Further Solving Quadratic Equations (Hidden Quadratics) (Edexcel A-Level Mathematics): Revision Notes

Some equations have a quadratic "hidden" inside them. Sometimes, maths problems look tricky but can be made simpler by changing how we see them. Imagine you have a big puzzle, but you can't solve it all at once. Instead, you break it into smaller, easier pieces. In maths, a tricky equation might have something like $\ x^4$ or $\ x^{2/3}$, but if we pretend $\ x^2$ is something simpler, like a different letter (say, $\ u)$, the equation becomes easier to solve.

So, if we have a problem like $\ x^4 - 5x^2 + 6 = 0$, we can think of $\ x^2 \ as \ u$), turning the problem into $\ u^2 - 5u + 6 = 0 .$

This new problem is easier to solve, like putting together a smaller puzzle. Once we solve it, we just remember that $\ u$ was really $\ x^2$, and we can figure out the original problem. This trick helps us solve what seemed like a really hard problem in a much simpler way.

📑Example: Solve $x^4 - 6x^2 + 9 = 0$

1. Substitute $u = x^2,\ so\ u^2 = x^4$

• Equation becomes: $u^2 - 6u + 9 = 0$

2. Rewrite and solve for u:

• $(u - 3)(u - 3) = 0$
• $u = 3$

3. Substitute back $u$ for $x$:

• $x^2 = 3$
• $x = \pm \sqrt{3}$

📑Example: Solve $x + 8\sqrt{x} - 20 = 0$

1. Substitute $u = \sqrt{x},\ so\ u^2 = x$

• Equation becomes: $u^2 + 8u - 20 = 0$

2. Rewrite and solve for u:

• $(u + 10)(u - 2) = 0$
• $u = -10$ (not valid since the square root must be positive)
• $u = 2$

3. Substitute back $u$ for $x$:

• $\sqrt{x} = 2$
• $x = 4$

📑Example: Solve $x^4 + 8x^2 + 12 = 0$

1. Substitute $u = x^2,\ so\ u^2 = x^4$

• Equation becomes: $u^2 + 8u + 12 = 0$

2. Rewrite and solve for $u$:

• $(u + 2)(u + 6) = 0$
• $u = -2$ (no real roots since a square number cannot be negative)
• $u = -6$ (no real roots since a square number cannot be negative)

• For the quadratic equation $x^4 + 8x^2 + 12 = 0$, there are no real roots because a square number cannot be negative.