5) Solve x2 – 2x + 3 = 0 by using the quadratic formula​

[tex]\begin{gathered}\frak{ \pink{Given : }}\sf{\;\;\; x^2 – 2x + 3 = \bf{0}}\end{gathered} [/tex]

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[tex]\begin{gathered}\frak{ \pink{To \: find : }}\sf{\;\;\; Value \: of \: x \: by \: using \: quadratic \: formula.}\end{gathered} [/tex]

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[tex]\begin{gathered}\frak{ \pink{Solution : }}\end{gathered} [/tex]

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Given that : x² – 2x + 3 = 0 and it is a quadratic equation in form of ax² + bx + c = 0.

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Here,

• a = 1
• b = – 2
• c = 3

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Quadratic formula :

• [tex]\underline{\boxed{\bf x = \dfrac{-b\pm \sqrt{b^2 – 4ac}}{2a} }} [/tex]

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Now, by substituting values :

[tex] \sf : \implies x = \dfrac{-(-2)\pm \sqrt{(-2)^2 – 4(1)(3)}}{2(1)} [/tex]

[tex] \sf : \implies x = \dfrac{2\pm \sqrt{4 – 4\times 1 \times 3}}{2\times 1} [/tex]

[tex] \sf : \implies x = \dfrac{2\pm \sqrt{4 – 12}}{2} [/tex]

[tex] \sf : \implies x = \dfrac{2\pm \sqrt{- 8}}{2} [/tex]

[tex] \sf : \implies x = \dfrac{2\pm \sqrt{2 \times 2 \times – 2}}{2} [/tex]

[tex] \sf : \implies x = \dfrac{2\pm 2 \sqrt{- 2}}{2} [/tex]

[tex] \sf : \implies x = \dfrac{\cancel{2}(1\pm \sqrt{- 2})}{\cancel{2}} [/tex]

[tex] \sf : \implies x = 1\pm\sqrt{- 2}[/tex]

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[tex]\: \:\:\: \: \: \: \: \: \: \: \: \: \: \: \: \pink{\underline{\boxed{\pmb{\frak{x = 1+\sqrt{- 2} \:or\: x = 1- \sqrt{- 2}}}}}}\:\bigstar[/tex]

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[tex] \underline{ \bf Hence, \: value \: of \: x \: is \: \pink{1+\sqrt{- 2} \: or \: 1- \sqrt{- 2}}}.[/tex]