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

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

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

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

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

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\$

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

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

Solve for x by finding the square root of each side

\$x + 1/4 = ± \sqrt(33)/4\$

⇒\$x = - 1/4 ± \sqrt(33)/4\$

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\$.

The solutions given for the quadratic equation are not applicable

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

The solutions given for the quadratic equation are not applicable

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

The solutions given for the quadratic equation are not applicable

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

The pair of solutions align with the provided quadratic equation

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

Option

D

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