
Solve: \[{x^2} - (\sqrt 2 + i)x + \sqrt 2 i = 0\]
Answer
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Hint:The quadratic equation will always have two roots. The nature of roots may be either real or imaginary. Here, we are given a quadratic equation and we need to find the roots of the given equation. So first we will solve this by opening the brackets and solving this, we will get the two roots. Also, we can solve this by using quadratic polynomial equations i.e. \[{x^2} - (\alpha + \beta )x + \alpha \beta = 0\] and comparing this equation with the given equation we will get the final output.
Complete step by step answer:
Given that, we are given a quadratic equation as below:
\[{x^2} - (\sqrt 2 + i)x + \sqrt 2 i = 0\]
Removing the brackets, we will get,
\[ \Rightarrow {x^2} - \sqrt 2 x - ix + \sqrt 2 i = 0\]
Rearranging this equation, we will get,
\[ \Rightarrow {x^2} - ix - \sqrt 2 x + \sqrt 2 i = 0\]
\[ \Rightarrow x(x - i) - \sqrt 2 (x - i) = 0\]
\[ \Rightarrow (x - \sqrt 2 )(x - i) = 0\]
\[ \therefore x = \sqrt 2 \] and \[x = i\]
Another Method: We know that, a quadratic polynomial, when equated to zero, becomes a quadratic equation.Thus, equation of the quadratic polynomial is
\[{x^2} - (\alpha + \beta )x + \alpha \beta = 0\] where \[\alpha ,\beta \] are the roots or zeros of the equation. We are given the quadratic equation as below
\[{x^2} - (\sqrt 2 + i)x + \sqrt 2 i = 0\]
Thus, comparing both the equations, we will get,
\[\alpha + \beta = \sqrt 2 + i\] and \[\alpha \beta = \sqrt 2 i\]
\[\therefore \alpha = \sqrt 2 \] and \[\beta = i\]
Hence, for the given quadratic equation \[{x^2} - (\sqrt 2 + i)x + \sqrt 2 i = 0\] we get two values as \[x = \sqrt 2 \]and \[x = i\].
Note: Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = \[a{x^2} + bx + c\] where \[a,b,c \in R\] and \[a \ne 0\]. If the roots of this equation are given by α and β, then there exists a relationship between the roots of the quadratic equation and the coefficients a, b and c. Thus, the sum of roots of a quadratic equation is given by the negative ratio of coefficient of x and x2. The product of roots is given by the ratio of the constant term and the coefficient of x2. We know that the graph of a quadratic function is represented using a parabola.
Complete step by step answer:
Given that, we are given a quadratic equation as below:
\[{x^2} - (\sqrt 2 + i)x + \sqrt 2 i = 0\]
Removing the brackets, we will get,
\[ \Rightarrow {x^2} - \sqrt 2 x - ix + \sqrt 2 i = 0\]
Rearranging this equation, we will get,
\[ \Rightarrow {x^2} - ix - \sqrt 2 x + \sqrt 2 i = 0\]
\[ \Rightarrow x(x - i) - \sqrt 2 (x - i) = 0\]
\[ \Rightarrow (x - \sqrt 2 )(x - i) = 0\]
\[ \therefore x = \sqrt 2 \] and \[x = i\]
Another Method: We know that, a quadratic polynomial, when equated to zero, becomes a quadratic equation.Thus, equation of the quadratic polynomial is
\[{x^2} - (\alpha + \beta )x + \alpha \beta = 0\] where \[\alpha ,\beta \] are the roots or zeros of the equation. We are given the quadratic equation as below
\[{x^2} - (\sqrt 2 + i)x + \sqrt 2 i = 0\]
Thus, comparing both the equations, we will get,
\[\alpha + \beta = \sqrt 2 + i\] and \[\alpha \beta = \sqrt 2 i\]
\[\therefore \alpha = \sqrt 2 \] and \[\beta = i\]
Hence, for the given quadratic equation \[{x^2} - (\sqrt 2 + i)x + \sqrt 2 i = 0\] we get two values as \[x = \sqrt 2 \]and \[x = i\].
Note: Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = \[a{x^2} + bx + c\] where \[a,b,c \in R\] and \[a \ne 0\]. If the roots of this equation are given by α and β, then there exists a relationship between the roots of the quadratic equation and the coefficients a, b and c. Thus, the sum of roots of a quadratic equation is given by the negative ratio of coefficient of x and x2. The product of roots is given by the ratio of the constant term and the coefficient of x2. We know that the graph of a quadratic function is represented using a parabola.
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