How do you solve #24+x^2=10x#?

3 Answers
Mar 25, 2018

You have to pass #10x# to the left hand and equal the quadratic equation to 0
24 + #x^2##-10x#=0
then you rearrenge it
#x^2##-10x#+24=0

Then you have to think about two numbers that when you times them you get as the answer 24
and when you add them -10

The numbers are -6 and -4
(-6)x(-4)=24
(-6) +(-4)=-10
The final working is :
#x^2##-10x#+24=#(x-6)(x-4)#

So the answers are:
#x-6=0#
#x=6#

#x-4=0#
#x=4#

Mar 25, 2018

Answer:

#x=6# or #x=4#

Explanation:

#24+x^2=10x#

Put into standard form, #color(violet)(ax^2+bx+c=0)#

#x^2-10x+24=0#

#darr#Factor using criss-cross method of factoring

#1color(white)(XX)#-6

#1color(white)(XX)#-4

#-4-6#

#=-10# #lArr# same number as our b-value in our rearranged equation.

#:.# #24+x^2=10x# is #color(orange)"(x-6)(x-4)"#

Further on, finding the x-intercepts of #(x-6)(x-4)=0#

#x-6=0# #color(white)(XXXXXX)# and #color(white)(XXXXXX)##x-4=0#

#x=6##color(white)(XXXXXXXXXXXXXXXXX)##x=4#

#:.# the zeros are #color(blue)6# and #color(blue)4#.

Mar 25, 2018

Answer:

#x=6 or x=4#

Explanation:

Here,

#24+x^2=10x#

#=>x^2-10x+24=0#

Now,

#(-6)(-4)=24 and (-6)+(-4)=-10#

So,

#x^2-6x-4x+24=0#

#=>x(x-6)-4(x-6)=0#

#=>(x-6)(x-4)=0#

#=>x-6=0 or x-4=0#

#=>x=6 or x=4#