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Lesson Example Discussion Quiz: Class Homework

Step-5

Title: Quadratic-Equations and Factors

Grade: 1300-a Lesson: S2-L1

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Step	Type	Explanation	Answer
1	Problem	Solve the quadratic equation \$2x^2 + x - 4 = 0\$ by completing the square.
2	Step	Rearrange the equation by shifting the constant term (- 4) to the opposite side	\$2x^2 + x = 4\$
3	Clue	Divide the entire equation by the coefficient of \$x^2\$ to ensure the leading coefficient becomes 1	\$x^2+(1/2)x = 2\$
4	Hint	Take half of the coefficient of the x term, square it, and add it to both sides of the equation:	\$x^2 + 2 (1/4)x + 1/16 = 2 + 1/16\$
5	Step	Square the left side of the equation and express it as a perfect square by factoring	\$(x + 1/4)^2 = 33/16\$
6	Step	Solve for x by finding the square root of each side	\$x + 1/4 = ± \sqrt(33)/4\$ ⇒\$x = - 1/4 ± \sqrt(33)/4\$
7	Step	Hence, completing the square for the quadratic equation \$2x^2 + x - 4 = 0\$ yields solutions: \$x = (- 1 + \sqrt(33))/4\$ and \$x = (- 1 - \sqrt(33))/4\$.
8	Choice.A	The solutions given for the quadratic equation are not applicable	\$x = (- 1 + \sqrt(33))/2\$ and \$x = (- 1 - \sqrt(33))/2\$
9	Choice.B	The solutions given for the quadratic equation are not applicable	\$x = ( 1 + \sqrt(33))/2\$ and \$x = ( 1 - \sqrt(33))/2\$
10	Choice.C	The solutions given for the quadratic equation are not applicable	\$x = ( 1 + \sqrt(33))/4\$ and \$x = ( 1 - \sqrt(33))/4\$
11	Choice.D	The pair of solutions align with the provided quadratic equation	\$x = (- 1 + \sqrt(33))/4\$ and \$x = (- 1 - \sqrt(33))/4\$
12	Answer	Option	D
13	Sumup	Can you briefly tell me what you’ve learned and understood in today’s lesson?

Step

Type

Explanation

Answer

Problem

Solve the quadratic equation \$2x^2 + x - 4 = 0\$ by completing the square.

Step

Rearrange the equation by shifting the constant term
(- 4) to the opposite side

\$2x^2 + x = 4\$

Clue

Divide the entire equation by the coefficient of \$x^2\$ to ensure the leading coefficient becomes 1

\$x^2+(1/2)x = 2\$

Hint

Take half of the coefficient of the x term, square it, and add it to both sides of the equation:

\$x^2 + 2 (1/4)x + 1/16 = 2 + 1/16\$

Step

Square the left side of the equation and express it as a perfect square by factoring

\$(x + 1/4)^2 = 33/16\$

Step

Solve for x by finding the square root of each side

\$x + 1/4 = ± \sqrt(33)/4\$

⇒\$x = - 1/4 ± \sqrt(33)/4\$

Step

Hence, completing the square for the quadratic equation \$2x^2 + x - 4 = 0\$ yields solutions: \$x = (- 1 + \sqrt(33))/4\$ and \$x = (- 1 - \sqrt(33))/4\$.

Choice.A

The solutions given for the quadratic equation are not applicable

\$x = (- 1 + \sqrt(33))/2\$ and \$x = (- 1 - \sqrt(33))/2\$

Choice.B

The solutions given for the quadratic equation are not applicable

\$x = ( 1 + \sqrt(33))/2\$ and \$x = ( 1 - \sqrt(33))/2\$

Choice.C

The solutions given for the quadratic equation are not applicable

\$x = ( 1 + \sqrt(33))/4\$ and \$x = ( 1 - \sqrt(33))/4\$

Choice.D

The pair of solutions align with the provided quadratic equation

\$x = (- 1 + \sqrt(33))/4\$ and \$x = (- 1 - \sqrt(33))/4\$

Answer

Option

Sumup

Can you briefly tell me what you’ve learned and understood in today’s lesson?