Solve b^2 - 4b^3 b = -2

Question

Solve the equation

b_{1} = - \frac{2 + 2 33}{4}, b_{2} = \frac{2 + 2 33}{4}

Alternative Form

b_{1} \approx - 0.918189, b_{2} \approx 0.918189

Evaluate

b^{2} - 4 b^{3} \times b = - 2

Multiply

More Steps

b^{2} - 4 b^{4} = - 2

Move the expression to the left side

b^{2} - 4 b^{4} - (- 2) = 0

If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses

b^{2} - 4 b^{4} + 2 = 0

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

t - 4 t^{2} + 2 = 0

Rewrite in standard form

- 4 t^{2} + t + 2 = 0

Multiply both sides

4 t^{2} - t - 2 = 0

Substitute a= 4,b= - 1 and c= - 2 into the quadratic formula t = \frac{- b \pm b ^{2} - 4 a c}{2 a}

t = \frac{1 \pm ( - 1 ) ^{2} - 4 \times 4 ( - 2 )}{2 \times 4}

Simplify the expression

t = \frac{1 \pm ( - 1 ) ^{2} - 4 \times 4 ( - 2 )}{8}

Simplify the expression

More Steps

t = \frac{1 \pm 33}{8}

Separate the equation into 2 possible cases

t = \frac{1 + 33}{8} t = \frac{1 - 33}{8}

Substitute back

b^{2} = \frac{1 + 33}{8} b^{2} = \frac{1 - 33}{8}

Solve the equation for b

More Steps

b = \frac{2 + 2 33}{4} b = - \frac{2 + 2 33}{4} b^{2} = \frac{1 - 33}{8}

Since the left-hand side is always positive or 0,and the right-hand side is always negative,the statement is false for any value of b

b = \frac{2 + 2 33}{4} b = - \frac{2 + 2 33}{4} b \in / R

Find the union

b = \frac{2 + 2 33}{4} b = - \frac{2 + 2 33}{4}

Solution

b_{1} = - \frac{2 + 2 33}{4}, b_{2} = \frac{2 + 2 33}{4}

Alternative Form

b_{1} \approx - 0.918189, b_{2} \approx 0.918189

Show Solution

Solve the equation

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