Question

(1) x2 + 3x + 1 = (x - 2)2
(2) (x + 2)3 = 2x (x2 - 1)
(3) x3 - 4x2 - x + 1 = (x - 2)3​

Answer 1

LHS ⇒x2+3x+1    ...(1)

RHS ⇒(x−2)2

⇒x2−4x+4    ...(2)

∴x2+3x+1=x2−4x+4

⇒3x+1=−4x+4

⇒7x−3=0

Therefore, it is not a quadratic equation.

We know that the general form of quadratic equation is ax2+bx+c=0.

The given equation is (x+2)3=2x(x2−1) can be simplified as follows: 

(x+2)3=2x(x2−1)⇒x3+23+(3×x×2)(x+2)=2x3−2x(∵(a+b)3=a3+b3+3ab(a+b))⇒x3+8+6x(x+2)=2x3−2x⇒x3+8+6x2+12x=2x3−2x⇒x3+8+6x2+12x−2x3+2x=0⇒−x3+6x2+14x+8=0

Since the variable x in the equation −x3+6x2+14x+8=0 has degree 

Answer 2

Step-by-step explanation:

LHS =x2+3x+1 ....(1)

RHS = (x-2)2

=x2-4x+4 ....(2)

x2+3x+1=x2-4x+4

=3x+1=-4x+4

=7x-3=0

Therefore, it is quadratic equation.

we know that

The general form of quadratic equation is ax2+bx+c

The given equation is (x+2)3=2x(x2-1) can be simplified as follow :

(x+2)3=2x(x2-1)

=x3+23+(3x*x2) (x+2)=2x3-2x.

=x3+8+6x(x+2)=2x3-2x

=x3+8+6x2+12x=12x3-2x

=x3+8+6x2+12x-2x3+2x=0

=-x3+6x2+14+8=0

since, the variable x in the equation =-x3+6x2+14+8=0 has degree