Question

Solve the following quadratic equation for all values of xx in simplest form.
4, left parenthesis, 5, x, minus, 10, right parenthesis, squared, minus, 34, equals, minus, 33
4(5x−10)
2
−34=
−33

Answer 1

To solve the equation

\[
4(5x - 10)^2 - 34 = -33,
\]

let's start by isolating the quadratic term. We can add 34 to both sides:

\[
4(5x - 10)^2 - 34 + 34 = -33 + 34,
\]

which simplifies to

\[
4(5x - 10)^2 = 1.
\]

Next, we divide both sides by 4:

\[
(5x - 10)^2 = \frac{1}{4}.
\]

Now, we take the square root of both sides. Remember to consider both the positive and negative roots:

\[
5x - 10 = \frac{1}{2} \quad \text{or} \quad 5x - 10 = -\frac{1}{2}.
\]

**Case 1:**

\[
5x - 10 = \frac{1}{2}
\]

Adding 10 to both sides gives:

\[
5x = \frac{1}{2} + 10.
\]

Converting 10 to a fraction:

\[
10 = \frac{20}{2} \Rightarrow 5x = \frac{1}{2} + \frac{20}{2} = \frac{21}{2}.
\]

Now, divide both sides by 5:

\[
x = \frac{21/2}{5} = \frac{21}{10}.
\]

**Case 2:**

\[
5x - 10 = -\frac{1}{2}.
\]

Adding 10 to both sides gives:

\[
5x = -\frac{1}{2} + 10.
\]

Converting 10 to a fraction:

\[
10 = \frac{20}{2} \Rightarrow 5x = -\frac{1}{2} + \frac{20}{2} = \frac{19}{2}.
\]

Now divide both sides by 5:

\[
x = \frac{19/2}{5} = \frac{19}{10}.
\]

Thus, the solutions for the quadratic equation are:

\[
x = \frac{21}{10} \quad \text{and} \quad x = \frac{19}{10}.
\]