Substitute t = x², solve quadratic, then find x.

Solve x⁴ - 5x² + 4 = 0 — Quadratic in Disguise

Question

Solve the equation:

x^4 - 5x^2 + 4 = 0

Find all real values of $x$ .

Solution — Step by Step

The powers are 4, 2, and 0. Notice that $x^4 = (x^2)^2$ . So if we call $t = x^2$ , the equation becomes a standard quadratic in $t$ .

This is the entire trick. Once you see it, the problem is straightforward.

Let $t = x^2$ . The equation becomes:

t^2 - 5t + 4 = 0

We’ve reduced a degree-4 equation to something we can factor in seconds.

We need two numbers that multiply to 4 and add to −5. That’s −1 and −4.

(t - 1)(t - 4) = 0

So $t = 1$ or $t = 4$ .

Remember $t = x^2$ , so now we solve two simple equations:

Case 1: $x^2 = 1 \Rightarrow x = \pm 1$

Case 2: $x^2 = 4 \Rightarrow x = \pm 2$

The four solutions are:

\boxed{x = -2,\ -1,\ 1,\ 2}

A degree-4 polynomial always has at most 4 roots — and we found exactly 4 real roots here, which is a satisfying check.

Why This Works

A polynomial like $ax^4 + bx^2 + c = 0$ is called a biquadratic equation (or quadratic in disguise). The key observation is that only even powers of $x$ appear — 4, 2, and 0. This means the substitution $t = x^2$ perfectly collapses the structure.

After solving for $t$ , you must back-substitute and solve $x^2 = t$ . This step gives two values of $x$ for each positive value of $t$ — that’s where all 4 solutions come from.

One thing worth checking: if your quadratic in $t$ gave a negative value of $t$ , you’d discard it (since $x^2 \geq 0$ always). Here both $t = 1$ and $t = 4$ are positive, so all four roots are real.

Alternative Method — Direct Factoring

Experienced students can skip the substitution entirely by factoring the original polynomial directly.

x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4)

Now apply difference of squares to each factor:

(x-1)(x+1)(x-2)(x+2) = 0

This gives $x = 1, -1, 2, -2$ directly.

In JEE Main, where time matters, train yourself to factor biquadratics directly. If the constant term is a small perfect square (like 4, 9, 16), the direct factoring route is usually faster than going through the substitution.

This method also makes it visually obvious why there are exactly 4 real roots — you’re looking at a product of 4 linear factors.

Common Mistake

The most common error: after finding $t = 1$ and $t = 4$ , students write $x = 1$ and $x = 4$ and stop. They forget that $t = x^2$ , not $x$ . From $x^2 = 4$ , you get $x = +2$ and $x = -2$ . Missing the negative roots costs you half the answer — and in JEE’s integer-type questions, a wrong count means zero marks.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Factoring

Common Mistake

Try These Next