Solve $2 x^2+x-4=0$
$
\begin{array}{l}
x^2+\frac{1}{2} x+-2=0 \\
x+\frac{1}{2} x+\square=2+\square \\
(x+\square)^2=
\end{array}$
Answer (1)
Divide the equation by 2: x 2 + \t f r a c 1 2 x − 2 = 0 becomes x 2 + \t f r a c 1 2 x = 2 .
Complete the square by adding ( \t f r a c 1 4 ) 2 = \t f r a c 1 16 to both sides: x 2 + \t f r a c 1 2 x + \t f r a c 1 16 = 2 + \t f r a c 1 16 .
Rewrite the left side as a squared term: ( x + \t f r a c 1 4 ) 2 = \t f r a c 33 16 .
Isolate x: x = \t f r a c − 1 \t p m \t s q r t 33 4 . The solutions are x = \t f r a c − 1 \t p m \t s q r t 33 4 .
Explanation
Problem Analysis We are given the quadratic equation 2 x 2 + x − 4 = 0 and asked to solve it by completing the square. The provided steps guide us through the process.
Divide by 2 First, divide the entire equation by 2 to make the coefficient of x 2 equal to 1: x 2 + 2 1 x − 2 = 0
Isolate x terms Next, isolate the terms with x on one side of the equation: x 2 + 2 1 x = 2
Complete the square To complete the square, we need to add a value to both sides of the equation such that the left side becomes a perfect square. The value to add is ( 2 1 × 2 1 ) 2 = ( 4 1 ) 2 = 16 1 . So we add 16 1 to both sides: x 2 + 2 1 x + 16 1 = 2 + 16 1
Rewrite as squared term Now, rewrite the left side as a squared term and simplify the right side: ( x + 4 1 ) 2 = 16 32 + 16 1 = 16 33
Take the square root Take the square root of both sides of the equation: x + 4 1 = ± 16 33 = ± 4 33
Isolate x Finally, isolate x to find the solutions: x = − 4 1 ± 4 33 = 4 − 1 ± 33 Thus, the solutions are x = 4 − 1 + 33 and x = 4 − 1 − 33 .
Final Answer The solutions to the quadratic equation 2 x 2 + x − 4 = 0 are x = 4 − 1 + 33 ≈ 1.186 and x = 4 − 1 − 33 ≈ − 1.686
Examples
Completing the square is a useful technique in many areas, such as finding the vertex of a parabola or solving optimization problems. For example, suppose you want to find the dimensions of a rectangular garden that maximize the area, given a fixed perimeter. By expressing the area in terms of one variable and completing the square, you can easily find the dimensions that yield the maximum area. This method is also fundamental in deriving the quadratic formula, a general solution for any quadratic equation.
