# Quadratic Equations - Class 10 : Notes

Notes for quadratic equations chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

(1) A polynomial of degree 2 is called a quadratic polynomial. The general form of a quadratic polynomial is $a{x^2} + bx + c$, where a, b, c are real number such that a $\ne$0 and x is a real variable.
For Example: ${x^2} + 5x + 3$, where $a = 1,b = 5,c = 3$ are real number. So given equation is quadratic polynomial.

(2) If $p(x) = a{x^2} + bx + c, a \ne 0$ is a quadratic polynomial and $\alpha$ is a real number, then $p(\alpha ) = a{\alpha ^2} + b\alpha + c$ is known as the value of the quadratic polynomial $p(\alpha )$
For Example: $p(\alpha ) = {\alpha ^2} + 5\alpha + 3$ in that equation if $\alpha=3$ then $p(\alpha ) = 27$. So $27$ is a value of quadratic polynomial

(3) A real number $\alpha$ is said to be a zero of quadratic polynomial $p(x) = a{x^2} + bx + c$, if $p(\alpha ) = 0.$
For Example: $p(x) = {x^2} + 6x + 5$
If $x =( - 5)$ then $p(x) =0$, So $-5$ is the zero of polynomial.

(4) If $p(x) = a{x^2} + bx + c$ is a quadratic polynomial, then $p(x) = 0$ i.e., $a{x^2} + bx + c = 0$, $a \ne 0$ is called a quadratic equation.
For Example: $p(x) = {x^2} - 8x + 16$ is a quadratic polynomial, then $p(x) = 0$ i.e., ${x^2} - 8x + 16 = 0$, $a \ne 0$ is called a quadratic equation.

(5) A real number $\alpha$ is said to be a root of the quadratic equation $a{x^2} + bx + c = 0$.
In other words,   $\alpha$ is a root of $a{x^2} + bx + c = 0$ if and only if $\alpha$ is a zero of the polynomial $p(x) = a{x^2} + bx + c$.
For Example: Suppose quadratic equation is $2{x^2} - x - 6 = 0$.
If we put $x = 2$ then $p(x)=0$, So $2$ is a root of that given equation so here $\alpha=2$.

(6) If $a{x^2} + bx + c = 0$, $a \ne 0$ is factorizable into a product of two linear factors,  then the roots of the quadratic equation $a{x^2} + bx + c = 0$ can be found by equating each factor to zero.
For Example: The Given equation is $a{x^2} + bx + c = 0$
Now, Solving the above equation using factorization method. $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$(3x + 1) (3x - 2) =0 $a{x^2} + bx + c = 0$ (3x + 1) = 0 or (3x - 2) = 0 $a{x^2} + bx + c = 0$3x = -1 or 3x = 2 $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$   or $a{x^2} + bx + c = 0$
Hence, $a{x^2} + bx + c = 0$  and $a{x^2} + bx + c = 0$ are the two roots of the given equation

(7) The roots of a quadratic equation can also be found by using the method of completing the square.
For Example:The Given equation is - $a{x^2} + bx + c = 0$
Dividing through out by 2 $a{x^2} + bx + c = 0$
Shifting the constant term to the right hand side. $a{x^2} + bx + c = 0$
Adding square of the half of coefficient of x on the both side. $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$
Taking square root of both sides- $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$
Hence x= 3, and x=1/2 are the two root of the given equation

(8) The roots of the quadratic equation $a{x^2} + bx + c = 0$, $a \ne 0$ can be found by using the quadratic formula ${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$ , provided that $\sqrt {{b^2} - 4ac} \ge 0$.
For Example: $\sqrt {{b^2} - 4ac} \ge 0$ the given equation in the form of $\sqrt {{b^2} - 4ac} \ge 0$,
Where a= √3, b=10 c= 8√3
Therefore, the discriminant- $\sqrt {{b^2} - 4ac} \ge 0$
D= (10)2 – 4 x √3 x (-8√3)
D= 100 + 96
D= 196
Since, D > 0
Therefore, the roots of the given equation are real and distinct.
The real roots α and β are given by, $\sqrt {{b^2} - 4ac} \ge 0$ $\sqrt {{b^2} - 4ac} \ge 0$ $\sqrt {{b^2} - 4ac} \ge 0$ $\sqrt {{b^2} - 4ac} \ge 0$
For, $\sqrt {{b^2} - 4ac} \ge 0$ $\sqrt {{b^2} - 4ac} \ge 0$ $\sqrt {{b^2} - 4ac} \ge 0$
Hence $\sqrt {{b^2} - 4ac} \ge 0$ and $\sqrt {{b^2} - 4ac} \ge 0$ are the two root of the given equation.

(9) Nature of the roots of quadratic equation $a{x^2} + bx + c = 0$, $a \ne 0$ depends upon the value of $D = {b^2} - 4ac$, which is known as the discriminate of the quadratic equation.
For Example: Value of $D$ can be (i) $D > 0$ , (ii) $D = 0$ (iii) $D < 0$.

(10) The quadratic equation $a{x^2} + bx + c = 0$ , $a \ne 0$ has:
(i) Two distinct real roots, if D ba-4ac 0 two equal roots i.e. coincident real roots if $D = {b^2} - 4ac > 0$
For Example:  16x2 = 24x + 1
16x2 – 24x – 1 = 0
The given equation is of the form of ax2 + bx + c = 0, where a = 16, b = -24, c = -1
Therefore, the discriminant- D = b2 – 4ac
D= (-24)2 – 4 x 16 x (-1)
D= 576 + 64
D= 640
Since, D > 0
Therefore, the roots of the given equation are real and distinct.
The real roots α and β are given by, $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$
For, $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$ $D = {b^2} - 4ac > 0$
Hence $D = {b^2} - 4ac > 0$  and $D = {b^2} - 4ac > 0$are the two root of the given equation.

(ii) Two equal roots i.e. coincident real roots, if $D = {b^2} - 4ac = 0$.
For Example: 2x2 - 2√6x + 3 = 0
The given equation is of the form of ax2 + bx + c = 0, where a = 2, b = - 2√6, c = 3
Therefore, the discriminant- D = b2 – 4ac
= (- 2√6)2 – 4 x 2 x 3
= 24 - 24
= 0
Since, D = 0
Therefore, the roots of the given equation are real.
The real and equal roots are given by $D = {b^2} - 4ac = 0$   and $D = {b^2} - 4ac = 0$. $D = {b^2} - 4ac = 0$ $D = {b^2} - 4ac = 0$

(iii) No real roots, if $D = {b^2} - 4ac < 0$.
For Example: The given equation is
x2 + x + 2 = 0
The given equation is of the form of ax2 + bx + c = 0, where a = 1, b = 1, c = 2
Therefore, the discriminant
D = b2 – 4ac
D= (1)2 – 4 x 1 x 2
D= 1 - 8
D= -7
Since, D < 0
Therefore, the given equation has not real roots.

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