Normally, when you "solve" an expression like this, it is because you have an equation (find the value of x that makes the expression equal to a given value).
Here, it appears that you are asked to find the "roots" of the expression (the values of x that make the expression equal to 0).
This is the same as if you were asked to solve
x² - 20x - 200 = 0
One sure way to find the roots is to use the "quadratic formula" which is nothing more than a recipe that we follow to find the roots.
For the recipe to work, the quadratic (an expression with a square - quadratic is from a Latin word for "square") must be written in this format:
ax² + bx + c
Yours already is, with a = 1, b = -20 and c = -200
The recipe to find the roots is:
x = [ -b +/- √( b^2 - 4ac ) ] / 2a
Just fill in the values:
x = [ +20 +/- √( 400 - 4(1)(-200) ) ] / 2
x = [ +20 +/- √(1200)] / 2
1200 = 3 * 400
√(1200) = √(400) * √(3) = 20√3
x = (20 +/- 20√(3))/2
distribute the division
x = 10 +/- 10√3
One root is found using the +
x = 10 + 10√3
The other root is found by using the -
x = 10 - 10√3
----
Of course, any of the other methods to solve quadratic equations should work, as long as you are looking for roots (the same as making the expression equal to zero)
For example, "completing the square"
a perfect square is the result of multiplying a "linear" factor by itself
For example: (x - k)^2
= x² - 2kx + k²
You are given x² - 20x - 200
If you make -20 x = - 2 kx
then this forces you to use k = 10
which forces you to make k² = 100
but you have -200; you get out of the problem by adding 300 to both sides
x² - 20x - 200 = 0
x² - 20 x + 100 = 300 (we added 300 to both sides)
The left side is now a perfect square (we know that because we have made it that way)
Answers & Comments
Verified answer
There are 2 methods you can use to find the answer...
1 is completing the square
x²-20x-200 = 0
x²-20x = 200
(x-10)² = 200 +10²
(x-10)² = 300
(x-10) = ±10√3
x = 10±10√3
2) the other method is the equation. Its easy to use.
x = (-b ± √(b²-4ac)) / 2a
a, b, c are the coefficients of x², x and the constant respectively
x = (20 ± √(20²+4x200)) / 2
x = 10 ± (√1200)/2
x = 10 ± 10√3
Normally, when you "solve" an expression like this, it is because you have an equation (find the value of x that makes the expression equal to a given value).
Here, it appears that you are asked to find the "roots" of the expression (the values of x that make the expression equal to 0).
This is the same as if you were asked to solve
x² - 20x - 200 = 0
One sure way to find the roots is to use the "quadratic formula" which is nothing more than a recipe that we follow to find the roots.
For the recipe to work, the quadratic (an expression with a square - quadratic is from a Latin word for "square") must be written in this format:
ax² + bx + c
Yours already is, with a = 1, b = -20 and c = -200
The recipe to find the roots is:
x = [ -b +/- √( b^2 - 4ac ) ] / 2a
Just fill in the values:
x = [ +20 +/- √( 400 - 4(1)(-200) ) ] / 2
x = [ +20 +/- √(1200)] / 2
1200 = 3 * 400
√(1200) = √(400) * √(3) = 20√3
x = (20 +/- 20√(3))/2
distribute the division
x = 10 +/- 10√3
One root is found using the +
x = 10 + 10√3
The other root is found by using the -
x = 10 - 10√3
----
Of course, any of the other methods to solve quadratic equations should work, as long as you are looking for roots (the same as making the expression equal to zero)
For example, "completing the square"
a perfect square is the result of multiplying a "linear" factor by itself
For example: (x - k)^2
= x² - 2kx + k²
You are given x² - 20x - 200
If you make -20 x = - 2 kx
then this forces you to use k = 10
which forces you to make k² = 100
but you have -200; you get out of the problem by adding 300 to both sides
x² - 20x - 200 = 0
x² - 20 x + 100 = 300 (we added 300 to both sides)
The left side is now a perfect square (we know that because we have made it that way)
(x - 10)² = 300
square root both sides:
x - 10 = ±√300
x = 10 ±√300
We can reduce √300 to √(100)*√(3) = 10√3
Leaving us with
x = 10 ± 10√3
20x 200
x²-20x-200
for quadratic formula ===> A=1,B= --20, C= --200
Hence discriminant =D =[--20]^2 - 4 [ 1][--200]
=400+800=1200
Hence √D=± √1200 = ± 20 √3
x= [ 20+ 20√3]/2 & x= [20--20√3]/2
0r x= [10+10√3] & x= [ 10--10√3]
delta' = 10² + 4.200 = 900
=> sqrt ( delta' ) = 10√3
=> x = 10±10√3