Question

5.
a student creates a fourth - degree trinomial with a leading coefficient
of 2 and a constant value of 5. the trinomial could be
(1) $2x^4 + 3x^2 + 5$ (3) $4x^2 - 3x + 5$
(2) $2x^4 + 5x + 3$ (4) $4x^3 - 5x^2 + 3$

6.
using the quadratic formula, solve $x^2 + 4x - 3 = 0$.
express your solution in simplest radical form. $x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$

Explanation:

Question 5

Step1: Recall trinomial, degree, leading coefficient, constant definitions

A trinomial has 3 terms. The degree of a polynomial is the highest exponent of \(x\). The leading coefficient is the coefficient of the highest - degree term. The constant term is the term without \(x\).

Step2: Analyze Option (1)

For \(2x^{4}+3x^{2}+5\):

• Number of terms: 3 (trinomial).
• Highest exponent: 4 (fourth - degree).
• Leading coefficient: 2 (coefficient of \(x^{4}\)).
• Constant term: 5. This satisfies all the given conditions.

Step3: Analyze Option (2)

For \(2x^{4}+5x + 3\):

• Number of terms: 3 (trinomial).
• Highest exponent: 4 (fourth - degree).
• Leading coefficient: 2.
• Constant term: 3 (not 5). So, this does not satisfy the condition of constant value 5.

Step4: Analyze Option (3)

For \(4x^{2}-3x + 5\):

• Number of terms: 3 (trinomial).
• Highest exponent: 2 (second - degree, not fourth - degree). So, this does not satisfy the condition of fourth - degree.

Step5: Analyze Option (4)

For \(4x^{3}-5x^{2}+3\):

• Number of terms: 3 (trinomial).
• Highest exponent: 3 (third - degree, not fourth - degree). So, this does not satisfy the condition of fourth - degree.

Step1: Identify \(a\), \(b\), \(c\) from the quadratic equation

For the quadratic equation \(x^{2}+4x - 3=0\), the general form of a quadratic equation is \(ax^{2}+bx + c = 0\). So, \(a = 1\), \(b = 4\), and \(c=-3\).

Step2: Substitute into the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute \(a = 1\), \(b = 4\), and \(c=-3\) into the formula:

$$
LATEXBLOCK0
$$

Step3: Simplify the radical

Simplify \(\sqrt{28}\). We know that \(28 = 4\times7\), so \(\sqrt{28}=\sqrt{4\times7}=2\sqrt{7}\).
Then, \(x=\frac{-4\pm2\sqrt{7}}{2}\).

Step4: Simplify the fraction

Factor out a 2 from the numerator: \(x=\frac{2(-2\pm\sqrt{7})}{2}\).
Cancel out the common factor of 2 in the numerator and the denominator: \(x=-2\pm\sqrt{7}\).

Answer:

(1) \(2x^{4}+3x^{2}+5\)

Question 6