How do you solve #\sqrt { 10y ^ { 2} + 23y } = \sqrt { 2y ^ { 2} + 3}#?

Answer 1

#1/8# and - 3#

Square both sides: #10y^2 + 23y = 2y^2 + 3# Bring the equation to standard form: #f(y) = 8y^2 + 23y - 3 = 0# Solve it by the new Transforming Method (Google): Transformed equation: #f'(y) = y^2 + 23y - 24 = 0#. Since a +b + c = 0, the 2 real roots of f'(y) are: 1 and #c/a = - 24# The 2 real roots of f(y) are: #y1 = 1/a = 1/8# and #y2 = - 24/a = - 24/8 = - 3#
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Answer 2

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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Answer from HIX Tutor

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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