How do you write the quadratic equation with integer coefficients that have the roots 7-3i and 7+3i?
In the quadratic equation
ax²+bx+c=0
b/a = -(sum of the two roots)
c/a = product of the two roots.
Assume a=1, then
b=-(7-3i+7+3i)=-14
c=(7-3i)*(7+3i)=49+9=58
So the quadratic equation is
x²-14x+58=0
Check by solving the equation and you should get back 7±3i as the roots.
x = 7 + 3i, and X = 7 - 3i.
We can use either of the 2 values of X:
X = 7 + 3i,
(X - 7) = 3i,
Square both sides:
X^2 --14X + 49 = 9(-1),
X^2 - 14X + 49 +9 = 0,
X^2 - 14X + 58 = 0.
To write the quadratic equation with integer coefficients that have the roots 7-3i and 7+3i, you can use the fact that complex roots always come in conjugate pairs.
Step 1: Identify the roots.
The given roots are:
Root 1: 7-3i
Root 2: 7+3i
Step 2: Write the quadratic equation.
To find the quadratic equation, you need to use the fact that the sum and product of the roots can help in determining the coefficients of the quadratic equation. Since the roots are 7-3i and 7+3i, the sum of the roots can be calculated as:
Sum of the roots = (Root 1) + (Root 2)
= (7-3i) + (7+3i)
= 14
The product of the roots can be calculated as:
Product of the roots = (Root 1) * (Root 2)
= (7-3i) * (7+3i)
= 49 - 9i^2
= 49 + 9
= 58
Step 3: Use the sum and product of the roots to write the quadratic equation.
The quadratic equation with integer coefficients that has the given roots can be written as:
(x - Root 1) * (x - Root 2) = 0
(x - 7-3i) * (x - 7+3i) = 0
(x - 7+3i)(x - 7-3i) = 0
(x - 7)^2 - (3i)^2 = 0
(x - 7)^2 - 9i^2 = 0
(x - 7)^2 - 9(-1) = 0
(x - 7)^2 + 9 = 0
So, the quadratic equation with integer coefficients that has the roots 7-3i and 7+3i is
(x - 7)^2 + 9 = 0.
To write the quadratic equation with integer coefficients that has the roots 7-3i and 7+3i, you can use the fact that complex roots of quadratic equations always come in conjugate pairs. In other words, if one root is a+bi, then the other root must be a-bi.
In this case, the roots are 7-3i and 7+3i. Since these two roots are conjugates of each other, we can use the property that the product of the roots of a quadratic equation is equal to the constant term divided by the leading coefficient.
Let's denote the quadratic equation as Ax^2 + Bx + C = 0, where A, B, and C are integers. Since the roots are 7-3i and 7+3i, the product of the roots is (7-3i)(7+3i).
To simplify this expression, we can use the difference of squares formula: (a-b)(a+b) = a^2 - b^2. In this case, a = 7 and b = 3i.
(7-3i)(7+3i) = 7^2 - (3i)^2 = 49 - 9i^2.
Now, remember that i^2 is equal to -1. So, substituting i^2 = -1 into our equation:
49 - 9i^2 = 49 - 9(-1) = 49 + 9 = 58.
Therefore, the product of the roots is 58.
Now, let's divide the product of the roots by the leading coefficient A to get the constant term C:
C = 58 / A.
Since we want to find a quadratic equation with integer coefficients, we need to ensure that C is an integer. To do that, we need to choose an appropriate value for A.
We can choose any integer value for A that divides 58 evenly. For simplicity, let's choose A = 1.
C = 58 / 1 = 58.
Now that we have the constant term C, we can find the sum of the roots by using the fact that the sum of the roots of a quadratic equation is equal to the opposite of the coefficient of the linear term (B) divided by the leading coefficient (A).
Since A = 1, the sum of the roots is -B/1 = -B.
The sum of the roots, in this case, is 7-3i + 7+3i = 14. And since the sum of the roots is equal to -B, we have:
14 = -B.
Solving for B, we find that B = -14.
Now we have A = 1, B = -14, and C = 58. Plugging these values into the general form of a quadratic equation, we get:
x^2 - 14x + 58 = 0.
Therefore, the quadratic equation with integer coefficients that has the roots 7-3i and 7+3i is x^2 - 14x + 58 = 0.