Math / Algebra

QuestionFind the real solutions of the equation. $6 x^{4}-x^{2}-1=0$
What is the solution set? Select the correct choice below and fill in any answer boxes in your choice. A. $\square$ (Use a comma to separate answers, as needed. Rationalize all denominators. Type an integer or a simplified fraction.) B. There are no real solutions.

Studdy Solution

STEP 1

1. We are given the equation $6x^4 - x^2 - 1 = 0$ .
2. We need to find the real solutions of this equation.

STEP 2

1. Perform a substitution to simplify the equation.
2. Solve the resulting quadratic equation.
3. Back-substitute to find the values of $x$ .
4. Determine the real solutions.

STEP 3

Perform a substitution to simplify the equation. Let $y = x^2$ . Then the equation becomes:
$6y^2 - y - 1 = 0$

STEP 4

Solve the quadratic equation $6y^2 - y - 1 = 0$ using the quadratic formula:
The quadratic formula is given by:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 6$ , $b = -1$ , and $c = -1$ .

STEP 5

Calculate the discriminant $b^2 - 4ac$ :
$(-1)^2 - 4(6)(-1) = 1 + 24 = 25$

STEP 6

Substitute the values into the quadratic formula:
$y = \frac{-(-1) \pm \sqrt{25}}{2 \times 6}$ $y = \frac{1 \pm 5}{12}$

STEP 7

Calculate the two possible values for $y$ :
$y = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2}$
$y = \frac{1 - 5}{12} = \frac{-4}{12} = -\frac{1}{3}$

STEP 8

Back-substitute to find the values of $x$ .
For $y = \frac{1}{2}$ :
$x^2 = \frac{1}{2}$ $x = \pm \sqrt{\frac{1}{2}}$ $x = \pm \frac{\sqrt{2}}{2}$
For $y = -\frac{1}{3}$ :
Since $x^2 = -\frac{1}{3}$ has no real solutions (as a square cannot be negative), we discard this solution.

STEP 9

Determine the real solutions. The real solutions are:
$x = \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$
The solution set is:
A. $\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$

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Studdy Solution

STEP 1

1. We are given the equation $6x^4 - x^2 - 1 = 0$ .
2. We need to find the real solutions of this equation.

STEP 2

1. Perform a substitution to simplify the equation.
2. Solve the resulting quadratic equation.
3. Back-substitute to find the values of $x$ .
4. Determine the real solutions.

STEP 3

Perform a substitution to simplify the equation. Let $y = x^2$ . Then the equation becomes:
$6y^2 - y - 1 = 0$

STEP 4

Solve the quadratic equation $6y^2 - y - 1 = 0$ using the quadratic formula:
The quadratic formula is given by:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 6$ , $b = -1$ , and $c = -1$ .

STEP 5

Calculate the discriminant $b^2 - 4ac$ :
$(-1)^2 - 4(6)(-1) = 1 + 24 = 25$

STEP 6

Substitute the values into the quadratic formula:
$y = \frac{-(-1) \pm \sqrt{25}}{2 \times 6}$ $y = \frac{1 \pm 5}{12}$

STEP 7

Calculate the two possible values for $y$ :
$y = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2}$
$y = \frac{1 - 5}{12} = \frac{-4}{12} = -\frac{1}{3}$

STEP 8

Back-substitute to find the values of $x$ .
For $y = \frac{1}{2}$ :
$x^2 = \frac{1}{2}$ $x = \pm \sqrt{\frac{1}{2}}$ $x = \pm \frac{\sqrt{2}}{2}$
For $y = -\frac{1}{3}$ :
Since $x^2 = -\frac{1}{3}$ has no real solutions (as a square cannot be negative), we discard this solution.

STEP 9

Determine the real solutions. The real solutions are:
$x = \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$
The solution set is:
A. $\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$

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Studdy Solution

STEP 1

1. We are given the equation 6x4−x2−1=0 6x^4 - x^2 - 1 = 0 6x4−x2−1=0.2. We need to find the real solutions of this equation.

STEP 2

1. Perform a substitution to simplify the equation.2. Solve the resulting quadratic equation.3. Back-substitute to find the values of x x x.4. Determine the real solutions.

STEP 3

Perform a substitution to simplify the equation. Let y=x2 y = x^2 y=x2. Then the equation becomes:6y2−y−1=0 6y^2 - y - 1 = 0 6y2−y−1=0

STEP 4

Solve the quadratic equation 6y2−y−1=0 6y^2 - y - 1 = 0 6y2−y−1=0 using the quadratic formula:The quadratic formula is given by:y=−b±b2−4ac2a y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} y=2a−b±b2−4ac​​where a=6 a = 6 a=6, b=−1 b = -1 b=−1, and c=−1 c = -1 c=−1.

STEP 5

Calculate the discriminant b2−4ac b^2 - 4ac b2−4ac:(−1)2−4(6)(−1)=1+24=25 (-1)^2 - 4(6)(-1) = 1 + 24 = 25 (−1)2−4(6)(−1)=1+24=25

STEP 6

Substitute the values into the quadratic formula:y=−(−1)±252×6 y = \frac{-(-1) \pm \sqrt{25}}{2 \times 6} y=2×6−(−1)±25​​ y=1±512 y = \frac{1 \pm 5}{12} y=121±5​

STEP 7

Calculate the two possible values for y y y:y=1+512=612=12 y = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2} y=121+5​=126​=21​y=1−512=−412=−13 y = \frac{1 - 5}{12} = \frac{-4}{12} = -\frac{1}{3} y=121−5​=12−4​=−31​

STEP 8

STEP 9

Determine the real solutions. The real solutions are:x=22,−22 x = \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} x=22​​,−22​​ The solution set is:A. 22,−22 \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} 22​​,−22​​

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1. We are given the equation $6x^4 - x^2 - 1 = 0$ .
2. We need to find the real solutions of this equation.

1. Perform a substitution to simplify the equation.
2. Solve the resulting quadratic equation.
3. Back-substitute to find the values of $x$ .
4. Determine the real solutions.

Perform a substitution to simplify the equation. Let $y = x^2$ . Then the equation becomes:
$6y^2 - y - 1 = 0$

Solve the quadratic equation $6y^2 - y - 1 = 0$ using the quadratic formula:
The quadratic formula is given by:
$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 6$ , $b = -1$ , and $c = -1$ .

Calculate the discriminant $b^2 - 4ac$ :
$(-1)^2 - 4(6)(-1) = 1 + 24 = 25$

Substitute the values into the quadratic formula:
$y = \frac{-(-1) \pm \sqrt{25}}{2 \times 6}$ $y = \frac{1 \pm 5}{12}$

Calculate the two possible values for $y$ :
$y = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2}$
$y = \frac{1 - 5}{12} = \frac{-4}{12} = -\frac{1}{3}$

Determine the real solutions. The real solutions are:
$x = \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$
The solution set is:
A. $\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$