Solve for $x$ in the equation $x^2+10 x+12=36$.
A. $x=-11$ or $x=1$
B. $x=-2$ or $x=12$
C. $x=-12$ or $x=2$
D. $x=-1$ or $x=11$
Answer (1)
Rewrite the equation in the standard quadratic form: x 2 + 10 x − 24 = 0 .
Complete the square: ( x + 5 ) 2 − 49 = 0 .
Isolate the squared term: ( x + 5 ) 2 = 49 .
Solve for x : x = − 5 ± 7 , which gives x = 2 or x = − 12 .
The solutions are x = − 12 or x = 2 .
Explanation
Rewrite the equation First, we need to rewrite the given equation x 2 + 10 x + 12 = 36 in the standard quadratic form. To do this, we subtract 36 from both sides of the equation: x 2 + 10 x + 12 − 36 = 0 x 2 + 10 x − 24 = 0
Complete the square Now, we will complete the square. To complete the square for a quadratic equation of the form x 2 + b x + c = 0 , we need to add and subtract ( 2 b ) 2 from the left side of the equation. In our case, b = 10 , so we have ( 2 10 ) 2 = 5 2 = 25 . Thus, we add and subtract 25 from the left side of the equation: x 2 + 10 x + 25 − 25 − 24 = 0 ( x 2 + 10 x + 25 ) − 49 = 0
Express as a squared term Next, we express the quadratic expression in parentheses as a squared term: ( x + 5 ) 2 − 49 = 0
Isolate the squared term Now, we isolate the squared term by adding 49 to both sides of the equation: ( x + 5 ) 2 = 49
Take the square root We take the square root of both sides of the equation, remembering to consider both positive and negative roots: x + 5 = ± 49 x + 5 = ± 7
Solve for x Finally, we solve for x by subtracting 5 from both sides of the equation: x = − 5 ± 7 This gives us two possible solutions for x :
x = − 5 + 7 = 2 x = − 5 − 7 = − 12
Final Answer Therefore, the solutions for x are x = 2 and x = − 12 .
So the answer is x = − 12 or x = 2 .
Examples
Completing the square is a useful technique in physics, especially when dealing with projectile motion. For example, if you want to find the maximum height of a ball thrown upwards, you can model the height as a quadratic function of time. By completing the square, you can rewrite the function in vertex form, which directly gives you the maximum height and the time at which it occurs. This method allows physicists and engineers to easily determine key parameters of motion without needing calculus.
