How do you solve #2x^2 + 10x = 0#?

2 Answers
Jan 24, 2017

Answer:

#x = 0#, or #x = -5#.

Explanation:

Typically you would use the quadratic formula, using #a,b,c# as the coefficients of #x^2#, #x#, and the constant respectively, to find that

#x = (-b +-sqrt(b^2 - 4ac))/(2a)#

However, notice that the constant is missing. In those cases, you can factor out a #x#, to transform the expression into a product that equals #0#:

#2x^2 + 10x = 0 => x(2x + 10) = 0#

When a product is equal to #0#, at least one of its factors must be #0#, or

#ab = 0 => a = 0# or #b= 0#.

So in our case, either

#x = 0#, or

#2x + 10 = 0 => 2x = -10 => x = -5#

Therefore, the roots to this equation, are #0# and #-5#.

Jan 24, 2017

Answer:

Factor #x# out of the expression and you get #x(2x+10)=0#
This can be true only if #x=0# or #x=-5#

Explanation:

Remember this technique for those equations that have no constant term (#c=0# in terms of the standard #ax^2+bx+c=0# form).

Since both the remaining terms have #x# in them, you factor this out, and immediately one root must be #x=0#.

Solve the monomial that remains in the product by setting it equal to zero. The idea is that the entire expression can only equal zero if one of the two factors #x# or #(2x+10)# is equal to zero.