How do you solve #2x^2+9x+10=0#?

2 Answers
Dec 22, 2016

Answer:

#-5/2 and -2#

Explanation:

Use improved quadratic formula (Socratic Search)
#y = 2x^2 + 9x + 10 = 0#
Both roots are negative. (rule of signs)
#D = d^2 = b^2 - 4ac = 81 - 80 = 1 # --> #d = +- 1#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = - 9/4 +- 1/4 = (-9 +- 1)/4#
#x1 = -10/4 = -5/2# and #x2 = -8/4 = -2#

Dec 22, 2016

Answer:

#x = -2" "# or #" "x = -5/2#

Explanation:

Given:

#2x^2+9x+10 = 0#

We can use an AC method.

Find a pair of factors of #AC=2*10=20# with sum #B=9#

The pair #5, 4# works.

Use this pair to split the middle term and factor by grouping:

#0 = 2x^2+9x+10#

#color(white)(0) = 2x^2+5x+4x+10#

#color(white)(0) = (2x^2+5x)+(4x+10)#

#color(white)(0) = x(2x+5)+2(2x+5)#

#color(white)(0) = (x+2)(2x+5)#

Hence solutions:

#x = -2" "# or #" "x = -5/2#