Question

( x-9)^2 = -x^2+ 15x+ 50

Answer 1

$( x-9)^2 = -x^2+ 15x+ 50\\\\x^2-18x+81+x^2-15x-50=0\\\\2x^2-33x+31=0\\\\D=(-33)^2-4\cdot2\cdot31=1089-248=841\\\\\sqrt D=\sqrt{841}=29\\\\x_1=\frac{33-29}{2\cdot2}=\frac{4}{4}=1\\\\x_2=\frac{33+29}{2\cdot2}=\frac{62}{4}=15,5$

Answer 2

To solve the equation (x-9)^2 = -x^2 + 15x + 50, follow these steps:

Step 1: Expand the squared term on the left side: (x - 9)^2 = (x - 9)(x - 9) = x^2 - 18x + 81

Step 2: Now the equation becomes: x^2 - 18x + 81 = -x^2 + 15x + 50

Step 3: Move all the terms to one side of the equation to set it equal to zero: x^2 + x^2 - 18x - 15x + 81 - 50 = 0

Step 4: Combine like terms: 2x^2 - 33x + 31 = 0

Step 5: The equation is now in standard quadratic form (ax^2 + bx + c = 0). To solve for x, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 2, b = -33, and c = 31: x = (33 ± √((-33)^2 - 4 * 2 * 31)) / 2 * 2 x = (33 ± √(1089 - 248)) / 4 x = (33 ± √841) / 4 x = (33 ± 29) / 4

Now, we'll have two solutions for x:

1. x = (33 + 29) / 4 x = 62 / 4 x = 15.5

2. x = (33 - 29) / 4 x = 4 / 4 x = 1

So the solutions to the equation are x = 15.5 and x = 1.