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Question: How do you write a quadratic equation in standard form with solutions \(2 + 3i,\;{\text{2 - 3i}}\)?...

How do you write a quadratic equation in standard form with solutions 2+3i,  2 - 3i2 + 3i,\;{\text{2 - 3i}}?

Explanation

Solution

Let us suppose the given two roots to be α\alpha and β\beta and then find the product and sum of the roots and then place in the standard formula, x2−(α+β)x+αβ=0{x^2} - (\alpha + \beta )x + \alpha \beta = 0and then simplify for the resultant required value.

Complete step by step answer:
Let us assume that α=2+3i\alpha = 2 + 3iand β=2−3i\beta = 2 - 3i
Now, the product of roots can be given as –
αβ=(2+3i)(2−3i)\alpha \beta = (2 + 3i)(2 - 3i)
Simplify the above equation using the identity for the difference of two squares, a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b)
αβ=(2)2−(3i)2\alpha \beta = {(2)^2} - {(3i)^2}
Simplify the above equation –
αβ=4−9i2\alpha \beta = 4 - 9{i^2}
Place i2=(−1){i^2} = ( - 1)in the above equation
αβ=4−9(−1)\alpha \beta = 4 - 9( - 1)
Simplify the above equation –
αβ=4+9 αβ=13   ..... (A)  \alpha \beta = 4 + 9 \\\ \alpha \beta = 13\;{\text{ }}.....{\text{ (A)}} \\\
Now, find the sum of the roots
α+β=2+3i+2−3i\alpha + \beta = 2 + 3i + 2 - 3i
Like terms with the same value and the opposite sign cancels each other –
α+β=4\alpha + \beta = 4 ….. (B)
Now, place the values of the equation (A) and (B) in the standard form
x2−(α+β)x+αβ=0{x^2} - (\alpha + \beta )x + \alpha \beta = 0
x2−4x+13=0{x^2} - 4x + 13 = 0
This is the required solution.

Note: Always remember the standard equation and formula properly. Follow the given data and conditions carefully. A quadratic equation is an equation of second degree, it means at least one of the terms is squared. Standard equation is ax2+bx+c=0a{x^2} + bx + c = 0 where a,b,and c are constant and “a '' can never be zero and “x” is unknown. : Be careful regarding the sign convention. Always remember that the square of negative number or the positive number is always positive. Also, product of two negative numbers is always positive whereas, product of one positive and one negative number gives us the negative number.
Every quadratic polynomial has almost two roots. Remember, the quadratic equations having coefficients as the rational numbers has the irrational roots. Also, the quadratic equations whose coefficients are all the distinct irrationals but both the roots are the rational.