Answer :
To find the value of [tex]\( a^2 + \frac{1}{a^2} \)[/tex] for the quadratic equation [tex]\( a^2 - 6a + 1 = 0 \)[/tex]:
1. Identify the given quadratic equation: [tex]\( a^2 - 6a + 1 = 0 \)[/tex].
2. Use the quadratic formula to solve for [tex]\( a \)[/tex]:
The quadratic formula is:
[tex]\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, comparing [tex]\( a^2 - 6a + 1 = 0 \)[/tex] with [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\[ a = 1, \quad b = -6, \quad c = 1 \][/tex]
3. Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-6)^2 - 4(1)(1) = 36 - 4 = 32 \][/tex]
4. Find the roots using the quadratic formula:
[tex]\[ a = \frac{-(-6) \pm \sqrt{32}}{2(1)} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \][/tex]
Therefore, the two roots are:
[tex]\[ a_1 = 3 + 2\sqrt{2} \quad \text{and} \quad a_2 = 3 - 2\sqrt{2} \][/tex]
5. Calculate [tex]\( a^2 + \frac{1}{a^2} \)[/tex] for each root:
For [tex]\( a_1 = 3 + 2\sqrt{2} \)[/tex]:
[tex]\[ a_1^2 = (3 + 2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \][/tex]
[tex]\[ \frac{1}{a_1} = \frac{1}{3 + 2\sqrt{2}} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \][/tex]
[tex]\[ \frac{1}{a_1^2} = (3 - 2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2} \][/tex]
[tex]\[ a_1^2 + \frac{1}{a_1^2} = (17 + 12\sqrt{2}) + (17 - 12\sqrt{2}) = 34 \][/tex]
Similarly, for [tex]\( a_2 = 3 - 2\sqrt{2} \)[/tex]:
[tex]\[ a_2^2 = (3 - 2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2} \][/tex]
[tex]\[ \frac{1}{a_2} = \frac{1}{3 - 2\sqrt{2}} = \frac{1}{3 - 2\sqrt{2}} \times \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} = \frac{3 + 2\sqrt{2}}{9 - 8} = 3 + 2\sqrt{2} \][/tex]
[tex]\[ \frac{1}{a_2^2} = (3 + 2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \][/tex]
[tex]\[ a_2^2 + \frac{1}{a_2^2} = (17 - 12\sqrt{2}) + (17 + 12\sqrt{2}) = 34 \][/tex]
Hence, the value of [tex]\( a^2 + \frac{1}{a^2} \)[/tex] for both roots is [tex]\( 34 \)[/tex].
1. Identify the given quadratic equation: [tex]\( a^2 - 6a + 1 = 0 \)[/tex].
2. Use the quadratic formula to solve for [tex]\( a \)[/tex]:
The quadratic formula is:
[tex]\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, comparing [tex]\( a^2 - 6a + 1 = 0 \)[/tex] with [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\[ a = 1, \quad b = -6, \quad c = 1 \][/tex]
3. Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = (-6)^2 - 4(1)(1) = 36 - 4 = 32 \][/tex]
4. Find the roots using the quadratic formula:
[tex]\[ a = \frac{-(-6) \pm \sqrt{32}}{2(1)} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \][/tex]
Therefore, the two roots are:
[tex]\[ a_1 = 3 + 2\sqrt{2} \quad \text{and} \quad a_2 = 3 - 2\sqrt{2} \][/tex]
5. Calculate [tex]\( a^2 + \frac{1}{a^2} \)[/tex] for each root:
For [tex]\( a_1 = 3 + 2\sqrt{2} \)[/tex]:
[tex]\[ a_1^2 = (3 + 2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \][/tex]
[tex]\[ \frac{1}{a_1} = \frac{1}{3 + 2\sqrt{2}} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \][/tex]
[tex]\[ \frac{1}{a_1^2} = (3 - 2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2} \][/tex]
[tex]\[ a_1^2 + \frac{1}{a_1^2} = (17 + 12\sqrt{2}) + (17 - 12\sqrt{2}) = 34 \][/tex]
Similarly, for [tex]\( a_2 = 3 - 2\sqrt{2} \)[/tex]:
[tex]\[ a_2^2 = (3 - 2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2} \][/tex]
[tex]\[ \frac{1}{a_2} = \frac{1}{3 - 2\sqrt{2}} = \frac{1}{3 - 2\sqrt{2}} \times \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} = \frac{3 + 2\sqrt{2}}{9 - 8} = 3 + 2\sqrt{2} \][/tex]
[tex]\[ \frac{1}{a_2^2} = (3 + 2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \][/tex]
[tex]\[ a_2^2 + \frac{1}{a_2^2} = (17 - 12\sqrt{2}) + (17 + 12\sqrt{2}) = 34 \][/tex]
Hence, the value of [tex]\( a^2 + \frac{1}{a^2} \)[/tex] for both roots is [tex]\( 34 \)[/tex].
