Question

which of the following are solutions to $2x^2 - 8x - 90$? select all that apply.
2
-5
-7
9
15
5
7
eoc style question
which of the following is a possible solution to $3x^2 - 6x + 528 = 633$?
a. $x = -7$
b. $x = 5$
c. $x = 14$
d. $x = 7$
eoc style question
find both solutions of the quadratic $-2x^2 - 2x = -40$
$x = \square$
$x = \square$

Explanation:

First Question: Solutions to \(2x^2 - 8x - 90 = 0\) (assuming it's set to zero)

Step 1: Simplify the equation

Divide the entire equation by 2: \(x^2 - 4x - 45 = 0\)

Step 2: Factor the quadratic

We need two numbers that multiply to -45 and add to -4. The numbers are -9 and 5. So, \((x - 9)(x + 5) = 0\)

Step 3: Solve for \(x\)

Set each factor equal to zero: \(x - 9 = 0\) or \(x + 5 = 0\), so \(x = 9\) or \(x = -5\)

Step 1: Simplify the equation

Subtract 633 from both sides: \(3x^2 - 6x + 528 - 633 = 0\) → \(3x^2 - 6x - 105 = 0\)
Divide by 3: \(x^2 - 2x - 35 = 0\)

Step 2: Factor the quadratic

Find two numbers that multiply to -35 and add to -2. The numbers are -7 and 5. So, \((x - 7)(x + 5) = 0\)

Step 3: Solve for \(x\)

Set each factor equal to zero: \(x - 7 = 0\) or \(x + 5 = 0\), so \(x = 7\) or \(x = -5\). From the options, \(x = 7\) (option d) is a solution.

Step 1: Rewrite the equation

Multiply both sides by -1: \(2x^2 + 2x = 40\)
Subtract 40 from both sides: \(2x^2 + 2x - 40 = 0\)
Divide by 2: \(x^2 + x - 20 = 0\)

Step 2: Factor the quadratic

Find two numbers that multiply to -20 and add to 1. The numbers are 5 and -4. So, \((x + 5)(x - 4) = 0\)

Step 3: Solve for \(x\)

Set each factor equal to zero: \(x + 5 = 0\) or \(x - 4 = 0\), so \(x = -5\) or \(x = 4\)

Answer:

-5, 9

Second Question: Solution to \(3x^2 - 6x + 528 = 633\)