
What is the sum of the squares of the roots of \[\begin{array}{*{20}{c}}
{{x^2} - 3x + 1}& = &0
\end{array}\]
A. 5
B. 7
C. 9
D. 10
Answer
163.2k+ views
Hint:
In this question, we have given a quadratic equation whose roots are \[\alpha \] and \[\beta \]. To get the answer, first of all, we will determine the sum and product of the roots and then we will apply the formula of \[{\left( {\alpha + \beta } \right)^2}\]and hence, we will get our desired answer.
Formula used:
Sum of the roots:
\[\begin{array}{*{20}{c}}{\alpha + \beta }& = &{ - \frac{b}{a}}\end{array}\]
Product of the roots:
\[\begin{array}{*{20}{c}}{\alpha \beta }& = &{\frac{c}{a}}\end{array}\]
Square formula:
\[\begin{array}{*{20}{c}}{{{\left( {\alpha + \beta } \right)}^2}}& = &{{\alpha ^2} + {\beta ^2} + 2\alpha \beta }\end{array}\]
Complete step by step solution:
We have been provided in the question that the equation
\[{x^2} - 3x + 1 = 0\]
And we are asked to determine the sum of the squares of the roots of
\[{x^2} - 3x + 1 = 0\]
And the roots of this equation are \[\alpha \] and \[\beta \] respectively.
First, we have to determine the roots of the equation, we get
Therefore, we know the sum of the roots. Consequently, we can write it as,
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha + \beta }& = &3\end{array}\] ------- (1)
And the product of the roots is,
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha \beta }& = &1\end{array}\] ------ (2)
Now, we have to use the below formula to get the desired answer.
Therefore, we can write it as
\[ \Rightarrow \begin{array}{*{20}{c}}{{{\left( {\alpha + \beta } \right)}^2}}& = &{{\alpha ^2} + {\beta ^2} + 2\alpha \beta }\end{array}\]
Now, we have to rewrite it as,
\[\begin{array}{*{20}{c}}{ \Rightarrow {\alpha ^2} + {\beta ^2}}& = &{{{\left( {\alpha + \beta } \right)}^2} - }\end{array}2\alpha \beta \]
Now put the value of equations (1) and (2) in the above expression, and we will get
\[\begin{array}{*{20}{c}}{ \Rightarrow {\alpha ^2} + {\beta ^2}}& = &{{{\left( 3 \right)}^2} - }\end{array}2 \times \left( 1 \right)\]
Now, we will simplify the above expression.
\[\begin{array}{*{20}{c}}{ \Rightarrow {\alpha ^2} + {\beta ^2}}& = &{9 - 2}\end{array}\]
On simplifying the right side of the equation, we get
\[\begin{array}{*{20}{c}}{ \Rightarrow {\alpha ^2} + {\beta ^2}}& = &7\end{array}\]
Therefore, the sum of the squares of the roots of \[{x^2} - 3x + 1 = 0\]⟹ \[{\alpha ^2} + {\beta ^2} = 7\]
Hence, the correct option is B
Note:
In this problem, we have solved the given quadratic equation by using a square formula by finding the sum and product of the roots of the given equation. Students are likely to make mistakes in these types of problems because it includes some formulas to remember and one should keep in mind that signs are important in solving these types of problems as it leads to wrong solutions if not taken into consideration.
In this question, we have given a quadratic equation whose roots are \[\alpha \] and \[\beta \]. To get the answer, first of all, we will determine the sum and product of the roots and then we will apply the formula of \[{\left( {\alpha + \beta } \right)^2}\]and hence, we will get our desired answer.
Formula used:
Sum of the roots:
\[\begin{array}{*{20}{c}}{\alpha + \beta }& = &{ - \frac{b}{a}}\end{array}\]
Product of the roots:
\[\begin{array}{*{20}{c}}{\alpha \beta }& = &{\frac{c}{a}}\end{array}\]
Square formula:
\[\begin{array}{*{20}{c}}{{{\left( {\alpha + \beta } \right)}^2}}& = &{{\alpha ^2} + {\beta ^2} + 2\alpha \beta }\end{array}\]
Complete step by step solution:
We have been provided in the question that the equation
\[{x^2} - 3x + 1 = 0\]
And we are asked to determine the sum of the squares of the roots of
\[{x^2} - 3x + 1 = 0\]
And the roots of this equation are \[\alpha \] and \[\beta \] respectively.
First, we have to determine the roots of the equation, we get
Therefore, we know the sum of the roots. Consequently, we can write it as,
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha + \beta }& = &3\end{array}\] ------- (1)
And the product of the roots is,
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha \beta }& = &1\end{array}\] ------ (2)
Now, we have to use the below formula to get the desired answer.
Therefore, we can write it as
\[ \Rightarrow \begin{array}{*{20}{c}}{{{\left( {\alpha + \beta } \right)}^2}}& = &{{\alpha ^2} + {\beta ^2} + 2\alpha \beta }\end{array}\]
Now, we have to rewrite it as,
\[\begin{array}{*{20}{c}}{ \Rightarrow {\alpha ^2} + {\beta ^2}}& = &{{{\left( {\alpha + \beta } \right)}^2} - }\end{array}2\alpha \beta \]
Now put the value of equations (1) and (2) in the above expression, and we will get
\[\begin{array}{*{20}{c}}{ \Rightarrow {\alpha ^2} + {\beta ^2}}& = &{{{\left( 3 \right)}^2} - }\end{array}2 \times \left( 1 \right)\]
Now, we will simplify the above expression.
\[\begin{array}{*{20}{c}}{ \Rightarrow {\alpha ^2} + {\beta ^2}}& = &{9 - 2}\end{array}\]
On simplifying the right side of the equation, we get
\[\begin{array}{*{20}{c}}{ \Rightarrow {\alpha ^2} + {\beta ^2}}& = &7\end{array}\]
Therefore, the sum of the squares of the roots of \[{x^2} - 3x + 1 = 0\]⟹ \[{\alpha ^2} + {\beta ^2} = 7\]
Hence, the correct option is B
Note:
In this problem, we have solved the given quadratic equation by using a square formula by finding the sum and product of the roots of the given equation. Students are likely to make mistakes in these types of problems because it includes some formulas to remember and one should keep in mind that signs are important in solving these types of problems as it leads to wrong solutions if not taken into consideration.
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