Solve 121x^2>=25 - ScanMath

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

Solve for x

x \in (- \infty, - \frac{5}{11}] \cup [\frac{5}{11}, + \infty)

Evaluate

121 x^{2} \geq 25

Move the expression to the left side

121 x^{2} - 25 \geq 0

Rewrite the expression

121 x^{2} - 25 = 0

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

121 x^{2} = 0 + 25

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

121 x^{2} = 25

Divide both sides

\frac{121 x ^{2}}{121} = \frac{25}{121}

Divide the numbers

x^{2} = \frac{25}{121}

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

x = \pm \frac{25}{121}

Simplify the expression

More Steps

x = \pm \frac{5}{11}

Separate the equation into 2 possible cases

x = \frac{5}{11} x = - \frac{5}{11}

Determine the test intervals using the critical values

x < - \frac{5}{11} - \frac{5}{11} < x < \frac{5}{11} x > \frac{5}{11}

Choose a value form each interval

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

To determine if x < - \frac{5}{11} is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

More Steps

x < - \frac{5}{11} is the solution x_{2} = 0 x_{3} = 2

To determine if - \frac{5}{11} < x < \frac{5}{11} is the solution to the inequality,test if the chosen value x = 0 satisfies the initial inequality

More Steps

x < - \frac{5}{11} is the solution - \frac{5}{11} < x < \frac{5}{11} is not a solution x_{3} = 2

To determine if x > \frac{5}{11} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < - \frac{5}{11} is the solution - \frac{5}{11} < x < \frac{5}{11} is not a solution x > \frac{5}{11} is the solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

x \leq - \frac{5}{11} is the solution x \geq \frac{5}{11} is the solution

Solution

x \in (- \infty, - \frac{5}{11}] \cup [\frac{5}{11}, + \infty)

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