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