Solve 14h^2-4h 7h^3-8

Evaluate

14 h^{2} - 4 h \times 7 h^{3} - 8

To find the roots of the expression,set the expression equal to 0

14 h^{2} - 4 h \times 7 h^{3} - 8 = 0

Multiply

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14 h^{2} - 28 h^{4} - 8 = 0

Solve the equation using substitution t = h^{2}

14 t - 28 t^{2} - 8 = 0

Rewrite in standard form

- 28 t^{2} + 14 t - 8 = 0

Multiply both sides

28 t^{2} - 14 t + 8 = 0

Substitute a= 28,b= - 14 and c= 8 into the quadratic formula t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

t = \frac{14 \pm ( - 14 ) ^{2} - 4 \times 28 \times 8}{2 \times 28}

Simplify the expression

t = \frac{14 \pm ( - 14 ) ^{2} - 4 \times 28 \times 8}{56}

Simplify the expression

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t = \frac{14 \pm - 700}{56}

Simplify the radical expression

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t = \frac{14 \pm 10 7 \times i}{56}

Separate the equation into 2 possible cases

t = \frac{14 + 10 7 \times i}{56} t = \frac{14 - 10 7 \times i}{56}

Simplify the expression

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t = \frac{1}{4} + \frac{5 7}{28} i t = \frac{14 - 10 7 \times i}{56}

Simplify the expression

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t = \frac{1}{4} + \frac{5 7}{28} i t = \frac{1}{4} - \frac{5 7}{28} i

Substitute back

h^{2} = \frac{1}{4} + \frac{5 7}{28} i h^{2} = \frac{1}{4} - \frac{5 7}{28} i

Solve the equation for h

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Substitute back

h^{2} = \frac{1}{4} + \frac{5 7}{28} i

Simplify

h = \frac{1}{4} + \frac{5 7}{28} i

Rewrite the complex number in polar form

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h = \frac{14}{7} cos arctan \frac{\frac{5 7}{28}}{\frac{1}{4}} + i sin arctan \frac{\frac{5 7}{28}}{\frac{1}{4}}

Calculate the nth roots of a complex r (cos (θ) + i \times sin (θ),using n z = n r (cos \frac{θ + 2 kπ}{n} + i sin \frac{θ + 2 kπ}{n})

h = \frac{14}{7} \times cos \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 kπ}{2} + i sin \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 kπ}{2}

Simplify

h = \frac{4 686}{7} cos \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 kπ}{2} + i sin \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 kπ}{2}

Since n = 2,substitute k = 0, 1 into the expression

h_{1} = \frac{4 686}{7} cos \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 \times 0 \times π}{2} + i sin \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 \times 0 \times π}{2} h_{2} = \frac{4 686}{7} cos \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 \times 1 \times π}{2} + i sin \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 \times 1 \times π}{2}

Calculate

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h_{1} = \frac{4 686}{7} cos \frac{arctan ( \frac{5 7}{7} )}{2} + i sin \frac{arctan ( \frac{5 7}{7} )}{2} h_{2} = \frac{4 686}{7} cos \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 \times 1 \times π}{2} + i sin \frac{arctan ( \frac{\frac{5 7}{28}}{\frac{1}{4}} ) + 2 \times 1 \times π}{2}

Calculate

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h_{1} = \frac{4 686}{7} cos \frac{arctan ( \frac{5 7}{7} )}{2} + i sin \frac{arctan ( \frac{5 7}{7} )}{2} h_{2} = \frac{4 686}{7} cos \frac{arctan ( \frac{5 7}{7} ) + 2 π}{2} + i sin \frac{arctan ( \frac{5 7}{7} ) + 2 π}{2}

Calculate

h_{1} = \frac{4 686 \times cos ( \frac{a r c t a n ( \frac{5 7}{7} )}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( \frac{5 7}{7} )}{2} )}{7} i h_{2} = \frac{4 686 \times cos ( \frac{a r c t a n ( \frac{5 7}{7} ) + 2 π}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( \frac{5 7}{7} ) + 2 π}{2} )}{7} i

h_{1} = \frac{4 686 \times cos ( \frac{a r c t a n ( \frac{5 7}{7} )}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( \frac{5 7}{7} )}{2} )}{7} i h_{2} = \frac{4 686 \times cos ( \frac{a r c t a n ( \frac{5 7}{7} ) + 2 π}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( \frac{5 7}{7} ) + 2 π}{2} )}{7} i h^{2} = \frac{1}{4} - \frac{5 7}{28} i

Solve the equation for h

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Substitute back

h^{2} = \frac{1}{4} - \frac{5 7}{28} i

Simplify

h = \frac{1}{4} - \frac{5 7}{28} i

Rewrite the complex number in polar form

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h = \frac{14}{7} (cos (arctan (- \frac{5 7}{7}) + 2 π) + i sin (arctan (- \frac{5 7}{7}) + 2 π))

Calculate the nth roots of a complex r (cos (θ) + i \times sin (θ),using n z = n r (cos \frac{θ + 2 kπ}{n} + i sin \frac{θ + 2 kπ}{n})

h = \frac{14}{7} \times cos \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 kπ}{2} + i sin \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 kπ}{2}

Simplify

h = \frac{4 686}{7} cos \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 kπ}{2} + i sin \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 kπ}{2}

Since n = 2,substitute k = 0, 1 into the expression

h_{1} = \frac{4 686}{7} cos \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 \times 0 \times π}{2} + i sin \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 \times 0 \times π}{2} h_{2} = \frac{4 686}{7} cos \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 \times 1 \times π}{2} + i sin \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 \times 1 \times π}{2}

Calculate

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h_{1} = \frac{4 686}{7} cos \frac{arctan ( - \frac{5 7}{7} ) + 2 π}{2} + i sin \frac{arctan ( - \frac{5 7}{7} ) + 2 π}{2} h_{2} = \frac{4 686}{7} cos \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 \times 1 \times π}{2} + i sin \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 \times 1 \times π}{2}

Calculate

h_{1} = \frac{4 686}{7} cos \frac{arctan ( - \frac{5 7}{7} ) + 2 π}{2} + i sin \frac{arctan ( - \frac{5 7}{7} ) + 2 π}{2} h_{2} = \frac{4 686}{7} cos \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 π}{2} + i sin \frac{arctan ( - \frac{5 7}{7} ) + 2 π + 2 π}{2}

Calculate

h_{1} = - \frac{4 686 \times cos ( - \frac{a r c t a n ( - \frac{5 7}{7} )}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( - \frac{5 7}{7} ) + 2 π}{2} )}{7} i h_{2} = \frac{4 686 \times cos ( \frac{a r c t a n ( - \frac{5 7}{7} ) + 2 π + 2 π}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( - \frac{5 7}{7} ) + 2 π + 2 π}{2} )}{7} i

h_{1} = \frac{4 686 \times cos ( \frac{a r c t a n ( \frac{5 7}{7} )}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( \frac{5 7}{7} )}{2} )}{7} i h_{2} = \frac{4 686 \times cos ( \frac{a r c t a n ( \frac{5 7}{7} ) + 2 π}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( \frac{5 7}{7} ) + 2 π}{2} )}{7} i h_{1} = - \frac{4 686 \times cos ( - \frac{a r c t a n ( - \frac{5 7}{7} )}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( - \frac{5 7}{7} ) + 2 π}{2} )}{7} i h_{2} = \frac{4 686 \times cos ( \frac{a r c t a n ( - \frac{5 7}{7} ) + 2 π + 2 π}{2} )}{7} + \frac{4 686 \times sin ( \frac{a r c t a n ( - \frac{5 7}{7} ) + 2 π + 2 π}{2} )}{7} i

Solution

h \in / R

Simplify the expression