Check whether the following are quadratic equations : (i) `(x-2)^2+1=2x-3` (ii) `x(x+1)+8=(x+2)(x 2)`(iii) `x(2x+3)=x^2+1` (iv) `(x+2)^3=x^3-4`
(i) LHS = (x – 2)2 + 1 = x2 – 4x + 4 + 1 = x2 – 4x + 5
Therefore, (x – 2)2 + 1 = 2x – 3 can be rewritten as
x2 – 4x + 5 = 2x – 3
i.e., x2 – 6x + 8 = 0
It is of the form ax2 + bx + c = 0.
Therefore, the given equation is a quadratic equation.
(ii) Since x(x + 1) + 8 = x2 + x + 8 and (x + 2)(x – 2) = x2 – 4
Therefore, x2 + x + 8 = x2 – 4
i.e., x + 12 = 0
It is not of the form ax2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(iii) Here, LHS = x (2x + 3) = 2x2 + 3x
So, x (2x + 3) = x2 + 1 can be rewritten as
2x2 + 3x = x2 + 1
Therefore, we get x2 + 3x – 1 = 0
It is of the form ax2 + bx + c = 0.
So, the given equation is a quadratic equation.
(iv) Here, LHS = (x + 2)3 = x3 + 6x2 + 12x + 8
Therefore, (x + 2)3 = x3 – 4 can be rewritten as
x3 + 6x2 + 12x + 8 = x3 – 4
i.e., 6x2 + 12x + 12 = 0 or, x2 + 2x + 2 = 0
It is of the form ax2 + bx + c = 0.
So, the given equation is a quadratic equation.
(i) `(x-2)^2+1 = 2x-3`
`=>x^2+4-4x+1 = 2x-3`
`=>x^2-6x+8 = 0`
As coefficient of `x^2` is not `0`, so it is a quadratic equation.
(ii)`x(x+1)+8 = (x+2)(x-2)`
`=>x^2+x+8 = x^2-4`
`=>x+12 = 0`
As coefficient of `x^2` is `0`, so it is not a quadratic equation.
(iii) `x(2x+3) = x^2+1`
`=>2x^2+3x = x^2+1`
`=>x^2+3x-1 = 0`
As coefficient of `x^2` is not `0`, so it is a quadratic equation.
(iv) `(x+2)^3 = x^3-4`
`=>x^3+2^3+3(2)(x)(x+2) = x^3-4`
`=>x^3+8+6x^2+12x = x^3-4`
`=>6x^2+12x+12 = 0`
`=>x^2+2x+2 = 0`
As coefficient of `x^2` is not `0`, so it is a quadratic equation.