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