The sum of the roots for 2x^2 + (k-3)x +1 =0 is equal to the product of the roots for 5x^2 - 3x + (k+1) =0 Algebraically determine the value(s) of k.

Sum of roots for 2x^2 + (k-3)x +1 =0 is equal to -b/a where b is the coefficient of x and a is the coefficient of x^2. -b/1 = [-(k-3)]/2. Product of roots for 5x^2 - 3x + (k+1) =0 is equal to c/a where c is the constant, a is the coefficient of x^2. so c/a = (k+1)/5 so solve: [-(k-3)]/2 = (k+1)/5

Just to show where the expressions for the sum of roots \(x_1 + x_2 = -\frac{b}{a}\) and product of roots \(x_1 x_2 = \frac{c}{a}\) come from, in case it's not immediately obvious, we can show that these two identities always work out through manipulation of the solutions given by quadratic formula. \[ \quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{Therefore, } x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \text{ and } \\ \qquad \qquad \qquad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}. \\ \\ \quad \text{The sum of the roots, } x_1 + x_2 \text{, is then given by substitution: } \\ \qquad x_1 + x_2 \\ \qquad = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} \\ \qquad = \frac{\color{green}{-b} + \cancel{\sqrt{b^2 - 4ac}} \color{Green}{- b} - \cancel{\sqrt{b^2 - 4ac}}}{2a} \\ \qquad = \frac{-\cancel{2}b}{\cancel{2}a} \\ \qquad = -\frac{b}{a} \\ \\ \\ \quad \text{Similarly, for the product of roots, } x_1 x_2 \text{:} \\ \qquad x_1 x_2 \\ \qquad = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \frac{-b - \sqrt{b^2 - 4ac}}{2a} \quad \text{Difference of squares} \\ \qquad = \frac{(-b)^2 - \left(\sqrt{b^2 - 4ac}\right)^2}{4a^2} \quad \text{Cancel the radical and square} \\ \qquad = \frac{b^2 - (b^2 - 4ac)}{4a^2} \qquad = \frac{\cancel{b^2 - b^2} + 4ac}{4a^2} \\ \qquad = \frac{\cancel{4a}c}{\cancel{4}a^{\cancel{2}}} \\ \qquad = \frac{c}{a} \\ \]

Thank you guys!! That's exactly what I did, I just wanted to make sure I was on the right track :D

You're welcome. :)

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