How do you solve #x^2+x=20#?
2 Answers
Explanation:
Here are a few methods:
Method 1 - Fishing for factors
Note that
#x(x+1) = 20#
So we want to find a number
It should not take too much guessing to find:
#color(blue)(4)(color(blue)(4)+1) = (4)(5) = 20#
So
This is a quadratic equation, so should normally have two solutions, so what is the other?
With a bit of thought, we find:
#((color(blue)(-5))((color(blue)(-5))+1) = (-5)(-4) = 20#
So
Method 2 - Completing the square
Given:
#x^2+x=20#
Note that in general:
#(x+p)^2 = x^2+2px+p^2#
So to make
#x^2+x+1/4 = 81/4#
Both sides are now perfect squares:
#(x+1/2)^2 = (9/2)^2#
Hence:
#x+1/2 = +-9/2#
Subtracting
#x = -1/2+-9/2#
That is:
#x = 4" "# or#" "x = -5#
Method 3 - Quadratic formula
Given:
#x^2+x=20#
Subtract
#x^2+x-20 = 0#
This is in the standard form:
#ax^2+bx+c = 0#
with
It has solutions given by the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#color(white)(x) = (-1+-sqrt((-1)^2-4(1)(-20)))/(2*1)#
#color(white)(x) = (-1+-sqrt(1+80))/2#
#color(white)(x) = (-1+-9)/2#
That is:
#x = 4" "# or#" "x = -5#
Explanation:
Method 4. Solving by new Transforming Method (Google Search)
Find 2 real roots, that have opposite signs (ac < 0), knowing the sum (- b = - 1) and the product (c = - 20).
They are: