Solve $x^2+\frac{1}{2} x+\frac{1}{16}=\frac{4}{9}$.
Factor the perfect-square trinomial on the left side of the equation.
$(x+\square)^2=\frac{4}{9}$
Answer (2)
Recognize the left side as a perfect square trinomial and factor it: ( x + 4 1 ) 2 = 9 4 .
Take the square root of both sides: x + 4 1 = ± 3 2 .
Solve for x in both cases: x = − 4 1 + 3 2 and x = − 4 1 − 3 2 .
Simplify to find the two solutions: x = 12 5 and x = − 12 11 .
x = 12 5 , x = − 12 11
Explanation
Analyze the problem We are given the equation x 2 + 2 1 x + 16 1 = 9 4 . Our goal is to solve for x . Notice that the left side of the equation is a perfect square trinomial, which can be factored.
Factor the perfect square trinomial We can rewrite the left side of the equation as a perfect square. We have x 2 + 2 1 x + 16 1 . This can be written as ( x + 4 1 ) 2 because ( x + 4 1 ) 2 = x 2 + 2 ( 4 1 ) x + ( 4 1 ) 2 = x 2 + 2 1 x + 16 1 . So the equation becomes ( x + 4 1 ) 2 = 9 4 .
Take the square root of both sides Now, we take the square root of both sides of the equation: ( x + 4 1 ) 2 = ± 9 4 . This simplifies to x + 4 1 = ± 3 2 .
Isolate x Next, we solve for x by subtracting 4 1 from both sides: x = − 4 1 ± 3 2 . This gives us two possible solutions for x .
Solve for x (first solution) For the first solution, we have x = − 4 1 + 3 2 . To combine these fractions, we need a common denominator, which is 12. So, x = − 12 3 + 12 8 = 12 5 .
Solve for x (second solution) For the second solution, we have x = − 4 1 − 3 2 . Again, we need a common denominator, which is 12. So, x = − 12 3 − 12 8 = − 12 11 .
State the solutions Therefore, the two solutions for x are 12 5 and − 12 11 .
Examples
Understanding how to solve quadratic equations like this is crucial in many fields, such as physics and engineering. For example, when calculating the trajectory of a projectile, you often need to solve a quadratic equation to find the time it takes for the projectile to reach a certain height or distance. Similarly, in electrical engineering, quadratic equations are used to analyze circuits and determine the values of components. Mastering these algebraic techniques provides a solid foundation for tackling more complex problems in various scientific and technical disciplines.
The equation x 2 + 2 1 x + 16 1 = 9 4 can be factored as ( x + 4 1 ) 2 = 9 4 . Solving this gives the two solutions: x = 12 5 and x = − 12 11 .
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