Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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. Please come back anytime for the latest information and answers to your questions. Your visit means a lot to us. Don't hesitate to return 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.