Solve 8j j^3-2j^2-16

Question

Simplify the expression

8 j^{4} - 2 j^{2} - 16

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2 (4 j^{4} - j^{2} - 8)

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j_{1} = - \frac{2 + 2 129}{4}, j_{2} = \frac{2 + 2 129}{4}

Alternative Form

j_{1} \approx - 1.242871, j_{2} \approx 1.242871

Evaluate

8 j \times j^{3} - 2 j^{2} - 16

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

8 j \times j^{3} - 2 j^{2} - 16 = 0

Multiply

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8 j^{4} - 2 j^{2} - 16 = 0

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

8 t^{2} - 2 t - 16 = 0

Substitute a= 8,b= - 2 and c= - 16 into the quadratic formula t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

t = \frac{2 \pm ( - 2 ) ^{2} - 4 \times 8 ( - 16 )}{2 \times 8}

Simplify the expression

t = \frac{2 \pm ( - 2 ) ^{2} - 4 \times 8 ( - 16 )}{16}

Simplify the expression

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t = \frac{2 \pm 516}{16}

Simplify the radical expression

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t = \frac{2 \pm 2 129}{16}

Separate the equation into 2 possible cases

t = \frac{2 + 2 129}{16} t = \frac{2 - 2 129}{16}

Simplify the expression

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t = \frac{1 + 129}{8} t = \frac{2 - 2 129}{16}

Simplify the expression

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t = \frac{1 + 129}{8} t = \frac{1 - 129}{8}

Substitute back

j^{2} = \frac{1 + 129}{8} j^{2} = \frac{1 - 129}{8}

Solve the equation for j

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j = \frac{2 + 2 129}{4} j = - \frac{2 + 2 129}{4} j^{2} = \frac{1 - 129}{8}

Solve the equation for j

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j = \frac{2 + 2 129}{4} j = - \frac{2 + 2 129}{4} j = \frac{2 129 - 2}{4} i j = - \frac{2 129 - 2}{4} i

Calculate

j = \frac{2 + 2 129}{4} j = - \frac{2 + 2 129}{4}

Solution

j_{1} = - \frac{2 + 2 129}{4}, j_{2} = \frac{2 + 2 129}{4}

Alternative Form

j_{1} \approx - 1.242871, j_{2} \approx 1.242871

Show Solution