Question

If \[a\] and \[b\] are the roots of the equation \[{x^2} - 6x + 6 = 0\], then the value of \[{a^2} + {b^2}\] are:
A.6
B.12
C.24
D.36

Answer 1

Hint: Here, we have to find the sum of square of the roots of an equation. First, we have to find the roots of a quadratic equation. Then we have to find the sum of squares of the roots of the quadratic equation. A quadratic equation is a polynomial equation with the highest power as 2.

Formula used: Roots of the quadratic equation is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
where a, b, c are the coefficients of \[{x^2},x\]and the constant respectively.
The square of the sum of two numbers a, b is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
The square of the difference of two numbers a, b is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]

Complete step-by-step answer:
We are given with the quadratic equation \[{x^2} - 6x + 6 = 0\].
The quadratic equation is of the form \[a{x^2} + bx + c = 0\].
Comparing with the general quadratic equation, we have
a\[ = \]1; b \[ = \]-6; c\[ = \]6;
Now, we have to find the roots of the quadratic equation.
Roots of the quadratic equation is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting the values of a\[ = \]1; b \[ = \]-6; c\[ = \]6, we have
\[ \Rightarrow x = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(1)(6)} }}{{2(1)}}\]
The multiplication of two negative integers is negative.
Simplifying the equation, we have
\[ \Rightarrow x = \dfrac{{ + 6 \pm \sqrt {36 - 24} }}{2}\]
\[ \Rightarrow x = \dfrac{{6 \pm \sqrt {12} }}{2}\]
Simplifying the equation, we have
\[ \Rightarrow x = \dfrac{{6 \pm \sqrt {4 \times 3} }}{2}\]
\[ \Rightarrow x = \dfrac{{6 \pm 2\sqrt 3 }}{2}\]
\[ \Rightarrow x = \dfrac{{2(3 \pm \sqrt 3 )}}{2}\]
\[ \Rightarrow x = 3 \pm \sqrt 3 \]
\[ \Rightarrow x = 3 + \sqrt 3 ;x = 3 - \sqrt 3 \]
The roots of the quadratic equation are \[a = 3 + \sqrt 3 ;b = 3 - \sqrt 3 \];
Now, we have to find the sum of square of the roots of the equation
Since a, b are the roots of the equation, we have
\[ \Rightarrow {a^2} + {b^2} = {\left( {3 + \sqrt 3 } \right)^2} + {\left( {3 - \sqrt 3 } \right)^2}\]
The square of the sum of two numbers a, b is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
The square of the difference of two numbers a, b is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
\[ \Rightarrow {a^2} + {b^2} = \left( {{3^2} + {{(\sqrt 3 )}^2} + 2 \cdot 3 \cdot \sqrt 3 } \right) + \left( {{3^2} + {{(\sqrt 3 )}^2} - 2 \cdot 3 \cdot \sqrt 3 } \right)\]
Simplifying the terms, we have
\[ \Rightarrow {a^2} + {b^2} = \left( {9 + 3 + 2 \cdot 3 \cdot \sqrt 3 } \right) + \left( {9 + 3 - 2 \cdot 3 \cdot \sqrt 3 } \right)\]
\[ \Rightarrow {a^2} + {b^2} = \left( {9 + 3 + 2 \cdot 3 \cdot \sqrt 3 + 9 + 3 - 2 \cdot 3 \cdot \sqrt 3 } \right)\]
Adding and subtracting the terms, we have
\[ \Rightarrow {a^2} + {b^2} = 24\]
Therefore, \[{a^2} + {b^2} = 24\]
Hence, option C is the correct option.

Note: We can calculate the sum of square of roots by an alternate method. The quadratic equation is of the form \[{x^2} - \] (Sum of roots)\[x + \]Product of roots. So, in the quadratic equation \[{x^2} - 6x + 6 = 0\], we have the sum of roots \[a + b = 6\] and the product of roots \[ab = 6\].
\[ \Rightarrow \] Sum of roots \[a + b = 6\]
Squaring on both the sides, we have
\[ \Rightarrow \] \[{(a + b)^2} = {6^2}\]
The square of the sum of two numbers a, b is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Now, by using the algebraic identity, we have
\[ \Rightarrow {a^2} + {b^2} + 2ab = 36\]
Now, substituting the product of roots, we have
\[ \Rightarrow {a^2} + {b^2} + 2(6) = 36\]
Multiplying and rewriting the equation, we have
\[ \Rightarrow {a^2} + {b^2} = 36 - 12\]
\[ \Rightarrow {a^2} + {b^2} = 24\]