Solve , precalculus help answer asap please

Question

solve the following:
a) 3(2z+3)2-9(2z+3)-30 = 0
b)0.49k2-36m2 = 0

Answer 1

3(2z+3)² - 9(2z+3) - 30 = 0

Let x = 2z + 3 and substitute that in to make a quadratic, then factor:

3x² - 9x - 30 = 0
3(x² - 3x - 10) = 0
3(x - 5)(x + 2) = 0

Now, substitute 2z + 3 = x back in and simplify:

3(2z + 3 - 5)(2z + 3 + 2) = 0
3(2z - 2)(2z + 5) = 0

So, 2z - 2 = 0 OR 2z + 5 = 0.

 

That means:

 

2z - 2 = 0

2z = 2

z = 1

 

OR

 

2z + 5 = 0

2z = -5

z = -5/2

 

z = 1 OR -5/2

0.49k² - 36m² = 0

This expression is a difference of squares, which can be factored using this formula:

x² - y² = (x + y)(x - y)

√(0.49k²) = 0.7k and √(36m²) = 6m, so applying the formula to the original expression gives you:

(0.7k + 6m)(0.7k - 6m) = 0

That means either 0.7k + 6m = 0 OR 0.7k - 6m = 0. You need to know whether you're solving for m or k.

 

If you're solving for m...

 

0.7k + 6m = 0

6m = -0.7k

m = -7/60k

 

OR

 

0.7k - 6m = 0

-6m = -0.7k

m = 7/60k

 

That means m = -7/60k OR 7/60k.

 

If you're solving for k...

 

0.7k + 6m = 0

0.7k = -6m

k = -60/7m

 

OR

 

0.7k - 6m = 0

0.7k = 6m

k = 60/7m

 

That means k = -60/7m OR 60/7m.