I'm doing quadratic equations x^2 +bx+c =0. I know how to questions like x^2 +17x +42 =0 but this one says, j^2-9jk-10k^2. I have no clue on what I need to do

Just include "k" in the constants:

(j-10k)(j+k)

I'm doing quadratic equations x^2 +bx+c =0.I'm supposed to factor each polynomial. I know how to questions like x^2 +8x +12 =0 but this one says, j^2-9jk-10k^2. I'm confused

To solve quadratic equations like j^2-9jk-10k^2=0, you need to use the quadratic formula or factorization method.

First, let's identify the coefficients of the equation:
a = coefficient of j^2 = 1
b = coefficient of jk = -9
c = coefficient of k^2 = -10

Now, let's discuss two methods to solve this quadratic equation:

1. Quadratic Formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a

Now, substituting the values from our equation, we get:
j = [9k ± √((-9k)^2 - 4(1)(-10k^2))] / (2 * 1)

Simplifying this further, we have:
j = [9k ± √(81k^2 + 40k^2)] / 2
j = [9k ± √(121k^2)] / 2
j = [9k ± 11k] / 2

This gives us two possible solutions:
j1 = (9k + 11k) / 2 = 20k / 2 = 10k
j2 = (9k - 11k) / 2 = -2k / 2 = -k

So, the solutions to the given quadratic equation are:
j = 10k or j = -k

2. Factorization Method:
To solve the equation by factorization, we need to break down the equation into two binomial factors. We can write the equation as:

(j + 2k)(j - 5k) = 0

For the product of two factors to be zero, either (j + 2k) = 0 or (j - 5k) = 0.

So, the solutions are:
j + 2k = 0 => j = -2k
j - 5k = 0 => j = 5k

Therefore, the solutions to the given quadratic equation are:
j = -2k or j = 5k

These are the methods you can use to solve quadratic equations like j^2-9jk-10k^2=0. Try using both methods to verify your answer.