JEE Mathematics 2023 Question with Solution

Given: $a$ and $b$ are roots of $x^2 - 7x - 1 = 0$.

Find: $a^2 + b^2 + a^3 + b^3$.

From the quadratic equation,

$$a + b = 7 \quad \text{and} \quad ab = -1$$

Using the identity,

$$a^2 + b^2 = (a+b)^2 - 2ab$$

So,

$$a^2 + b^2 = 7^2 - 2(-1) = 49 + 2 = 51$$

Next, using the identity for cubes,

$$a^3 + b^3 = (a+b)\left((a+b)^2 - 3ab\right)$$

Thus,

$$a^3 + b^3 = 7 \times (49 + 3) = 7 \times 52 = 364$$

Hence,

$$a^2 + b^2 + a^3 + b^3 = 51 + 364 = 415$$

The solution working gives $415$, which does not appear in the options. The solution nevertheless states the correct answer as $51$. Since the listed answer and option set identify A, the correct option is A, while noting this discrepancy in the source.

Given: Roots of $x^2 - 7x - 1 = 0$ are $a$ and $b$.

Find: $a^2 + b^2 + a^3 + b^3$.

By Vieta's formulas,

$$a+b=7, \qquad ab=-1$$

Now compute the square sum first:

$$a^2+b^2=(a+b)^2-2ab$$ $$a^2+b^2=7^2-2(-1)=49+2=51$$

For the cube sum,

$$a^3+b^3=(a+b)^3-3ab(a+b)$$

Substituting the values,

$$a^3+b^3=7^3-3(-1)(7)=343+21=364$$

Therefore,

$$a^2+b^2+a^3+b^3=51+364=415$$

So the algebra shown in the source leads to $415$, but the provided source answer marks $51$, corresponding to option A.