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Use the quadratic formal to solve the equation

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3 Answers

(m-4)(3m-10)=3(m-2)+18
 
3m2 -10m -12m +40 = 3m - 6 + 18
 
3m2 -22m + 40 = 3m + 12
 
3m2 - 25m + 28 = 0
 
which is in the form ax2 + bx +c = 0
 
In this case a = 3; b = -25 and c = 28
 
m = [-b ± (b2 - 4ac)½]/2a 
 
m = {25 ± [(-25)2 - 4(3)(28)]½}/(2)(3)
 
m = {25  ± [625 - 336]½}/6
 
m = [25 ± 17]/6
 
m = [25 - 17]/6 = 8/6 = 1.333 and m = [25 + 17]/6 = 7
 
CHECK: Substituting m = 7 back into the original equations:
 
Does (7 - 4)[(3(7) - 10] = 3(7 - 2) + 18?
 
(3)(11) = 3(5) + 18?
 
YES!
 
Why don't you try substituting m = 1.3333 back into the original equations to see if it also "works."
(m-4)(3m-10)=3(m-2)+18  expand to get a quadratic equation as
3m²-22m+40=3m-6+18
3m²-25m+28=0 which, by quadratic formula, gives
m=(-(-25)±√(625-4(3)(28)))/6
m=(25±√(625-336))/6=(25±√289)/6=(25±17)/6=7 and 8/6 or 7, and 4/3
Hi Again Theresa;
(m-4)(3m-10)=3(m-2)+18
Let's convert this into a quadratic equation...
Let's foil the equation on the left side...
(m-4)(3m-10)
FIRST...3m2
OUTER...-10m
INNER...-12m
LAST...40
3m2-22m+40=3(m-2)+18
3m2-22m+40=3m-6+18=3m+12
Let's subtract 12 from both sides...
-12+3m2-22m+40=3m+12-12
3m2-22m+28=3m
Let's subtract 3m from both sides...
-3m+3m2-22m+28=3m-3m
3m2-25m+28=0
Let's factor...
We know that...
FIRST...must be (3m)(m)
INNER and OUTER....-3(?)m+-(??)m
LAST....(-7)(-4) or (-14)(-2)
(3m-4)(m-7)=0
Either
3m-4=0
or m-7=0
m=4/3
or
m=7
 
Let's check our work...
The original equation was...
(m-4)(3m-10)=3(m-2)+18
m=4/3
[(4/3)-4][3(4/3)-10]=3[(4/3)-2]+18
[-8/3][4-10]=3(-2/3)+18
[-8/3][-6]=-2+18
16=16
 
(m-4)(3m-10)=3(m-2)+18
m=7
(7-4)[3(7)-10]=3(7-2)+18
3(21-10)=3(5)+18
33=33