Answer :
Let us start by considering the quadratic equation given: [tex]\(4x^2 - 6x + 3 = 0\)[/tex]. We are told that [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex] are the roots of this equation.
The roots of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\(4x^2 - 6x + 3 = 0\)[/tex], the coefficients are [tex]\(a = 4\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = 3\)[/tex]. Plugging these values into the quadratic formula, we will get:
[tex]\[ \alpha, \beta = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4} \][/tex]
[tex]\[ = \frac{6 \pm \sqrt{36 - 48}}{8} \][/tex]
[tex]\[ = \frac{6 \pm \sqrt{-12}}{8} \][/tex]
Since the discriminant [tex]\((36 - 48 = -12)\)[/tex] is negative, the roots will be complex. Let’s express [tex]\(\sqrt{-12}\)[/tex] as [tex]\(2i\sqrt{3}\)[/tex]:
[tex]\[ \alpha, \beta = \frac{6 \pm 2i\sqrt{3}}{8} \][/tex]
[tex]\[ = \frac{3 \pm i\sqrt{3}}{4} \][/tex]
So:
[tex]\[ \alpha = \frac{3 + i\sqrt{3}}{4} \][/tex]
[tex]\[ \beta = \frac{3 - i\sqrt{3}}{4} \][/tex]
Next, we need to find the value of [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex]. Start by squaring [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
[tex]\[ \alpha^2 = \left(\frac{3 + i\sqrt{3}}{4}\right)^2 = \frac{(3 + i\sqrt{3})^2}{16} = \frac{9 + 6i\sqrt{3} - 3}{16} = \frac{6 + 6i\sqrt{3}}{16} = \frac{3 + 3i\sqrt{3}}{8} \][/tex]
[tex]\[ \beta^2 = \left(\frac{3 - i\sqrt{3}}{4}\right)^2 = \frac{(3 - i\sqrt{3})^2}{16} = \frac{9 - 6i\sqrt{3} - 3}{16} = \frac{6 - 6i\sqrt{3}}{16} = \frac{3 - 3i\sqrt{3}}{8} \][/tex]
To find [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex]:
[tex]\[ \frac{1}{\alpha^2} = \frac{8}{3 + 3i\sqrt{3}}, \quad \frac{1}{\beta^2} = \frac{8}{3 - 3i\sqrt{3}} \][/tex]
Multiplying numerator and denominator by the conjugate for each term:
[tex]\[ \frac{1}{\alpha^2} = \frac{8}{3 + 3i\sqrt{3}} \times \frac{3 - 3i\sqrt{3}}{3 - 3i\sqrt{3}} = \frac{8(3 - 3i\sqrt{3})}{(3 + 3i\sqrt{3})(3 - 3i\sqrt{3})} = \frac{8(3 - 3i\sqrt{3})}{9 + 27} = \frac{8(3 - 3i\sqrt{3})}{36} = \frac{2 - 2i\sqrt{3}}{9} \][/tex]
Similarly,
[tex]\[ \frac{1}{\beta^2} = \frac{8}{3 - 3i\sqrt{3}} \times \frac{3 + 3i\sqrt{3}}{3 + 3i\sqrt{3}} = \frac{8(3 + 3i\sqrt{3})}{(3 - 3i\sqrt{3})(3 + 3i\sqrt{3})} = \frac{8 \times (3 + 3i\sqrt{3})}{36} = \frac{2 + 2i\sqrt{3}}{9} \][/tex]
Adding these two results:
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{2 - 2i\sqrt{3}}{9} + \frac{2 + 2i\sqrt{3}}{9} = \frac{2 - 2i\sqrt{3} + 2 + 2i\sqrt{3}}{9} = \frac{4}{9} = \frac{4}{3} \][/tex]
Therefore, the value of [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex] is [tex]\(\boxed{\frac{4}{3}}\)[/tex].
The roots of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the equation [tex]\(4x^2 - 6x + 3 = 0\)[/tex], the coefficients are [tex]\(a = 4\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = 3\)[/tex]. Plugging these values into the quadratic formula, we will get:
[tex]\[ \alpha, \beta = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4} \][/tex]
[tex]\[ = \frac{6 \pm \sqrt{36 - 48}}{8} \][/tex]
[tex]\[ = \frac{6 \pm \sqrt{-12}}{8} \][/tex]
Since the discriminant [tex]\((36 - 48 = -12)\)[/tex] is negative, the roots will be complex. Let’s express [tex]\(\sqrt{-12}\)[/tex] as [tex]\(2i\sqrt{3}\)[/tex]:
[tex]\[ \alpha, \beta = \frac{6 \pm 2i\sqrt{3}}{8} \][/tex]
[tex]\[ = \frac{3 \pm i\sqrt{3}}{4} \][/tex]
So:
[tex]\[ \alpha = \frac{3 + i\sqrt{3}}{4} \][/tex]
[tex]\[ \beta = \frac{3 - i\sqrt{3}}{4} \][/tex]
Next, we need to find the value of [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex]. Start by squaring [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]:
[tex]\[ \alpha^2 = \left(\frac{3 + i\sqrt{3}}{4}\right)^2 = \frac{(3 + i\sqrt{3})^2}{16} = \frac{9 + 6i\sqrt{3} - 3}{16} = \frac{6 + 6i\sqrt{3}}{16} = \frac{3 + 3i\sqrt{3}}{8} \][/tex]
[tex]\[ \beta^2 = \left(\frac{3 - i\sqrt{3}}{4}\right)^2 = \frac{(3 - i\sqrt{3})^2}{16} = \frac{9 - 6i\sqrt{3} - 3}{16} = \frac{6 - 6i\sqrt{3}}{16} = \frac{3 - 3i\sqrt{3}}{8} \][/tex]
To find [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex]:
[tex]\[ \frac{1}{\alpha^2} = \frac{8}{3 + 3i\sqrt{3}}, \quad \frac{1}{\beta^2} = \frac{8}{3 - 3i\sqrt{3}} \][/tex]
Multiplying numerator and denominator by the conjugate for each term:
[tex]\[ \frac{1}{\alpha^2} = \frac{8}{3 + 3i\sqrt{3}} \times \frac{3 - 3i\sqrt{3}}{3 - 3i\sqrt{3}} = \frac{8(3 - 3i\sqrt{3})}{(3 + 3i\sqrt{3})(3 - 3i\sqrt{3})} = \frac{8(3 - 3i\sqrt{3})}{9 + 27} = \frac{8(3 - 3i\sqrt{3})}{36} = \frac{2 - 2i\sqrt{3}}{9} \][/tex]
Similarly,
[tex]\[ \frac{1}{\beta^2} = \frac{8}{3 - 3i\sqrt{3}} \times \frac{3 + 3i\sqrt{3}}{3 + 3i\sqrt{3}} = \frac{8(3 + 3i\sqrt{3})}{(3 - 3i\sqrt{3})(3 + 3i\sqrt{3})} = \frac{8 \times (3 + 3i\sqrt{3})}{36} = \frac{2 + 2i\sqrt{3}}{9} \][/tex]
Adding these two results:
[tex]\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{2 - 2i\sqrt{3}}{9} + \frac{2 + 2i\sqrt{3}}{9} = \frac{2 - 2i\sqrt{3} + 2 + 2i\sqrt{3}}{9} = \frac{4}{9} = \frac{4}{3} \][/tex]
Therefore, the value of [tex]\(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)[/tex] is [tex]\(\boxed{\frac{4}{3}}\)[/tex].
