Determining values of m in quadratic equations?

Question

How do you determine the value(s) of m in these equations, so that the two roots are equal?:
x^2 - 6m + m=0
And also so that 1 root is triple the other?:
3x^2 - 4x + n=0

Answer 1

is eq x^2-5m=0 or x^2-6mx+m=0
i am solving for x^2-5m=0
since roots r equal
then
D must be 0
a=1
b=0
c=-5m
d=b^2-4ac
d=-4*1*-5m
20m=0
m=0

2)a=3
b=-4
c=n
now x& y be roots
thenxy=c/a=n/3
x+y=-b/a=4/3
given x=3y
3y+y=4/3
4y=4/3
y=1/3
x=1
xy=n/3
therefore n=1
or
if we put y=3x
then n=1/3

Answer 2

How do you determine the value(s) of m in these equations, so that the two roots are equal?:
x^2 - 6x + m=0

If m = (-6/2)^2 = 9, then
x^2 - 6x + m = x^2 - 6x + 9 = (x-3)^2
Therefore, the two roots are equal.


And also so that 1 root is triple the other?:
3x^2 - 4x + n=0


Let 3x^2 - 4x + n = 3(x-x1)(x-3x1) = 3x^2 - 12x1x + 9x1^2
(reason: x2 = 3x1)

Compare coefficients,
x1 = 4/12 = 1/3
x2 = 3x1 = 1

Therefore, n = 9x1^2 = 1

Check:
3x^ - 4x + 1 = 0

3(x^2 - 4/3 x + 1/3) = 3(x - 1/3)(x - 1)

Answer 3

I think the first one might be x^2 - 6mx + m=0
you can apply the rules for the roots
the sum of the 2 roots equals -b/a where b would be (-6m) and a equals 1
and the product of the roots equals c/a where c is m and a is 1 again
so...
X1+X2 (the roots)= 6m
this means that 2*X1=6m
X1=6/m/2=3/m
and...
X1*X2=m
X1^2=m
(3/m)^2=m
9/m^2=m
9=m^3
the cubic root of nine that equals 2.008=9^(1/3) thats m

For the second one you can apply the same rules of the product and the addition of roots

Answer 4

x^2-8x+13 in case you're in calculus, locate the spinoff it fairly is 2x-8. x=4, and verify to work out that that could be a optimal.... Wait a 2d that could be a minimum. this means that the optimal fee may be infinity. even with the certainty that extremely, you will possibly desire to easily see that a > 0 subsequently the parabola opens up. Then the optimal fee may be infinity. For the 2d: x^3+7x^2+14x+8=0 That one is messy, yet use the rational roots attempt. The roots may be + or - a million,2,4,8/a million. So the roots might desire to be plus or minus a million,2,4,8. i might start up plugging in ranging from +a million. that would not artwork. -a million... -a million works. Then divide: x^3+7x^2+14x+8 / x-a million or use synthetic branch. you will locate that the different 2 roots are -2 and -4. Sorry, i can't fairly teach long branch or synthetic like this :/

Answer 5

for the roots to be equal b^2=4ac
so equating 36=4m
m=9

2.sum of the roots=4/3
product=n/3
n/3=1/3
n=1
check
x=[4+/-rt(16-12)]/6
=6/6 or 2/6
checks