68. The positive root of the equation

x⁴ + x² – 2 = 0 is             .                                

How to Find the Positive Root of x⁴ + x² − 2 = 0

At first glance, the equation x⁴ + x² − 2 = 0 appears to be a fourth-degree equation, which may seem difficult to solve directly. However, a closer observation reveals that only even powers of x are present. Such equations are called biquadratic equations. The key idea is to transform the equation into a quadratic equation using an appropriate substitution. Once the quadratic equation is solved, we substitute back to obtain the values of x. This substitution technique is one of the most powerful methods in algebra and is widely used in competitive examinations.

Correct Answer

Positive Root = 1

Understanding the Structure of the Equation

The given equation is

x⁴ + x² − 2 = 0

Notice that the powers of x are

• x⁴
• x²
• constant term

Since only even powers occur, the equation is not treated as an ordinary quartic equation. Instead, it behaves like a quadratic equation in the variable x². Recognizing this pattern immediately simplifies the problem and saves considerable time during competitive examinations.

Step 1: Apply a Suitable Substitution

Let

y = x²

Then

x⁴ = (x²)² = y²

Substituting these into the original equation gives

y² + y − 2 = 0

The fourth-degree equation has now been reduced to a simple quadratic equation.

Step 2: Solve the Quadratic Equation

The quadratic equation is

y² + y − 2 = 0

This quadratic can be solved either by factorization or by using the quadratic formula. Since it factors easily, factorization is the quickest approach.

We look for two numbers whose product is −2 and whose sum is 1.

These numbers are

2 and −1.

Hence,

y² + y − 2 = (y + 2)(y − 1)

Therefore,

(y + 2)(y − 1) = 0

Thus, the two possible values of y are

y = 1

or

y = −2

Step 3: Substitute Back to Find x

Recall that

y = x²

Therefore,

x² = 1

or

x² = −2

The equation x² = 1 gives

x = ±1

The equation x² = −2 has no real solution because the square of a real number can never be negative.

Hence, the only real roots are

x = −1 and x = 1.

Step 4: Identify the Positive Root

The question specifically asks for the positive root.

Among the two real roots, only

x = 1

is positive.

Therefore, the required answer is

1.

Mathematical Verification

Substitute x = 1 into the original equation.

1⁴ + 1² − 2

= 1 + 1 − 2

= 0 ✔

Thus, x = 1 satisfies the equation.

Similarly, substituting x = −1 gives

(−1)⁴ + (−1)² − 2

= 1 + 1 − 2

= 0 ✔

Both are valid real roots, but only 1 is positive.

Alternative Method Using the Difference of Squares

Observe that

x⁴ + x² − 2

can also be factorized directly as

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Since

x² − 1 = (x − 1)(x + 1)

the complete factorization becomes

(x − 1)(x + 1)(x² + 2) = 0

The factor x² + 2 never becomes zero for real numbers because it is always positive.

Hence, the only real roots are

x = ±1.

This method is shorter and is particularly useful if the factorization is immediately recognized.

Related Practice Example

Solve

x⁴ − 5x² + 4 = 0.

Let

y = x².

Then

y² − 5y + 4 = 0

= (y − 4)(y − 1)

Hence,

y = 4 or y = 1.

Therefore,

x = ±2 or ±1.

This example demonstrates the same substitution technique used in the present question.

Key Takeaways

Whenever a polynomial contains only even powers of the variable, first check whether it can be transformed into a quadratic equation by substituting x² = y. This technique significantly simplifies the problem and allows higher-degree equations to be solved using familiar quadratic methods. Always substitute back carefully and check whether the problem asks for positive roots, negative roots, or all real roots.

Final Answer

The equation

x⁴ + x² − 2 = 0

has the real roots

x = −1 and x = 1.

Therefore, the required positive root is

1.