Find the Roots (Zeros) 2x^4-9x^2+4

$2 x^{4} - 9 x^{2} + 4$

Step 1

Set $2 x^{4} - 9 x^{2} + 4$ equal to $0$ .

$2 x^{4} - 9 x^{2} + 4 = 0$

Step 2

Solve for $x$ .

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$2 u^{2} - 9 u + 4 = 0$

$u = x^{2}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 4 = 8$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 u$ .

$2 u^{2} - 9 u + 4 = 0$

Rewrite $- 9$ as $- 1$ plus $- 8$

$2 u^{2} + (- 1 - 8) u + 4 = 0$

Apply the distributive property.

$2 u^{2} - 1 u - 8 u + 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 u^{2} - 1 u) - 8 u + 4 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (2 u - 1) - 4 (2 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

$(2 u - 1) (u - 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 u - 1 = 0$

$u - 4 = 0$

Set $2 u - 1$ equal to $0$ and solve for $u$ .

Set $2 u - 1$ equal to $0$ .

$2 u - 1 = 0$

Solve $2 u - 1 = 0$ for $u$ .

Add $1$ to both sides of the equation.

$2 u = 1$

Divide each term in $2 u = 1$ by $2$ and simplify.

Divide each term in $2 u = 1$ by $2$ .

$\frac{2 u}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 u}{2} = \frac{1}{2}$

Divide $u$ by $1$ .

$u = \frac{1}{2}$

Set $u - 4$ equal to $0$ and solve for $u$ .

Set $u - 4$ equal to $0$ .

$u - 4 = 0$

Add $4$ to both sides of the equation.

$u = 4$

The final solution is all the values that make $(2 u - 1) (u - 4) = 0$ true.

$u = \frac{1}{2}, 4$

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = \frac{1}{2}$

${(x^{2})}^{1} = 4$

Solve the first equation for $x$ .

$x^{2} = \frac{1}{2}$

Solve the equation for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{\frac{1}{2}}$

Simplify $\pm \sqrt{\frac{1}{2}}$ .

Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{2}}$

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{2}}$

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Add $1$ and $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$x = \pm \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \pm \frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$x = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \pm \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

$x = \pm \frac{\sqrt{2}}{2^{1}}$

Evaluate the exponent.

$x = \pm \frac{\sqrt{2}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{\sqrt{2}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \frac{\sqrt{2}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Solve the second equation for $x$ .

${(x^{2})}^{1} = 4$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

The solution to $2 x^{4} - 9 x^{2} + 4 = 0$ is $x = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}, 2, - 2$ .

$x = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}, 2, - 2$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}, 2, - 2$

Decimal Form:

$x = 0.70710678 \dots, - 0.70710678 \dots, 2, - 2$

Step 4

Find the Roots (Zeros) 2x^4-9x^2+4

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