III. Solve the following quadratic equations not written in standard form and rational equations transformable into quadratic equations. 1. y(y-5) = 36 2. (x+3)(x-2) = x + 10 3. (k+ 5)² + (k - 2)² = 37
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1. y(y - 5) = 36
To solve this, expand the left side of the equation:
y^2 - 5y = 36
Move all terms to one side to get it in standard quadratic form:
y^2 - 5y - 36 = 0
Now, you can factor the quadratic:
(y - 9)(y + 4) = 0
Set each factor equal to zero and solve for y:
y - 9 = 0 --> y = 9
y + 4 = 0 --> y = -4
So, the solutions are y = 9 and y = -4.
2. (x + 3)(x - 2) = x + 10
Expand the left side:
x^2 + x - 6 = x + 10
Move all terms to one side:
x^2 + x - 6 - x - 10 = 0
Simplify:
x^2 - 16 = 0
Factor the quadratic:
(x - 4)(x + 4) = 0
Set each factor equal to zero and solve for x:
x - 4 = 0 --> x = 4
x + 4 = 0 --> x = -4
So, the solutions are x = 4 and x = -4.
3. (k + 5)^2 + (k - 2)^2 = 37
Expand and simplify the left side:
k^2 + 10k + 25 + k^2 - 4k + 4 = 37
Combine like terms:
2k^2 + 6k + 29 = 37
Subtract 37 from both sides:
2k^2 + 6k - 8 = 0
Divide the equation by 2 to simplify:
k^2 + 3k - 4 = 0
Factor the quadratic:
(k + 4)(k - 1) = 0
Set each factor equal to zero and solve for k:
k + 4 = 0 --> k = -4
k - 1 = 0 --> k = 1
So, the solutions are k = -4 and k = 1.
Im only in elem sorry can't help