Question

question 7 of 10
after being rearranged and simplified, which of the following equations could
be solved using the quadratic formula? check all that apply.

a. $5x + 4 = 3x^4 - 2$

b. $2x^2 + x^2 + x = 30$

c. $-x^2 + 4x + 7 = -x^2 - 9$

d. $9x + 3x^2 = 14 + x - 1$

Explanation:

Step1: Recall quadratic equation form

A quadratic equation is of the form \(ax^2 + bx + c = 0\), where \(a
eq0\). We need to check each option after rearranging.

Step2: Analyze Option A

Equation: \(5x + 4 = 3x^4 - 2\)
Rearrange: \(3x^4 - 5x - 6 = 0\). The highest power of \(x\) is 4, so it's a quartic (degree 4) equation, not quadratic.

Step3: Analyze Option B

Equation: \(2x^2 + x^2 + x = 30\)
Combine like terms: \(3x^2 + x - 30 = 0\). This is in the form \(ax^2+bx + c = 0\) with \(a = 3
eq0\), so it's a quadratic equation.

Step4: Analyze Option C

Equation: \(-x^2 + 4x + 7 = -x^2 - 9\)
Add \(x^2\) to both sides: \(4x + 7 = -9\)
Rearrange: \(4x + 16 = 0\). The highest power of \(x\) is 1, so it's a linear equation, not quadratic.

Step5: Analyze Option D

Equation: \(9x + 3x^2 = 14 + x - 1\)
Simplify right side: \(9x + 3x^2 = x + 13\)
Subtract \(x\) and 13 from both sides: \(3x^2 + 8x - 13 = 0\). This is in the form \(ax^2+bx + c = 0\) with \(a = 3
eq0\), so it's a quadratic equation.

Answer:

B. \(2x^2 + x^2 + x = 30\)
D. \(9x + 3x^2 = 14 + x - 1\)