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October 1, 2014

October 1, 2014

Posted by **Jayden Haddy** on Saturday, August 17, 2013 at 11:08pm.

m^2 x^2 + 2(2m-5)x + 8 = 0

a) If 2 is a root of the equation, find the possible values of m (there are four possible values)

b) For each value of m, find the other root

I have actually calculated the answer m=1 or m=-3 for (a), but I was told that there are four possible values of m. I was stuck on this part.

- maths -
**Reiny**, Sunday, August 18, 2013 at 12:21amsince 2 is a root,

4m^2 + 2(2m-5)(2) + 8 = 0

4m^2 + 8m - 20 + 8 = 0

4m^2 + 8m - 12 =

m^2 + 2m - 3 = 0

(m+3)(m-1) = 0

m = -3, or m = 1

if m = -3, we have

9x^2 -22x + 8 = 0

(x-2)(9x - 4) = 0

x = 2 (our given) or x = 4/9

if m = 1 , we have

x^2 -6x + 8 = 0

(x-2)(x-4) = 0

x=2 or x=4

Like you, I only get 2 values for m

and each value of m yields 2 answers. (one doubling up)

I also approached the problem using the sum and product of roots property.

let the roots be a and b, (we have a quadratic in x)

but we know one of them let b = 2

sum of roots = a+2

product of roots = 2a

sum of roots = -2(2m-5)/m^2

a+2 = (-4m + 10)/m^2 - 2

a = (-4m +10 - 2m^2)/m^2

product of roots = 8/m^2

2a = 8/m^2

a = 4/m^2

4/m^2 = (-4m + 10 - 2m^2)/m^2

4 = -4m + 10 - 2m^2

2m^2 + 4m - 6 = 0

m^2 + 2m - 3 = 0

(m+3)(m-1) = 0

m = -3 or m = 1

same as above, and only 2 values of m

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