Question

d. \\( z = -\frac{3}{5}, z = -\frac{2}{3} \\)\

1. solve \\( 10x^2 + 30x = -2x^2 - 30x - 75 \\) by factoring.\

\\( \text{a. } x = -\frac{5}{2}, x = 2 \\)\
\\( \text{b. } x = -\frac{5}{2} \\)\
\\( \text{c. } x = -\frac{5}{2}, x = \frac{5}{2} \\)\
\\( \text{d. } x = -\frac{2}{5} \\)\
short answer\
17 if a parabola with equation \\( y = ax^2 + bx + c \\) opens downward

Explanation:

Step1: Move all terms to left side

First, we need to get all terms on one side of the equation. Add \(2x^{2}\), \(30x\), and \(75\) to both sides of the equation \(10x^{2}+30x=-2x^{2}-30x - 75\).

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Step2: Simplify the equation

We can factor out a common factor of \(3\) from each term:
\(3(4x^{2}+20x + 25)=0\)

Step3: Factor the quadratic

Now, factor the quadratic \(4x^{2}+20x + 25\). Notice that it is a perfect square trinomial, since \((2x)^{2}=4x^{2}\), \(5^{2} = 25\), and \(2\times(2x)\times5=20x\). So, \(4x^{2}+20x + 25=(2x + 5)^{2}\).
Our equation becomes \(3(2x + 5)^{2}=0\)

Step4: Solve for \(x\)

Divide both sides by \(3\): \((2x + 5)^{2}=0\)
Take the square root of both sides: \(2x+5 = 0\)
Solve for \(x\): \(2x=-5\), so \(x=-\frac{5}{2}\)

Answer:

b. \(x = -\frac{5}{2}\)