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solve $2x^2 - 3x = 1$ using the quadratic formula $x = \\dfrac{-b \\pm …

Question

solve $2x^2 - 3x = 1$ using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. use the keypad to enter your answer in the box. you can find additional symbols by using the drop - down arrow at the top of the keypad. the solutions to the quadratic equation are $x = \square$ and $x = \square$.

Explanation:

Step1: Rewrite in standard form

Rewrite \(2x^2 - 3x = 1\) as \(2x^2 - 3x - 1 = 0\). Here, \(a = 2\), \(b = -3\), \(c = -1\).

Step2: Substitute into quadratic formula

Substitute \(a\), \(b\), \(c\) into \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\):
\(x=\frac{-(-3)\pm\sqrt{(-3)^2 - 4\times2\times(-1)}}{2\times2}=\frac{3\pm\sqrt{9 + 8}}{4}=\frac{3\pm\sqrt{17}}{4}\)

Answer:

\(x = \frac{3 + \sqrt{17}}{4}\) and \(x = \frac{3 - \sqrt{17}}{4}\)