# determine the values of n in each equations that 1 root is triple the other root 3x^2-4x+n=0

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3x^2 - 4x + n = 0

Let x1 and x2 be the roots;

==> x2= 3x1 .......(1)

Let us use viete's rule:

we know that:

x1+ x2= -b/a = 4/3

==> x1+ 3x1 = 4/3

==> 4x1 = 4/3

==> x1= 1/3

==> x2= 3*1/3 = 1

Also,

x1*x2= c/a = n/3

==> 1*1/3= n/3

**==> n= 1**

Since one root is triple the other, we assume x1 and 3x1 are the roots of the given equation 3x^2-4x+n = 0.

Then 3x^2 -4x+n = k(x-x1)(x-3x1)^2 should be an identity. So we choose k = 3 to make the coefficient of x^2 equal on both sides.

3x^2-4x +n = 3(x-x1)(x-3x1)

We expand the right side.

3x^2-4x+n = 3x^2- 3(x1+3x1)x +9x1^2

Equating the coefficients of x's on both sides, we get:

-4 = -(x1+3x1).

4x1 = 4

x1 = 1

Equating the constant terms, we get:

n = 9x1^2 .

n = 9*1^2.

n = 9