How do you solve #6x^2-21x+15=0#?

2 Answers
Mar 26, 2018

Answer:

x= #5/2# or #1#

Explanation:

Start by simplifying your equation by factoring out a 3:
#3(2x^2-7x+5)=0#
#2x^2-7x+5=0#

This equation cannot be factored with whole numbers, so you should use the quadratic formula:

#(-b+-sqrt(b^2-4ac))/(2a)#, knowing that #ax^2+bx+c#

So now:
#(-(-7)+-sqrt((-7)^2-4(2)(5)))/(2(2))#
#(7+-sqrt(49-4(2)(5)))/(4)#
#(7+-sqrt(49-40))/(4)#
#(7+-sqrt(9))/(4)#
#(7+-3)/(4)#
#10/4# or #4/4#=
#5/2# or #1#

x= #5/2# or #1#

Mar 26, 2018

Answer:

#x=21/12+-sqrt(54/96)#

Explanation:

In order to complete the square move the last term (term without #x#) to other side of equation
#x^2-21/6x=-15/6#

Then you want to find a piece that allows you to find a square square of the left hand side
i.e. #a^2+2ab+b^2=(a+b)^2#
or
#a^2-2ab+b^2=(a-b)^2#

In this equation #x=a#, #2ab=-21/6x# so as #x=a# we know that #2b=-21/6# so to complete the square we just need #b^2# so if we half and square #2b# we will get it so #b^2=(21/12)^2#

So if we add this term to both sides we get

#x^2-21/6x+(21/12)^2=-15/6+(21/12)^2#
Now the left hand side can be simplifed into merely #(a-b)^2#

#(x-21/12)^2=-15/6+441/144#

#(x-21/12)^2=-15/6+49/16#

Find a common multiple for 16 and 6 and add them together

#(x-21/12)^2=-240/96+294/96#

#(x-21/12)^2=54/96#

Square root both sides

#x-21/12=+-sqrt(54/96)#

#x=21/12+-sqrt(54/96)#