
If \[\alpha \] and \[\beta \] are the roots of \[{x^2} - 3x + 4 = 0\] then \[\dfrac{1}{\alpha } + \dfrac{1}{\beta }\] is equal to ?
Answer
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Hint: Here, we are given a quadratic equation, and we need to find the value of \[\dfrac{1}{\alpha } + \dfrac{1}{\beta }\] where \[\alpha \] and \[\beta \] are the roots of the equation. We will compare this given equation with the standard form of quadratic equation i.e. \[a{x^2} + bx + c = 0\] , where a, b and c are real numbers and \[a \ne 0\] and ‘x’ is an unknown variable. Then we will need to find the sum of the roots (\[\alpha + \beta = \dfrac{{ - b}}{a}\]) and product of the roots (\[\alpha \beta = \dfrac{c}{a}\]). After using this, we will substitute in the given \[\dfrac{1}{\alpha } + \dfrac{1}{\beta }\] value to find the final output.
Complete step by step answer:
We are given that, \[\alpha \] and \[\beta \] are the roots of the equation:
\[{x^2} - 3x + 4 = 0\]
\[ \Rightarrow {x^2} + ( - 3)x + 4 = 0\] -------- (1)
We will know the standard form of the quadratic equation formula: \[a{x^2} + bx + c = 0\] , where $a, b$ and $c$ are real numbers and \[a \ne 0\] . Also, $x$ is a variable or unknown.
Now, we will compare the equation (1) with the standard form of the quadratic equation formula, we will get,
\[a = 1\], \[b = - 3\] and \[c = 4\]
First, sum of the roots are:
\[\alpha + \beta = \dfrac{{ - b}}{a}\]
\[ \Rightarrow \alpha + \beta = \dfrac{{ - ( - 3)}}{1}\]
\[ \Rightarrow \alpha + \beta = 3\]
Next, the product of the roots are:
\[\alpha \beta = \dfrac{c}{a}\]
\[ \Rightarrow \alpha \beta = \dfrac{4}{1}\]
\[ \Rightarrow \alpha \beta = 4\]
Now, the value of
\[\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\beta + \alpha }}{{\alpha \beta }}\]
\[ \Rightarrow \dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }}\]
\[\therefore \dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{3}{4}\]
Hence, for the given equation \[{x^2} - 3x + 4 = 0\] , the value of \[\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{3}{4}\].
Note: Quadratics or quadratic equations can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called an "Equation of Degree 2" (because of the "2" on the x). The solutions to the Quadratic Equation are where it is equal to zero. They are also called roots or sometimes zeroes. The roots of any polynomial are the solutions for the given equation.
Complete step by step answer:
We are given that, \[\alpha \] and \[\beta \] are the roots of the equation:
\[{x^2} - 3x + 4 = 0\]
\[ \Rightarrow {x^2} + ( - 3)x + 4 = 0\] -------- (1)
We will know the standard form of the quadratic equation formula: \[a{x^2} + bx + c = 0\] , where $a, b$ and $c$ are real numbers and \[a \ne 0\] . Also, $x$ is a variable or unknown.
Now, we will compare the equation (1) with the standard form of the quadratic equation formula, we will get,
\[a = 1\], \[b = - 3\] and \[c = 4\]
First, sum of the roots are:
\[\alpha + \beta = \dfrac{{ - b}}{a}\]
\[ \Rightarrow \alpha + \beta = \dfrac{{ - ( - 3)}}{1}\]
\[ \Rightarrow \alpha + \beta = 3\]
Next, the product of the roots are:
\[\alpha \beta = \dfrac{c}{a}\]
\[ \Rightarrow \alpha \beta = \dfrac{4}{1}\]
\[ \Rightarrow \alpha \beta = 4\]
Now, the value of
\[\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\beta + \alpha }}{{\alpha \beta }}\]
\[ \Rightarrow \dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{\alpha + \beta }}{{\alpha \beta }}\]
\[\therefore \dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{3}{4}\]
Hence, for the given equation \[{x^2} - 3x + 4 = 0\] , the value of \[\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{3}{4}\].
Note: Quadratics or quadratic equations can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called an "Equation of Degree 2" (because of the "2" on the x). The solutions to the Quadratic Equation are where it is equal to zero. They are also called roots or sometimes zeroes. The roots of any polynomial are the solutions for the given equation.
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