Question

The square of a number exceeds that number by 12. What are the two possible solutions?

Answer 1

Answer:

The two possible solutions are -3 or 4.

Step-by-step explanation:

$x^{2}=x+12\\x^{2} -x-12=0\\x^{2}-+3x-4x-12=0\\x(x+3)-4(x+3)=0\\(x-4)(x+3)=0\\x=-4/x=3$

Answer 2

Answer:

The correct answer is option C.

Step-by-step explanation:

Let the number be x.

The square of a number exceeds that number by 12 .Mathematically can be written as:

$x^2-x=12$

$x^2-x-12=0$

Using middle term splitting to factorize the expression:

$x^2-4x+3x-12=0$

$x(x-4)+(3)(x-4)=0$

$(x+3)(x-4)=0$

The two possible solutions are :

1) (x+3)=0

x = -3

2) (x-4)=0

x = 4