Which equation has the complex number 4-3i as a root?
(1) x^2+6x-25=0
(2) x^2-6x+25=0
(3) x^2+8x-25=0
(4) x^2-8x+25=0
I have no clue how to do it!!!PLEASE HELP!!!Please guide by step by step!!THANKS A LOT~!!!!
Look at the serds. To get a serd that is negative, you have to have a or c negative. That means only 1 or 3 can be correct. Now, which yields sqrt (b^2-4ac) to be 6i ?
I still don't get it
Sure! I can help you with this question. To determine which equation has the complex number 4-3i as a root, we can use the fact that complex roots always occur in conjugate pairs.
Step 1: Write the complex number in the form of a + bi.
4 - 3i can be written as (4 + (-3i)).
Step 2: Take the conjugate of the complex number.
The conjugate of (4 + (-3i)) is (4 + 3i).
Step 3: Write the equation using the complex number and its conjugate as the roots.
Since complex roots occur in conjugate pairs, the equation can be written as:
(x - (4 - 3i))(x - (4 + 3i)) = 0
Step 4: Simplify the equation.
Expanding the equation, we have:
(x - 4 + 3i)(x - 4 - 3i) = 0
(x - 4)^2 - (3i)^2 = 0
(x - 4)^2 - 9i^2 = 0
(x - 4)^2 + 9 = 0
Now, let's compare this equation to the given options:
(1) x^2 + 6x - 25 = 0
(2) x^2 - 6x + 25 = 0
(3) x^2 + 8x - 25 = 0
(4) x^2 - 8x + 25 = 0
Step 5: Compare the equations.
From the equation we obtained, (x - 4)^2 + 9 = 0, we can see that none of the given options match this equation. Therefore, the answer is none of the given options. The correct equation is not among the choices provided.
I hope this explanation was helpful to you! Let me know if you have any more questions.