x^{2} – 11x + k$ and $x^{2} – 14x + 2k$ Then x = $\mathbf {\alpha }$ will satisfy the given quadratic equations.

Therefore, $\mathbf {\alpha ^{2}\;-\;11\;\alpha \;+\;k\;=\;0}$ . . . . . . . . . . (1)

And, $\mathbf {\alpha ^{2}\;-\;14\;\alpha \;+\;2k\;=\;0}$ . . . . . . . . . . . . (2)

On Solving Equation (1) and Equation (2) we will get:

$\mathbf { \begin{vmatrix} \alpha ^{2} & \alpha & 1\\ 1 & -11 & k\\ 1 & -14 & 2 \end{vmatrix}\;=\;0}$

i.e. $\mathbf {\frac{\alpha ^{2}}{-22k\;+\;14k}=\frac{-\alpha }{2k\;-\;k}=\frac{1}{-14\;+\;11}}$

Therefore, $\mathbf {\alpha ^{2}\;=\;\frac{-22k\;+\;14k}{-3}\;=\;\frac{8}{3}}$ . . . . . . . (3)

And, $\mathbf {\alpha = \frac{2k\;-\;k}{-14\;+\;11}\;=\;\frac{k}{3}}$ . . . . . . . . . . . (4)

On Equating Equation (3) and Equation (4):

$\mathbf {\frac{8}{3}\;=\;\left ( \frac{k}{3} \right )^{2}}$

Therefore, the value of k = 24.

Q.5: How many real roots does the equation $x^{7} + 14x^{5} + 16x^{3} + 30x – 560 = 0$ will have?

Solution:

Since the given equation has 1 sign change so it can have a maximum 1 positive real roots. Now for f(-x), the equation has no sign change i.e. f (-x) = $-x^{7} – 14x^{5} – 16x^{3} – 30x – 560 = 0$. Hence, the equation will have zero negative real root.