Use the quadratic formal to solve the equation

Question

Use the quadratic formal to solve the equation (show all work)

 

(m-4)(3m-10)=3(m-2)+18

 

Answer 1

(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."

Answer 2

(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

Answer 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