Question

11 multiple choice 1 point
question 9
what are the solutions to the equation $2x^2 - 5x - 1 = 0$
$\bigcirc$ $x = \frac{-5\pm\sqrt{17}}{4}$
$\bigcirc$ $x = \frac{5\pm\sqrt{17}}{4}$
$\bigcirc$ $x = \frac{-5\pm\sqrt{33}}{4}$
$\bigcirc$ $x = \frac{5\pm\sqrt{33}}{4}$
12 essay 2 points
question 10
consider the equation $x^2 - 39 = 0$
a. does the quadratic formula work to solve this equation? explain or show how you know.
b. can you solve this equation using square roots? explain or show how you know.

Explanation:

Question 9

Step1: Identify quadratic coefficients

For $2x^2 -5x -1=0$, $a=2$, $b=-5$, $c=-1$

Step2: Apply quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute values:
$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-1)}}{2(2)}$

Step3: Simplify the expression

Calculate discriminant: $\sqrt{25 + 8}=\sqrt{33}$
Simplify numerator/denominator: $x=\frac{5\pm\sqrt{33}}{4}$

a. The quadratic formula works for all quadratic equations in the form $ax^2+bx+c=0$. For $x^2-39=0$, we can rewrite it as $1x^2+0x-39=0$ (so $a=1$, $b=0$, $c=-39$), which fits the standard quadratic form, so the quadratic formula is applicable.
b. Yes, this equation can be solved using square roots because it is a difference of squares (after rearranging to isolate $x^2$). Isolate $x^2$ first, then take the square root of both sides, remembering to include both positive and negative roots.

Answer:

$\boldsymbol{x = \frac{5\pm\sqrt{33}}{4}}$

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Question 10