How do you solve #5x^2+19x=4#?

1 Answer
Aug 31, 2016

Answer:

#x = 1/5 or x = -4#

Explanation:

As it is a quadratic equation # rArr # make = 0

#5x^2 + 19x -4 = 0#

Find factors of 5 and 4 which subtract to make 19.
Signs will be different - more positive.
Cross multiply and subtract

#color(white)(xxx)(5)" "(4)#

#color(white)(xxxx)color(magenta)(5)" "color(lime)(1 rarr 1xx1 =1)#
#color(white)(xxxx)color(lime)(1)" "color(magenta)(4 rarr 5xx4=20)" "20-1 = 19#

We have the right factors - now fill in the signs.

#color(white)(xxxx)color(red)(5" "-1) rarr 1xx-1 =-1#
#color(white)(xxxx)color(blue)(1" "+4) rarr 5xx+4=+20" "+20-1 = +19#

#color(red)((5x-1))color(blue)((x+4))=0#

Either factor could be equal to 0.

If #color(red)(5x-1 = 0, rarr 5x =1 rarr x = 1/5#

If #color(blue)( x+4=0 rarr x = -4#