Equations that become quadratic

A quadratic in disguise

Look at $x^4 - 13x^2 + 36 = 0$. It has a fourth power, so it is not a quadratic equation in $x$. But $x^4 = (x^2)^2$. Put $y = x^2$:

This is an ordinary quadratic in $y$. It factorises as $(y - 4)(y - 9) = 0$, so $y = 4$ or $y = 9$.

Now go back to $x$. From $x^2 = 4$ we get $x = \pm 2$, and from $x^2 = 9$ we get $x = \pm 3$. So the equation has four roots: $-3$, $-2$, $2$ and $3$.

An equation with only $x^4$, $x^2$ and a number is called biquadratic. The method always works:

1. Substitute. Put $y = x^2$.
2. Solve the quadratic in $y$.
3. Return to $x$. Each positive $y$ gives two roots, $\pm\sqrt{y}$. A zero $y$ gives $x = 0$. A negative $y$ gives no real root at all.

For example, $x^4 + 3x^2 + 2 = 0$ gives $y = -1$ or $y = -2$. No real number has a negative square, so this equation has no real roots.

Other things to replace

An expression that repeats. In $(x^2 + x)^2 - 8(x^2 + x) + 12 = 0$, put $y = x^2 + x$. Then $y^2 - 8y + 12 = 0$, so $y = 2$ or $y = 6$.

1. $x^2 + x = 2$ gives $x^2 + x - 2 = 0$, so $x = 1$ or $x = -2$.
2. $x^2 + x = 6$ gives $x^2 + x - 6 = 0$, so $x = 2$ or $x = -3$.

A number and its reciprocal. Solve $x + \dfrac{1}{x} = \dfrac{5}{2}$, with $x \neq 0$. Multiply every term by $2x$:

So $x = 2$ or $x = \dfrac{1}{2}$. The two roots are reciprocals of each other, which makes sense: swapping $x$ and $\dfrac{1}{x}$ does not change the left side.

A power of 2. Solve $4^x - 3 \times 2^x + 2 = 0$. Since $4^x = (2^x)^2$, put $y = 2^x$:

Then $2^x = 1$ gives $x = 0$, and $2^x = 2$ gives $x = 1$. Note that $2^x$ is always positive, so a negative $y$ would have to be rejected.

Square roots and false roots

Solve $\sqrt{x + 7} = x - 5$.

1. Square both sides. $x + 7 = x^2 - 10x + 25$, so $x^2 - 11x + 18 = 0$.
2. Solve. $(x - 2)(x - 9) = 0$, so $x = 2$ or $x = 9$.
3. Check in the first equation. For $x = 9$: $\sqrt{16} = 4$ and $9 - 5 = 4$. It works. For $x = 2$: $\sqrt{9} = 3$ but $2 - 5 = -3$. It fails.

So the only root is $x = 9$. Where did $x = 2$ come from? Squaring cannot tell $3$ from $-3$, because both square to $9$. Squaring solves $\sqrt{x + 7} = x - 5$ and $-\sqrt{x + 7} = x - 5$ together. A root of the second one is called extraneous. Always check.

Practise with new numbers

Try it in Python

The program solves three biquadratic equations with $y = x^2$. It keeps only the values of $y$ that are not negative.

Common mistakes

• Stopping at y. The question asks for $x$. After $y = 4$, go on to $x = \pm 2$.
• Taking only the positive square root. $x^2 = 9$ gives $x = 3$ and $x = -3$.
• Keeping a negative y. $x^2 = -1$ has no real root, so drop it.
• Not checking after squaring. A root of the squared equation may fail the first one.
• Forgetting that x cannot be 0 when the equation has $\dfrac{1}{x}$ in it.

Remember these

• Look for an expression that appears squared and also on its own. Call it $y$.
• Solve the quadratic in $y$, then return to $x$.
• $x^2 = y$ gives two roots if $y > 0$, one if $y = 0$, and none if $y < 0$.
• After squaring both sides, check every root in the first equation.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which substitution turns $x^4 - 7x^2 + 12 = 0$ into a quadratic, and what are the four roots?

2. How many real roots has $x^4 + 5x^2 + 6 = 0$?

3. Solve $\sqrt{x + 2} = x$.

4. In $9^x - 4 \times 3^x + 3 = 0$, what should $y$ be?

5. Why do the roots of $x + \dfrac{1}{x} = k$ always come in a pair $r$ and $\dfrac{1}{r}$?