Math

Question Determine if (1,29)(-1,29) is a solution to the quadratic equation y=2x25x+22y=-2x^2-5x+22. Show your work.

Studdy Solution

STEP 1

Assumptions
1. We are given the equation y=2x25x+22y = -2x^2 - 5x + 22.
2. We need to check if the point (1,29)(-1, 29) is a solution to the given equation.

STEP 2

To determine if (1,29)(-1, 29) is a solution to the equation, we need to substitute x=1x = -1 into the equation and check if the resulting value of yy is equal to 2929.

STEP 3

Substitute x=1x = -1 into the equation.
y=2(1)25(1)+22y = -2(-1)^2 - 5(-1) + 22

STEP 4

Calculate the square of 1-1.
(1)2=1(-1)^2 = 1

STEP 5

Replace (1)2(-1)^2 with 11 in the equation.
y=2(1)5(1)+22y = -2(1) - 5(-1) + 22

STEP 6

Multiply 2-2 by 11.
2×1=2-2 \times 1 = -2

STEP 7

Replace 2(1)-2(1) with 2-2 in the equation.
y=25(1)+22y = -2 - 5(-1) + 22

STEP 8

Multiply 5-5 by 1-1.
5×1=5-5 \times -1 = 5

STEP 9

Replace 5(1)-5(-1) with 55 in the equation.
y=2+5+22y = -2 + 5 + 22

STEP 10

Add the numbers together to find the value of yy.
y=2+5+22=3+22y = -2 + 5 + 22 = 3 + 22

STEP 11

Complete the addition.
y=3+22=25y = 3 + 22 = 25

STEP 12

Compare the calculated value of yy with the yy-coordinate of the given point (1,29)(-1, 29).

STEP 13

Since the calculated value of yy is 2525 and the yy-coordinate of the point is 2929, they do not match.

STEP 14

Conclude that (1,29)(-1, 29) is not a solution to the equation y=2x25x+22y = -2x^2 - 5x + 22 because when x=1x = -1, the value of yy should be 2525, not 2929.
The point (1,29)(-1, 29) is not a solution to the equation y=2x25x+22y = -2x^2 - 5x + 22.

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