Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Solve [tex]\(2x^2 + x - 4 = 0\)[/tex].

Given:
[tex]\[
2x^2 + x - 4 = 0
\][/tex]

Complete the square:
[tex]\[
\begin{array}{l}
2 \left(x^2 + \frac{1}{2}x \right) = 2 \left(\frac{1}{2}\right) - 4 \\
2 \left(x^2 + \frac{1}{2}x + \left(\frac{1}{4}\right)\right) = 2 \left(\frac{1}{4}\right) - 4 \\
2 \left(x + \frac{1}{4}\right)^2 = 2 \left(\frac{1}{4}\right) - 4
\end{array}
\][/tex]

Solve for [tex]\(x\)[/tex]:
[tex]\[
\begin{array}{l}
\left(x + \frac{1}{4}\right)^2 = \frac{2 \left(\frac{1}{4}\right) - 4}{2} \\
\left(x + \frac{1}{4}\right)^2 = \frac{\frac{1}{2} - 4}{2} \\
\left(x + \frac{1}{4}\right)^2 = \frac{-\frac{7}{2}}{2} \\
\left(x + \frac{1}{4}\right)^2 = -\frac{7}{4} \\
x + \frac{1}{4} = \pm \sqrt{-\frac{7}{4}} \\
x = -\frac{1}{4} \pm \sqrt{-\frac{7}{4}}
\end{array}
\][/tex]

Sagot :

Certainly! Let's solve the quadratic equation [tex]\(2x^2 + x - 4 = 0\)[/tex] by completing the square, step-by-step.

### Step 1: Normalize the quadratic coefficient
First, we want to make the coefficient of [tex]\(x^2\)[/tex] equal to 1. We do this by dividing the entire equation by 2.
[tex]\[ \frac{2x^2 + x - 4}{2} = 0 \][/tex]
[tex]\[ x^2 + \frac{1}{2}x - 2 = 0 \][/tex]

### Step 2: Move the constant term to the other side
Next, we move the constant term [tex]\(-2\)[/tex] to the right side of the equation.
[tex]\[ x^2 + \frac{1}{2}x = 2 \][/tex]

### Step 3: Complete the square
To complete the square, we need to add and subtract the same value on the left-hand side such that it forms a perfect square trinomial. We take half the coefficient of [tex]\(x\)[/tex] (which is [tex]\(\frac{1}{2}\)[/tex]), square it, and add it to both sides.
[tex]\[ \left( \frac{1}{2} \cdot \frac{1}{2} \right)^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16} \][/tex]
Now we add [tex]\(\frac{1}{16}\)[/tex] to both sides:
[tex]\[ x^2 + \frac{1}{2}x + \frac{1}{16} = 2 + \frac{1}{16} \][/tex]

### Step 4: Rewrite the left side as a binomial squared
The left-hand side now forms a perfect square trinomial:
[tex]\[ \left( x + \frac{1}{4} \right)^2 = 2 + \frac{1}{16} \][/tex]

### Step 5: Simplify the right side
Combine the terms on the right side:
[tex]\[ 2 + \frac{1}{16} = \frac{32}{16} + \frac{1}{16} = \frac{33}{16} \][/tex]
So we have:
[tex]\[ \left( x + \frac{1}{4} \right)^2 = \frac{33}{16} \][/tex]

### Final Result
Let's summarize our completed square form:
[tex]\[ \left( x + \frac{1}{4} \right)^2 = \frac{33}{16} \][/tex]

This is the result after solving the quadratic equation by completing the square.