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Step-by-Step Solution
Step 1: Identify the sum and product of the roots
For the quadratic equation
x^{2} + 60^{\frac{1}{4}}\,x + a = 0 ,
let its roots be \alpha and \beta . By Vieta's formulas:
\alpha + \beta = -\,60^{\frac{1}{4}} \quad \text{and} \quad \alpha\,\beta = a.
Step 2: Express \alpha^{4} + \beta^{4} in terms of \alpha + \beta and \alpha \beta
We are given \alpha^{4} + \beta^{4} = -\,30 . Recall the identity:
\alpha^{4} + \beta^{4} = (\alpha^{2} + \beta^{2})^{2} - 2\,\alpha^{2}\,\beta^{2}.
Since \alpha\,\beta = a , we have \alpha^{2}\,\beta^{2} = a^{2} . Therefore:
\alpha^{4} + \beta^{4} = (\alpha^{2} + \beta^{2})^{2} - 2\,a^{2}.
Step 3: Relate \alpha^{2} + \beta^{2} to the given information
Another standard identity gives:
\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\,\alpha \beta.
Substituting \alpha + \beta = -\,60^{\frac{1}{4}} and \alpha \beta = a :
\alpha^{2} + \beta^{2}
= \bigl(-\,60^{\frac{1}{4}}\bigr)^{2} - 2\,a
= 60^{\frac{1}{2}} - 2\,a.
Step 4: Use the given condition \alpha^{4} + \beta^{4} = -\,30
We now write:
(\alpha^{2} + \beta^{2})^{2} - 2\,a^{2} = -\,30.
Substitute \alpha^{2} + \beta^{2} = 60^{\frac{1}{2}} - 2\,a :
\bigl(60^{\frac{1}{2}} - 2\,a\bigr)^{2} - 2\,a^{2} = -\,30.
Expand \bigl(60^{\frac{1}{2}} - 2\,a\bigr)^{2} :
\bigl(60^{\frac{1}{2}} - 2\,a\bigr)^{2}
= 60 + 4\,a^{2} - 4\,60^{\frac{1}{2}}\,a.
Therefore:
60 + 4\,a^{2} - 4\,60^{\frac{1}{2}}\,a - 2\,a^{2} = -\,30.
Simplify:
2\,a^{2} - 4\,60^{\frac{1}{2}}\,a + 60 + 30 = 0,
2\,a^{2} - 4\,60^{\frac{1}{2}}\,a + 90 = 0.
Notice 60^{\frac{1}{2}} = \sqrt{60} = 2\sqrt{15} , so the equation becomes
2\,a^{2} - 8\,\sqrt{15}\,a + 90 = 0.
Divide by 2 to simplify:
a^{2} - 4\,\sqrt{15}\,a + 45 = 0.
Step 5: Solve for a and find the product of all possible values
Use the quadratic formula:
a
= \frac{4\,\sqrt{15} \pm \sqrt{\bigl(4\,\sqrt{15}\bigr)^{2} - 4\cdot 1\cdot 45}}{2}
= \frac{4\,\sqrt{15} \pm \sqrt{16\cdot 15 - 180}}{2}
= \frac{4\,\sqrt{15} \pm \sqrt{240 - 180}}{2}
= \frac{4\,\sqrt{15} \pm \sqrt{60}}{2}.
Since \sqrt{60} = 2\,\sqrt{15} , the expression becomes:
a
= \frac{4\,\sqrt{15} \pm 2\,\sqrt{15}}{2}.
Hence the two values of a are:
a_1 = \frac{4\,\sqrt{15} + 2\,\sqrt{15}}{2} = 3\,\sqrt{15}, \;
a_2 = \frac{4\,\sqrt{15} - 2\,\sqrt{15}}{2} = \sqrt{15}.
Their product is:
a_1 \times a_2 = 3\,\sqrt{15}\,\times\,\sqrt{15} = 3 \times 15 = 45.
Final Answer
The product of all possible values of a is 45 .
