The values of x such that \(2x^2 - 6x + 5 = 0\)
are \( m+ni\) and \(m-ni,\) where \(m\) and \(n\) are positive. What is \(m\cdot n?\)
+118608
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-b \over2a} \pm {\sqrt{b^2-4ac} \over 2a}\\~\\ m=\frac{-b}{2a}=\frac{--6}{2*2}=\frac{3}{2}\\ ni={\sqrt{b^2-4ac} \over 2a}={\sqrt{(-6)^2-4*2*5} \over 2*2}={\sqrt{36-40} \over 4}={2i \over 4}={1\over 2}i\\\ mn=\frac{3}{2}\times {1\over 2}=\frac{3}{4}\)
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+128406 In the form ax^2 + bx + c = 0
The sum of the roots will be = -b/a
And the product of the roots will be = c/a
Sum of the roots = 2m
Product of the roots = m^2 + n^2
So
2m = 6/2 ⇒ m = 3/2
And
m^2 + n^2 = 5/2 ⇒ (9/4) + n^2 = 5/2 ⇒ n^2 = 10/4 - 9/4 = 1/4 ⇒ n = 1/2
So
m * n = (3/2)(1/2) = 3/4
