Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
To find the solutions of the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex], we can follow a systematic, step-by-step approach.
1. Substitute [tex]\( u = x^2 \)[/tex]:
This substitution reduces the quartic equation to a quadratic form. We substitute [tex]\( u = x^2 \)[/tex], transforming the equation into:
[tex]\[ u^2 - 5u - 14 = 0 \][/tex]
2. Solve the quadratic equation [tex]\( u^2 - 5u - 14 = 0 \)[/tex]:
To solve this quadratic equation, we use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -14 \)[/tex].
First, we calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-5)^2 - 4(1)(-14) = 25 + 56 = 81 \][/tex]
Now, we find the roots:
[tex]\[ u_1 = \frac{5 + \sqrt{81}}{2 \cdot 1} = \frac{5 + 9}{2} = \frac{14}{2} = 7 \][/tex]
[tex]\[ u_2 = \frac{5 - \sqrt{81}}{2 \cdot 1} = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \][/tex]
3. Back-substitute [tex]\( u = x^2 \)[/tex]:
We revert back to [tex]\( x \)[/tex] from [tex]\( u \)[/tex] by solving the equations [tex]\( x^2 = u_1 \)[/tex] and [tex]\( x^2 = u_2 \)[/tex].
- For [tex]\( x^2 = 7 \)[/tex]:
[tex]\[ x = \pm \sqrt{7} \][/tex]
- For [tex]\( x^2 = -2 \)[/tex]:
Since [tex]\( x \)[/tex] must be a complex number to satisfy this equation, we have:
[tex]\[ x = \pm \sqrt{-2} = \pm i \sqrt{2} \][/tex]
Thus, the solutions to the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] are:
[tex]\[ x = \pm \sqrt{7}, \pm i \sqrt{2} \][/tex]
The correct solution is:
[tex]\[ x= \pm \sqrt{7} \, \text{and} \, x= \pm i \sqrt{2} \][/tex]
1. Substitute [tex]\( u = x^2 \)[/tex]:
This substitution reduces the quartic equation to a quadratic form. We substitute [tex]\( u = x^2 \)[/tex], transforming the equation into:
[tex]\[ u^2 - 5u - 14 = 0 \][/tex]
2. Solve the quadratic equation [tex]\( u^2 - 5u - 14 = 0 \)[/tex]:
To solve this quadratic equation, we use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = -14 \)[/tex].
First, we calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-5)^2 - 4(1)(-14) = 25 + 56 = 81 \][/tex]
Now, we find the roots:
[tex]\[ u_1 = \frac{5 + \sqrt{81}}{2 \cdot 1} = \frac{5 + 9}{2} = \frac{14}{2} = 7 \][/tex]
[tex]\[ u_2 = \frac{5 - \sqrt{81}}{2 \cdot 1} = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \][/tex]
3. Back-substitute [tex]\( u = x^2 \)[/tex]:
We revert back to [tex]\( x \)[/tex] from [tex]\( u \)[/tex] by solving the equations [tex]\( x^2 = u_1 \)[/tex] and [tex]\( x^2 = u_2 \)[/tex].
- For [tex]\( x^2 = 7 \)[/tex]:
[tex]\[ x = \pm \sqrt{7} \][/tex]
- For [tex]\( x^2 = -2 \)[/tex]:
Since [tex]\( x \)[/tex] must be a complex number to satisfy this equation, we have:
[tex]\[ x = \pm \sqrt{-2} = \pm i \sqrt{2} \][/tex]
Thus, the solutions to the equation [tex]\( x^4 - 5x^2 - 14 = 0 \)[/tex] are:
[tex]\[ x = \pm \sqrt{7}, \pm i \sqrt{2} \][/tex]
The correct solution is:
[tex]\[ x= \pm \sqrt{7} \, \text{and} \, x= \pm i \sqrt{2} \][/tex]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.