Use completing the square to solve for [tex]$x$[/tex] in the equation [tex](x-12)(x+4)=9[/tex]
[tex]x=-1[/tex] or 15
[tex]x=1[/tex] or 7
[tex]x=4 \pm \sqrt{41}[/tex]
[tex]x=4 \pm \sqrt{73}[/tex]

By expanding and rearranging the given equation, we complete the square and find the solutions to be x = 4 ± 73 ​ . The working demonstrates the steps to solve a quadratic equation using this method. Thus, the final answer is x = 4 ± 73 ​ .
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Answered by Anonymous | 2025-07-04

Expand the given equation ( x − 12 ) ( x + 4 ) = 9 to get x 2 − 8 x − 48 = 9 .
Rewrite the equation as x 2 − 8 x = 57 .
Complete the square by adding ( 2 − 8 ​ ) 2 = 16 to both sides: x 2 − 8 x + 16 = 57 + 16 , which simplifies to ( x − 4 ) 2 = 73 .
Take the square root of both sides and solve for x : x = 4 ± 73 ​ .

The solutions are x = 4 ± 73 ​ ​
Explanation

Understanding the Problem We are given the equation ( x − 12 ) ( x + 4 ) = 9 and asked to solve for x by completing the square. This method involves transforming the quadratic equation into a perfect square trinomial, making it easier to solve.

Expanding the Equation First, expand the left side of the equation: ( x − 12 ) ( x + 4 ) = x 2 − 12 x + 4 x − 48 = x 2 − 8 x − 48

Rewriting the Equation Now, rewrite the equation as: x 2 − 8 x − 48 = 9

Isolating the Quadratic and Linear Terms Move the constant term to the right side: x 2 − 8 x = 9 + 48
x 2 − 8 x = 57

Completing the Square To complete the square, we need to add a value to both sides of the equation that will make the left side a perfect square trinomial. This value is ( 2 b ​ ) 2 , where b is the coefficient of the x term. In this case, b = − 8 , so we add ( 2 − 8 ​ ) 2 = ( − 4 ) 2 = 16 to both sides: x 2 − 8 x + 16 = 57 + 16

Expressing as a Perfect Square Rewrite the left side as a squared term: ( x − 4 ) 2 = 73

Taking the Square Root Take the square root of both sides: x − 4 = ± 73 ​

Solving for x Solve for x :
x = 4 ± 73 ​

Final Answer Therefore, the solutions for x are x = 4 + 73 ​ and x = 4 − 73 ​ .


Examples
Completing the square is a useful technique in physics, particularly in mechanics. For example, when analyzing projectile motion under gravity, completing the square can help determine the maximum height reached by the projectile. The equation describing the height of the projectile as a function of time often involves quadratic terms, and completing the square allows us to rewrite the equation in a form that directly reveals the vertex of the parabola, which corresponds to the maximum height and the time at which it is achieved. This method simplifies the process of finding key parameters of the motion.

Answered by GinnyAnswer | 2025-07-04