Find m such that x^2+(mx+3)^2-3=0 has equal roots.

Here we're presented with a quadratic equation which needs to be expanded and then rewritten in descending powers of x:

1x^2 + m^2x^2 + 6mx + 9 - 3 = 0.

Let's group like terms:  1x^2 + m^2x^2 + 6mx + 6 = 0.

The first 2 terms can be rewritten as a single term:  (1+m^2)x^2, and so we now have:

(1+m^2)x^2 + 6mx + 6 = 0.

We must now calculate the discriminant and set the resulting expression = to 0, as a preliminary to finding the value of m for which the given quadratic has equal roots:

discriminant:  (6m)^2 - 4(1+m^2)(6) = 0

Then 36m^2 - 24(1+m^2) = 0, which simplifies to 12m^2 - 24 = 0.

Then 12 m^2 = 24; m^2 = 2, and m = √2.

When the discriminant is zero, as it is here when m = √2, then the given quadratic has two equal roots.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: