Question

Which of the following are quadratic equation?
A. x(x+1)+8=(x+2)(x-2)
B. (x+2)*3(cube)=x*3(cube)-4

Answer 1

Answer:

Option B

Step-by-step explanation:

(x+2)(cube) = x(cube)-4

x*3+2*3+3x × 2(x+2)= x*3-4

8+6x(x+2)=-4

8+6x*2+12x+4=0

6x*2+12x+12=0

Answer 2

(Note:- Quadratic eq. are those eq. which has highest power as 2.)

a)

$x(x + 1) + 8(x + 2) = (x + 2)(x - 2) \\ \\ solve \: it \\ \\ \cancel{{x}^{2}} + x + 8x + 16 = \cancel{{x}^{2}} - {2}^{2} \\ x + 8x + 16 + 4 = 0 \\ 9x + 20 = 0 1$

$so \: it \: is \: not \: a \: qadratic \: eq. \: bcz \: highest \: power \: is \: 1$

b)

$(x + 2 {)}^{3} = {x}^{3} - 4 \\ open \: identity \: for \: (x + 2 {)}^{3} \\ = > {x}^{3} + {2}^{3} + 3(x)(2)(x + 2) = {x}^{3} - 4\\ = > \cancel{ {x}^{3} } + 8 + 6 {x}^{2 } + 12x = \cancel{ {x}^{3} } - 4 \\ = > 6 {x}^{2} + 12x + 8 + 4 = 0 \\ 6 {x}^{2} + 12x + 12 = 0$

$\small{ in \: this \: heighest \: power \: is \: 2 \: so \: it \: is \: a \: quadratic \: eq.}$

$so \: answer \: is \: \boxed{(b)(x + 2 {)}^{3} = {x}^{3} - 4}$