Solve 1-6⋅|7x^4|<-4

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

x \in (- \infty, - \frac{4 5 \times 4 2 ^{3}}{42}) \cup (\frac{4 5 \times 4 2 ^{3}}{42}, + \infty)

Evaluate

1 - 6 7 x^{4} < - 4

Simplify

More Steps

1 - 42 x^{4} < - 4

Move the expression to the left side

1 - 42 x^{4} - (- 4) < 0

Subtract the numbers

More Steps

5 - 42 x^{4} < 0

Rewrite the expression

5 - 42 x^{4} = 0

Move the constant to the right-hand side and change its sign

- 42 x^{4} = 0 - 5

Removing 0 doesn't change the value,so remove it from the expression

- 42 x^{4} = - 5

Change the signs on both sides of the equation

42 x^{4} = 5

Divide both sides

\frac{42 x ^{4}}{42} = \frac{5}{42}

Divide the numbers

x^{4} = \frac{5}{42}

Take the root of both sides of the equation and remember to use both positive and negative roots

x = \pm 4 \frac{5}{42}

Simplify the expression

More Steps

x = \pm \frac{4 5 \times 4 2 ^{3}}{42}

Separate the equation into 2 possible cases

x = \frac{4 5 \times 4 2 ^{3}}{42} x = - \frac{4 5 \times 4 2 ^{3}}{42}

Determine the test intervals using the critical values

x < - \frac{4 5 \times 4 2 ^{3}}{42} - \frac{4 5 \times 4 2 ^{3}}{42} < x < \frac{4 5 \times 4 2 ^{3}}{42} x > \frac{4 5 \times 4 2 ^{3}}{42}

Choose a value form each interval

x_{1} = - 2 x_{2} = 0 x_{3} = 2

To determine if x < - \frac{4 5 \times 4 2 ^{3}}{42} is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

More Steps

x < - \frac{4 5 \times 4 2 ^{3}}{42} is the solution x_{2} = 0 x_{3} = 2

To determine if - \frac{4 5 \times 4 2 ^{3}}{42} < x < \frac{4 5 \times 4 2 ^{3}}{42} is the solution to the inequality,test if the chosen value x = 0 satisfies the initial inequality

More Steps

x < - \frac{4 5 \times 4 2 ^{3}}{42} is the solution - \frac{4 5 \times 4 2 ^{3}}{42} < x < \frac{4 5 \times 4 2 ^{3}}{42} is not a solution x_{3} = 2

To determine if x > \frac{4 5 \times 4 2 ^{3}}{42} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < - \frac{4 5 \times 4 2 ^{3}}{42} is the solution - \frac{4 5 \times 4 2 ^{3}}{42} < x < \frac{4 5 \times 4 2 ^{3}}{42} is not a solution x > \frac{4 5 \times 4 2 ^{3}}{42} is the solution

Solution

x \in (- \infty, - \frac{4 5 \times 4 2 ^{3}}{42}) \cup (\frac{4 5 \times 4 2 ^{3}}{42}, + \infty)

Show Solution