# Quadratic Equations - Class 10 : Notes

Quadratic Equations - Class 10 : Notes

(1) A polynomial of degree 2 is called a quadratic polynomial. The general form of a quadratic polynomial is ax2+bx+c, where a, b, c are real number such that a ≠0 and x is a real variable.
For Example: x2+5x+3, where a=1,b=5,c=3 are real number. So given equation is quadratic polynomial.

(2) If p(x)=ax2+bx+c,a≠0 is a quadratic polynomial and α is a real number, then p(α)=aα2+bα+c is known as the value of the quadratic polynomial p(α)
For Example: p(α)=α2+5α+3 in that equation if α=3 then p(α)=27. So 27 is a value of quadratic polynomial

(3) A real number α is said to be a zero of quadratic polynomial p(x)=ax2+bx+c, if p(α)=0.
For Example: p(x)=x2+6x+5
If x=(−5) then p(x)=0, So −5 is the zero of polynomial.

(4) If p(x)=ax2+bx+c is a quadratic polynomial, then p(x)=0 i.e., ax2+bx+c=0, a≠0 is called a quadratic equation.
For Example: p(x)=x2−8x+16 is a quadratic polynomial, then p(x)=0 i.e., x2−8x+16=0, a≠0 is called a quadratic equation.

(5) A real number α is said to be a root of the quadratic equation ax2+bx+c=0.
In other words,   α is a root of ax2+bx+c=0 if and only if α is a zero of the polynomial p(x)=ax2+bx+c.
For Example: Suppose quadratic equation is 2x2−x−6=0.
If we put x=2 then p(x)=0, So 2 is a root of that given equation so here α=2.

(6) If ax2+bx+c=0, a≠0 is factorizable into a product of two linear factors,  then the roots of the quadratic equation ax2+bx+c=0 can be found by equating each factor to zero.
For Example: The Given equation is $9x^{2}-3x-2=0$
Now, Solving the above equation using factorization method.
$\Rightarrow&space;9x^{2}-6x+3x-2$
$\Rightarrow3x(3x-2)+1(3x-2)$
$\Rightarrow$(3x + 1) (3x - 2) =0
$\Rightarrow$ (3x + 1) = 0 or (3x - 2) = 0
$\Rightarrow$3x = -1 or 3x = 2
$\Rightarrow$$x&space;=&space;-\frac{1}{3}$   or $x&space;=&space;\frac{2}{3}$
Hence,$x&space;=&space;-\frac{1}{3}$  and  $x&space;=&space;\frac{2}{3}$ 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 -
$2x^{2}-7x+3=0$
Dividing through out by 2
$\Rightarrow&space;x^{2}-\frac{7}{2}x+\frac{3}{2}=0$
Shifting the constant term to the right hand side.
$\Rightarrow x^{2}-\frac{7}{2}x=-\frac{3}{2}$
Adding square of the half of coefficient of x on the both side.
$\Rightarrow x^{2}-\frac{7}{2}x + (\frac{7}{4})^{2}=-\frac{3}{2}+ (\frac{7}{4})^{2}$
$\Rightarrow x^{2}-2 \frac{7}{4}x + (\frac{7}{4})^{2}=-\frac{3}{2}+ (\frac{49}{16})$
$\Rightarrow&space;(x-\frac{7}{4})^{2}=\frac{-24+49}{16}$
$\Rightarrow (x-\frac{7}{4})^{2}=\frac{25}{16}$
Taking square root of both sides-
$\Rightarrow (x-\frac{7}{4})=\sqrt{}\frac{25}{16}$
$\Rightarrow (x-\frac{7}{4})=\pm \frac{5}{4}$
$\Rightarrow x= \frac{5}{4}+\frac{7}{4} or x= -\frac{5}{4}+\frac{7}{4}$
$\Rightarrow x= \frac{12}{4} or x= \frac{2}{4}$
$\Rightarrow x= 3 or x= \frac{1}{2}$
Hence x= 3, and x=1/2 are the two root of the given equation

(8) The roots of the quadratic equation ax2+bx+c=0, a≠0 can be found by using the quadratic formula −b±b2−4ac√2a , provided that b2−4ac−−−−−−−√≥0.
For Example: $\sqrt{3}x^{2}+10x-8\sqrt{3}=0$ the given equation in the form of $ax^{2}+bx-c=0$,
Where a= √3, b=10 c= 8√3
Therefore, the discriminant- $D=b^{2}-4ac$
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,
$\alpha&space;=\frac{-b+&space;\sqrt{D}}{2a}$
$\alpha =\frac{-10+ \sqrt{196}}{2\sqrt{3}}$$\alpha =\frac{-10+ 14}{2\sqrt{3}}$
$\alpha&space;=\frac{4}{2\sqrt{3}}&space;=&space;\frac{2}{\sqrt{3}}$
For, $\beta =\frac{-b-\sqrt{D}}{2a}=\frac{-10-\sqrt{196}}{2\sqrt{3}}$
$\beta =\frac{-10-14}{2\sqrt{3}}=\frac{-24}{2\sqrt{3}}$
$\beta =\frac{-12}{\sqrt{3}}=\frac{-4x\sqrt{3}x\sqrt{3}}{\sqrt{3}}=-4\sqrt{3}$
Hence $\alpha&space;=&space;\frac{2}{\sqrt{3}}$ and $\beta&space;=-4\sqrt{3}$ are the two root of the given equation.

(9) Nature of the roots of quadratic equation ax2+bx+c=0, a≠0 depends upon the value of D=b2−4ac, 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 ax2+bx+c=0 , a≠0 has:
(i) Two distinct real roots, if D ba-4ac 0 two equal roots i.e. coincident real roots if D=b2−4ac>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,
$\alpha&space;=\frac{-b+&space;\sqrt{D}}{2a}$
$\alpha =\frac{-(-24)+ \sqrt{640}}{2x16}$
$\alpha =\frac{24+ \sqrt{64x10}}{32}$

$\alpha =\frac{24+ 8\sqrt{10}}{32}$
$\alpha =8(\frac{3+ \sqrt{10}}{32})$
$\alpha =(\frac{3+ \sqrt{10}}{4})$
For, $\beta =\frac{-b- \sqrt{D}}{2a}$
$\beta&space;=\frac{-(-24)-&space;\sqrt{640}}{2x16}$
$\beta&space;=\frac{24-&space;\sqrt{64x10}}{32}$
$\beta&space;=\frac{24-&space;8\sqrt{10}}{32}$
$\beta&space;=8(\frac{3-&space;\sqrt{10}}{32})$
$\beta&space;=(\frac{3-&space;\sqrt{10}}{4})$
Hence$\alpha =(\frac{3+ \sqrt{10}}{4})$  and  $\beta&space;=(\frac{3-&space;\sqrt{10}}{4})$are the two root of the given equation.

(ii) Two equal roots i.e. coincident real roots, if D=b2−4ac=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 $\frac{-b}{2a}$   and  $\frac{-b}{2a}$.
$\frac{-b}{2a}=&space;\frac{-(-2\sqrt{6})}{2x2}=&space;\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}$
$\frac{\sqrt{3}\sqrt{3}}{\sqrt{2}\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$

(iii) No real roots, if D=b2−4ac<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.