1. Solve the following system of equations algebraically using the quadratic formula. y = -x^2 + 2x + 9 and y = -5x^2 + 10x + 12 2. The sum of the first number and the square of a second number is 18. The different between the square of the second number and twice the first number is 12. a) Write a system of equations to represent this situation. b) Solve the system to determine the numbers
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Solve the system: `y = -x^2 + 2x + 9 and y = -5x^2 + 10x + 12.`
`-x^2 + 2x + 9 = -5x^2 + 10x + 12` Bring everything to right side.
`4x^2 - 8x - 3 = 0`
Use quadratic formula. `(8+-sqrt(8^2 - 4(4)(-3)))/(2*4)`
`(8+-sqrt(112))/8` = `(8+-4sqrt(7))/8` = `(2+-sqrt(7))/2` = 2.323 and -0.323
x = 2.323, -0.323. Substitute into either equation to solve for y.
`y = -x^2 + 2x + 9`
`y = -(2.323)^2 + 2(2.323) + 9`
`y = 8.25`
`y = -x^2 + 2x + 9`
`y = -(-0.323)^2 + 2(-0.323) + 9`
`y = 8.25`
#1) The solutions are (2.323, 8.25) and (-0.323, 8.25).
#2) Let the first number be "x", and the second number be "y".
The system of equations would be:
`x + y^2 = 18` and `y^2 - 2x = 12.`
Solve for y^2. The 2 equations would be:
`y^2 = 18 - x` and `y^2 = 2x + 12` .
Therefore, `18 - x = 2x + 12.`
Solve for x.
`3x = 6 `
`x = 2`
Substitute to find y.
`x + y^2 = 18`
`2 + y^2 = 18`
`y^2 = 16`
`y = +-4`
There are 2 possible solutions that work. The 2 numbers could be 2 and 4, or 2 and -4.
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