Math, asked by Arvindsidar, 1 year ago

quadratic equation X square - 5 x + 3 is equal to zero

Answers

Answered by katyayanikml
4
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Answered by Inflameroftheancient
2

Hey there!

Solving it by applying quadratic formula for equation "x^2 - 5x + 3 = 0"

For a required quadratic equation of \bf{ax^2 + bx + c = 0} the solutions for which can be represented by the quadratic formula of :

\boxed{\bf{x_{1, \: 2} = \dfrac{- b +- \sqrt{b^2 - 4ac}}{2a}}} \\

Here, a = 1,  b = - 5,  c = 3.

Solving for positive and negative values respectively:

\bf{x_{1} = \dfrac{- (- 5) - \sqrt{(- 5)^2 - 4(1)(3)}}{2(1)}} \\

\bf{x_{1} = \dfrac{5 - \sqrt{25 - 12}}{2 \times 1}} \\

\bf{x_{1} = \dfrac{5 - \sqrt{13}}{2 \times 1}} \\

\bf{x_{1} = \dfrac{5 - \sqrt{13}}{2}} \\

\bf{\therefore \quad x_{1} = \dfrac{5 - \sqrt{13}}{2}}

For the second solution in Positive form of equation:

\bf{x_{2} = \dfrac{- (- 5) + \sqrt{(- 5)^2 - 4(1)(3)}}{2(1)}} \\

\bf{x_{2} = \dfrac{5 + \sqrt{25 - 12}}{2 \times 1}} \\

\bf{x_{2} = \dfrac{5 + \sqrt{13}}{2 \times 1}} \\

\bf{x_{2} = \dfrac{5 + \sqrt{13}}{2}} \\

\bf{\therefore \quad x_{2} = \dfrac{5 + \sqrt{13}}{2}}

Therefore the final solutions for this quadratic equations are:

\boxed{\underline{\bf{x_{1} = \dfrac{5- \sqrt{13}}{2}}}}

\boxed{\underline{\bf{x_{2} = \dfrac{5 + \sqrt{13}}{2}}}}

Hope this helps you and solves the doubts for getting the equation solved by quadratic method!

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