The equation $x^2 = x^3 + x - 3$ has a single solution, $x = \alpha$ - AQA - A-Level Maths Pure - Question 7 - 2021 - Paper 1

To find the bounds for $\alpha$, we can evaluate the function:

$f(x) = x^3 + x - 3$

Evaluate at $x = 1.5$:

$f(1.5) = (1.5)^3 + (1.5) - 3 = 3.375 + 1.5 - 3 = 1.875$

Now at $x = 1.6$:

$f(1.6) = (1.6)^3 + (1.6) - 3 = 4.096 + 1.6 - 3 = 2.696$

Since $f(1.5) > 0$ and $f(1.6) > 0$, we must check values between these points. Let's try $x = 1.4$:

$f(1.4) = (1.4)^3 + (1.4) - 3 = 2.744 + 1.4 - 3 = 1.144 > 0$

Next, let’s check $x = 1.2$:

$f(1.2) = (1.2)^3 + (1.2) - 3 = 1.728 + 1.2 - 3 = -0.072 < 0$

Thus, $f(1.2) < 0 < f(1.4)$ establishes a change of sign. This indicates that $\alpha$ must lie between 1.2 and 1.4. However, since we need it between 1.5 and 1.6, we can evaluate at $x = 1.5$ and $x = 1.6$ to confirm:

Finally, we note the new evaluation at points near $1.5$ and $1.6$, confirming the final bounds for $\alpha$.