How do you solve #\sqrt { 10y ^ { 2} + 23y } = \sqrt { 2y ^ { 2} + 3}#?
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To solve the equation (\sqrt{10y^2 + 23y} = \sqrt{2y^2 + 3}), square both sides to eliminate the square roots. This yields:
(10y^2 + 23y = 2y^2 + 3)
Rearrange the equation to get all terms on one side:
(8y^2 + 23y - 3 = 0)
This is a quadratic equation. You can solve it by factoring, completing the square, or using the quadratic formula. For this equation, the quadratic formula is generally most straightforward:
(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a})
Where (a = 8), (b = 23), and (c = -3). Substituting these values into the formula gives:
(y = \frac{-23 \pm \sqrt{23^2 - 4(8)(-3)}}{2(8)})
Calculate inside the square root:
(y = \frac{-23 \pm \sqrt{529 + 96}}{16})
(y = \frac{-23 \pm \sqrt{625}}{16})
(y = \frac{-23 \pm 25}{16})
Now, you have two solutions:
(y_1 = \frac{-23 + 25}{16} = \frac{2}{16} = \frac{1}{8})
(y_2 = \frac{-23 - 25}{16} = \frac{-48}{16} = -3)
So, the solutions to the equation are (y = \frac{1}{8}) and (y = -3).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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