Respuesta :

Answer:

I know you could easily solve this just looking at it.

But if you want the algebraic solution:

x + x^2 = 30

x^2 + x -30 = 0

a = 1  

b = 1

c = -30

Using the quadratic formula:

x = [ -b +- sqr root (b^2 - 4ac) ] / 2a

x = [-1 +- sqr root (1 - 4 * 1 -30) ] / 2*1

x = [-1 +- sqr root (1 + 120) ] / 2

x = -1 +- sqr root (121) / 2

x1 = (-1 + 11) / 2 = 10 / 2 = 5

x2 = (-1 -11) / 2 = -12 / 2 = -6

Answers are 5 and -6

5 + 5^2 = 30

-6 + (-6)^2  = 30

-6 +36  = 30

Step-by-step explanation:

If I add the square of a number  to the number itself, I get 30, then the number can be 5 or -6

Solution:

Given that if i add the square of a number  to the number itself, I get 30

We have to find the number

Let the number be "x"

Square of number + number = 30

Hence we get,

[tex]x^2 + x = 30\\\\x^2 + x - 30 = 0[/tex]

Let us factorize the expression to get the value of "x"

[tex]x^2 + x - 30 = 0[/tex]

"x" can be written as "-5x + 6x"

[tex]x^2 -5x + 6x - 30 = 0[/tex]

Now "30" can be written as [tex]6 \times 5[/tex]

[tex]x^2 - 5x + 6x - (6 \times 5) = 0[/tex]

Take "x" as common from first two terms and "6" as common from next two terms

[tex]x(x - 5) + 6(x - 5) = 0\\\\(x - 5)(x + 6) = 0[/tex]

Equating to zero we get,

x - 5 = 0 or x + 6 = 0

x = 5 or x = -6

Hence the number can be 5 or -6

ACCESS MORE