JEE Mathematics 2023 Question with Solution

Given: $\alpha, \beta$ are the roots of $x^2 + 60^{\frac{1}{4}} x + a = 0$.

Find: The product of all possible values of $a$ given that $\alpha^4 + \beta^4 = -30$.

From the quadratic equation,

$$\alpha + \beta = -60^{\frac{1}{4}}, \qquad \alpha\beta = a$$

Using the identity,

$$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2$$

and

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

So,

$$\left(( -60^{\frac{1}{4}} )^2 - 2a\right)^2 - 2a^2 = -30$$

Since $(-60^{\frac{1}{4}})^2 = 60^{\frac{1}{2}}$, this becomes

$$\left(60^{\frac{1}{2}} - 2a\right)^2 - 2a^2 = -30$$

Expanding and simplifying,

$$60 - 4\cdot 60^{\frac{1}{2}} a + 4a^2 - 2a^2 = -30$$ $$2a^2 - 4\cdot 60^{\frac{1}{2}} a + 90 = 0$$

For this quadratic in $a$, the product of all possible values of $a$ is

$$\frac{90}{2} = 45$$

Therefore, the product of all possible values of $a$ is $45$.

Given: $\alpha + \beta = -60^{\frac{1}{4}}$ and $\alpha\beta = a$.

Find: Product of all admissible values of $a$.

Start from the condition

$$\alpha^4 + \beta^4 = -30$$

Write it as

$$(\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 = -30$$

Now,

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

Hence,

$$\alpha^2 + \beta^2 = \left(-60^{\frac{1}{4}}\right)^2 - 2a = 60^{\frac{1}{2}} - 2a$$

Also,

$$\alpha^2\beta^2 = a^2$$

Substituting,

$$(60^{\frac{1}{2}} - 2a)^2 - 2a^2 = -30$$ $$60 - 4\cdot 60^{\frac{1}{2}}a + 4a^2 - 2a^2 = -30$$ $$2a^2 - 4\cdot 60^{\frac{1}{2}}a + 90 = 0$$

If the two possible values of $a$ are roots of this equation, then by Vieta's formula their product is

$$\frac{90}{2} = 45$$

Therefore, the required product is $45$.