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
Now, Solving the above equation using factorization method.

(3x + 1) (3x - 2) =0
(3x + 1) = 0 or (3x - 2) = 0
3x = -1 or 3x = 2
or
Hence,  and   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 -

Dividing through out by 2

Shifting the constant term to the right hand side.

Adding square of the half of coefficient of x on the both side.

Taking square root of both sides-

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:  the given equation in the form of ,
Where a= √3, b=10 c= 8√3
Therefore, the discriminant-
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,

For,

Hence  and  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,

For,

Hence  and  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    and  .

(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.

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