QUADRATIC EQUATIONS
. Stating whether the given equation is a
Quadratic Equation or not ?
(x + 1)2 = 2 (x – 3)
Q.) Check whether the following are quadratic equations :
i) (x + 1)2 = 2 (x – 3)
Sol :
x2
+ 2x
=
2x
– 6
∴
x2
+ 7
=
0
∴
The given equation is in the form of ax2 + bx + c = 0.
So it is a quadratic equation.
+ 1
(a + b)2 = a2 + 2ab + b2
Arrange equation such that we get RHS = 0
∴
Represent Middle term as 0x
Middle term is missing
x2
+ 0x
+ 7
= 0
Highest index of variable is 2
x2
=
0
+ 6
∴
+ 1
EX 4.1 1(I)
homework
Q.) Check whether the following are quadratic equations :
(ii) x² – 2x = (– 2) (3 – x)
Sol :
x² – 2x = (– 2) (3 – x)
∴
x² – 2x
=
– 6
+ 2x
∴
x²
– 4x
+ 6
=
0
The given equation is in the form of ax2 + bx + c = 0.
So it is a quadratic equation.
Highest index of variable is 2
Arrange equation such that we get RHS = 0
∴
x² – 2x
=
+ 6
0
- 2x
EX 4.1 1(II)
Q.) Check whether the following are quadratic equations :
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
Sol :
x
(x + 1)
∴
(x – 2)(x + 1) = (x – 1) (x + 3)
– 2
(x + 1)
=
x
(x + 3)
– 1
(x + 3)
x2
∴
+ x
– 2x
– 2
=
x2
+ 3x
=
2x
– 3
∴
– x
– 2
∴
– 3x
+ 1
=
0
∴
3x
– 1
=
0
The given equation is not in the form of ax2 + bx + c = 0.
So it is not a quadratic equation.
– x
– 3
Arrange Equation
such that we get
RHS as 0
=
- 2
+ 3
∴
– x
– 2x
0
dividing
throughout by -1
EX 4.1 1(III)