Precalculus Examples

${(x^{2} - 17)}^{2} - 16 (x^{2} - 17) + 64 = 0$

Step 1

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

$u^{2} - 50 u + 625 = 0$

$u = x^{2}$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite $625$ as $25^{2}$ .

$u^{2} - 50 u + 25^{2} = 0$

Step 2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$50 u = 2 \cdot u \cdot 25$

Step 2.3

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 25 + 25^{2} = 0$

Step 2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 25$ .

${(u - 25)}^{2} = 0$

Step 3

Set the $u - 25$ equal to $0$ .

$u - 25 = 0$

Step 4

Add $25$ to both sides of the equation.

$u = 25$

Step 5

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

$x^{2} = 25$

Step 6

Solve the equation for $x$ .

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Step 6.1

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

$x = \pm \sqrt{25}$

Step 6.2

Simplify $\pm \sqrt{25}$ .

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Step 6.2.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 6.2.2

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

$x = \pm 5$

Step 6.3

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

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Step 6.3.1

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

$x = 5$

Step 6.3.2

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

$x = - 5$

Step 6.3.3

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

$x = 5, - 5$