Question

quadratic equation √7y2-6y-13√7 ke mool

Answer 1

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Answer 2

Hey there!

Solving it by applying quadratic formula for equation "_/7y^2 - 6y - 13_/7 = 0"

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

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

Here, a = _/7,  b = - 6,  c = 13_/7.

Solving for positive and negative values respectively:

$\bf{x_{1} = \dfrac{- (- 6) + \sqrt{(- 6)^2 - 4(\sqrt{7})(13 \sqrt{7})}}{2(\sqrt{7})}}$

$\bf{x_{1} = \dfrac{6 + \sqrt{400}}{2 \sqrt{7}}}$

$\bf{x_{1} = \dfrac{6 + 20}{2 \sqrt{7}}}$

$\bf{x_{1} = \dfrac{26}{2 \sqrt{7}}}$

$\bf{x_{1} = \dfrac{13}{\sqrt{7}}}$

Rationalize the terms by a conjugate of "_/7":

$\bf{x_{1} = \dfrac{13 \sqrt{7}}{\sqrt{7} \sqrt{7}}}$

$\bf{x_{1} = \dfrac{13 \sqrt{7}}{7}}$

For the second solution in negative form of equation:

$\bf{x_{2} = \dfrac{- (- 6) - \sqrt{(- 6)^2 - 4(\sqrt{7})(13 \sqrt{7})}}{2(\sqrt{7})}}$

$\bf{x_{2} = \dfrac{6 - \sqrt{400}}{2 \sqrt{7}}}$

$\bf{x_{2} = \dfrac{6 - 20}{2 \sqrt{7}}}$

$\bf{x_{2} = \dfrac{- 14}{2 \sqrt{7}}}$

$\bf{x_{2} = - \dfrac{7}{\sqrt{7}}}$

Apply the radical rule for a number "a" that is:

$\bf{\sqrt[n]{a} = a^{\frac{1}{n}}}$

Therefore,

$\bf{x_{2} = \frac{7}{7^{\frac{1}{2}}}$

$\bf{x_{2} = 7^{\frac{1}{2}}}$

$\bf{x_{2} = - \sqrt{7}}$

Therefore the final solutions for this quadratic equations are:

$\boxed{\underline{\bf{x_{1} = \dfrac{13 \sqrt{7}}{7}}}}$

$\boxed{\underline{\bf{x_{2} = - \sqrt{7}}}}$

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