How do you solve #4x^4 - 16x^2 + 15 = 0#?

Question

Kennedy Adams · Accepted Answer

#+-sqrt(5/2)#
#+-sqrt(3/2)# for real coefficient equation equation of n‐th degree exist n roots so this equations exists 3 possible answers 1. two pairs of the complex conjugate of #a+bi# & #a-bi# 2. a pair of the complex conjugate of #a+bi# & #a-bi# and two real roots 3. four real roots #4x^4-16x^2+15=0# first I guess I can use "Cross method" to factorizative this equation it can be seen as below #(2x^2-5)(2x^2-3)=0# so there are four real roots #+-sqrt(5/2)# #+-sqrt(3/2)#

Isaiah Anthony · Answer

undefined To solve the equation 4x^4 - 16x^2 + 15 = 0, you can use substitution. Let y = x^2. Then the equation becomes a quadratic equation in terms of y: 4y^2 - 16y + 15 = 0. Factor or use the quadratic formula to solve for y. After finding the values of y, substitute them back into y = x^2 and solve for x to get the solutions.

HIX Tutor · Answer

undefined Great question! Let's solve the equation step by step.

1. **Look at the equation:** We have $4x^4 - 16x^2 + 15 = 0$.

2. **Make a substitution:** Let’s call $y = x^2$. So, $x^4$ becomes $y^2$. The equation now looks like this: 
   $$4y^2 - 16y + 15 = 0.$$

3. **Now, we have a quadratic equation:** This is $4y^2 - 16y + 15 = 0$.

4. **Use the quadratic formula:** The formula is:
   $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$
   where $a = 4$, $b = -16$, and $c = 15$.

5. **Calculate the discriminant:** First, find $b^2 - 4ac$:
   $$
   (-16)^2 - 4(4)(15) = 256 - 240 = 16.
   $$

6. **Insert into the formula:** Now calculate $y$:
   $$
   y = \frac{16 \pm \sqrt{16}}{2 \cdot 4} = \frac{16 \pm 4}{8}.
   $$

7. **Find the two possible values of $y$:**
   - First, $y = \frac{16 + 4}{8} = \frac{20}{8} = 2.5.$
   - Second, $y = \frac{16 - 4}{8} = \frac{12}{8} = 1.5.$

8. **Now, remember that $y = x^2$:** So we have:
   - $x^2 = 2.5$ 
   - $x^2 = 1.5$

9. **Take the square root of each:** 
   - For $x^2 = 2.5$, $x = \sqrt{2.5}$ or $x = -\sqrt{2.5}$.
   - For $x^2 = 1.5$, $x = \sqrt{1.5}$ or $x = -\sqrt{1.5}$.

10. **Final answer:** So the solutions are:
    - $x = \sqrt{2.5}, x = -\sqrt{2.5}, x = \sqrt{1.5}, x = -\sqrt{1.5}$.

Do you have any questions about the steps?

HIX Tutor · Answer

undefined Great question! Let’s solve it step by step.

1. **Substitute**: Notice that the equation has $x^4$ and $x^2$. Let’s use a substitution. Let’s say $y = x^2$. What happens to the equation if we substitute $y$ for $x^2$?

2. **Rewrite**: The equation $4x^4 - 16x^2 + 15 = 0$ changes to $4y^2 - 16y + 15 = 0$.

3. **Factor**: Now, let’s try to factor $4y^2 - 16y + 15$. Can you think of two numbers that multiply to $4 \times 15 = 60$ and add to $-16$?

4. **Find Numbers**: The numbers $-10$ and $-6$ work. So we can rewrite: $4y^2 - 10y - 6y + 15$.

5. **Group**: Now, group the terms: $(4y^2 - 10y) + (-6y + 15)$.

6. **Factor Again**: Factor each group:
   - From the first group, factor out $2y$: $2y(2y - 5)$.
   - From the second group, factor out $-3$: $-3(2y - 5)$.

7. **Combine**: Now we have:
   $$(2y - 5)(2y - 3) = 0$$.

8. **Solve for $y$**: Set each factor to zero:
   - $2y - 5 = 0$ gives $y = \frac{5}{2}$.
   - $2y - 3 = 0$ gives $y = \frac{3}{2}$.

9. **Back Substitute for $x$**: Remember that $y = x^2$. Now, set $x^2$ equal to each value:
   - For $y = \frac{5}{2}$, $x^2 = \frac{5}{2}$ gives $x = \pm\sqrt{\frac{5}{2}}$.
   - For $y = \frac{3}{2}$, $x^2 = \frac{3}{2}$ gives $x = \pm\sqrt{\frac{3}{2}}$.

10. **Final Answers**: So, the solutions for $x$ are:
   $$
   x = \sqrt{\frac{5}{2}}, \quad x = -\sqrt{\frac{5}{2}}, \quad x = \sqrt{\frac{3}{2}}, \quad x = -\sqrt{\frac{3}{2}}.
   $$

Do you see how that works?