Khan Academy $\newline$ Get Al Tutoring $\newline$ Donate [ㅈ $\newline$ $(y-k) y=\frac{1}{3}$ $\newline$ In the given equation, $k$ is a constant. One of the solutions to the equation is: $\newline$ $\frac{3+\sqrt{9+4\left(\frac{1}{3}\right)}}{2}$ $\newline$ What is the value of $k$ ?

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Q. Khan Academy

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Get Al Tutoring

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Donate [ㅈ

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(y-k) y=\frac{1}{3}

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In the given equation,

k

is a constant. One of the solutions to the equation is:

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\frac{3+\sqrt{9+4\left(\frac{1}{3}\right)}}{2}

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What is the value of

k

Identify equation and solution format: Identify the given equation and the solution format. $\newline$ The given equation is $(y-k)y = \frac{1}{3}$ , and one of the solutions is $\frac{3 + \sqrt{9 + 4(\frac{1}{3})}}{2}$ . This solution format suggests that the equation is a quadratic equation in the form of $y^2 - ky - \frac{1}{3} = 0$ .
Recognize quadratic formula: Recognize that the solution provided is in the form of the quadratic formula. $\newline$ The quadratic formula for the roots of the equation $ay^2 + by + c = 0$ is given by $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . The provided solution matches the positive part of this formula, which is $y = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ .
Compare solution to formula: Compare the given solution to the quadratic formula to find the value of $k$ . The given solution $\frac{3 + \sqrt{9 + 4(\frac{1}{3})}}{2}$ must match the form $\frac{-b + \sqrt{b^2 - 4ac}}{2a}$ . Since $a = 1$ and $c = -\frac{1}{3}$ , we can deduce that $b = k$ . Therefore, we have: $k = -(-b) = 3$
Verify value of k: Verify the value of k by substituting it back into the quadratic formula. $\newline$ Substitute $k = 3$ into the quadratic formula and check if it matches the given solution: $\newline$ $y = (3 + \sqrt{3^2 - 4(1)(-1/3)})/2$ $\newline$ $y = (3 + \sqrt{9 + 4/3})/2$ $\newline$ $y = (3 + \sqrt{9 + 4/3})/2$ $\newline$ Since this matches the given solution, our value for $k$ is correct.