# Chapter 9 - Section 9.5 - Graphs of Quadratic Equations - Exercise Set - Page 665: 5

$\{ 1, 3 \}$

#### Work Step by Step

Setting $y=0$ and solving for $x$ using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), then the $x-$intercept/s of the given equation, $y=x^2-4x+3 ,$ is/are \begin{array}{l}\require{cancel} x^2-4x+3=0 \\\\ x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)} \\\\ x=\dfrac{4\pm\sqrt{16-12}}{2} \\\\ x=\dfrac{4\pm\sqrt{4}}{2} \\\\ x=\dfrac{4\pm2}{2} .\end{array} Hence, the $x-$intercepts are $\{ 1, 3 \} .$

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