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Chapter 17: Chapter 17

Problem 477

Next question

Solve the equation \(2 \mathrm{x}^{2}-5 \mathrm{x}+3=0\)

Short Answer

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The solutions to the quadratic equation \(2x^2 - 5x + 3 = 0\) are \(x_1 = \frac{3}{2}\) and \(x_2 = 1\).
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Identify the coefficients a, b, and c

In the given quadratic equation \(2x^2 - 5x + 3 = 0\), we can identify the coefficients as: - \(a = 2\) - \(b = -5\) - \(c = 3\)

Step 2: Apply the quadratic formula

Recall, the quadratic formula is \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] Plug in the values of a, b, and c into the formula and simplify. \[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(3)}}{2(2)}\] \[x = \frac{5 \pm \sqrt{25 - 24}}{4}\]

Step 3: Calculate the discriminant and simplify

The discriminant is the expression inside the square root, \(b^2 - 4ac\). In our case, the discriminant is \(25 - 24 = 1\). We will now simplify the quadratic formula: \[x = \frac{5 \pm \sqrt{1}}{4}\]

Step 4: Calculate the values of x

Since the square root of 1 is 1, we can find the two possible values for x: \[x_1 = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}\] \[x_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1\] Therefore, the solutions to the quadratic equation \(2x^2 - 5x + 3 = 0\) are \(x_1 = \frac{3}{2}\) and \(x_2 = 1\).

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