Answer :
To solve the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] by factoring, let's go through the process step by step.
1. Substitute [tex]\( y = x^2 \)[/tex]:
This substitution simplifies the equation. Let [tex]\( y = x^2 \)[/tex], then the equation becomes:
[tex]\[ y^2 - 5y - 14 = 0 \][/tex]
2. Solve the quadratic equation [tex]\( y^2 - 5y - 14 = 0 \)[/tex]:
This is a standard quadratic equation. To solve it, we can use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -14 \)[/tex]. Plugging these values into the quadratic formula gives:
[tex]\[ y = \frac{5 \pm \sqrt{25 + 56}}{2} \][/tex]
[tex]\[ y = \frac{5 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ y = \frac{5 \pm 9}{2} \][/tex]
3. Calculate the solutions for [tex]\( y \)[/tex]:
There are two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{5 + 9}{2} = \frac{14}{2} = 7 \][/tex]
and
[tex]\[ y = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \][/tex]
4. Back-substitute [tex]\( y = x^2 \)[/tex] to find [tex]\( x \)[/tex]:
Recall that [tex]\( y = x^2 \)[/tex]. So we have:
[tex]\[ x^2 = 7 \][/tex]
and
[tex]\[ x^2 = -2 \][/tex]
Solving these equations for [tex]\( x \)[/tex]:
- For [tex]\( x^2 = 7 \)[/tex]:
[tex]\[ x = \pm \sqrt{7} \][/tex]
- For [tex]\( x^2 = -2 \)[/tex]:
[tex]\[ x = \pm \sqrt{-2} = \pm \sqrt{2}i \][/tex]
5. Combine all solutions:
The solutions to the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] are:
[tex]\[ x = \pm \sqrt{7} \quad \text{and} \quad x = \pm \sqrt{2}i \][/tex]
So, the correct answer is:
[tex]\[ x = \pm \sqrt{7} \quad \text{and} \quad x = \pm i \sqrt{2} \][/tex]
1. Substitute [tex]\( y = x^2 \)[/tex]:
This substitution simplifies the equation. Let [tex]\( y = x^2 \)[/tex], then the equation becomes:
[tex]\[ y^2 - 5y - 14 = 0 \][/tex]
2. Solve the quadratic equation [tex]\( y^2 - 5y - 14 = 0 \)[/tex]:
This is a standard quadratic equation. To solve it, we can use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -14 \)[/tex]. Plugging these values into the quadratic formula gives:
[tex]\[ y = \frac{5 \pm \sqrt{25 + 56}}{2} \][/tex]
[tex]\[ y = \frac{5 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ y = \frac{5 \pm 9}{2} \][/tex]
3. Calculate the solutions for [tex]\( y \)[/tex]:
There are two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{5 + 9}{2} = \frac{14}{2} = 7 \][/tex]
and
[tex]\[ y = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \][/tex]
4. Back-substitute [tex]\( y = x^2 \)[/tex] to find [tex]\( x \)[/tex]:
Recall that [tex]\( y = x^2 \)[/tex]. So we have:
[tex]\[ x^2 = 7 \][/tex]
and
[tex]\[ x^2 = -2 \][/tex]
Solving these equations for [tex]\( x \)[/tex]:
- For [tex]\( x^2 = 7 \)[/tex]:
[tex]\[ x = \pm \sqrt{7} \][/tex]
- For [tex]\( x^2 = -2 \)[/tex]:
[tex]\[ x = \pm \sqrt{-2} = \pm \sqrt{2}i \][/tex]
5. Combine all solutions:
The solutions to the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] are:
[tex]\[ x = \pm \sqrt{7} \quad \text{and} \quad x = \pm \sqrt{2}i \][/tex]
So, the correct answer is:
[tex]\[ x = \pm \sqrt{7} \quad \text{and} \quad x = \pm i \sqrt{2} \][/tex]
