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Worked Examples · Example 1

Q.Write the solution set of the equation x2+x−2=0x^2 + x - 2 = 0 in roster form.

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Factoring x2+x−2=0x^2 + x - 2 = 0 gives (x+2)(x−1)=0(x+2)(x-1) = 0, so x=−2x = -2 or x=1x = 1. The solution set in roster form is {−2,1}\{-2, 1\}.

Why "roster form" means listing the actual solutions

Roster form describes a set by listing its members directly inside curly braces, separated by commas. When the set in question is "the solution set of an equation," the members ARE the roots of that equation — so writing it in roster form means solving the equation first, then listing every root found.

Step-by-step solution

1. Set up the factoring.

We need x2+x−2=0x^2 + x - 2 = 0. To factor a quadratic of the form x2+bx+cx^2 + bx + c, look for two numbers that multiply to cc and add to bb. Here b=1b = 1 and c=−2c = -2.

The two numbers are 22 and −1-1: their product is 2×(−1)=−22 \times (-1) = -2 (matches cc), and their sum is 2+(−1)=12 + (-1) = 1 (matches bb).

2. Write the factored form.

x2+x−2=(x+2)(x−1)x^2 + x - 2 = (x + 2)(x - 1)

You can check this by expanding: (x+2)(x−1)=x2−x+2x−2=x2+x−2(x+2)(x-1) = x^2 - x + 2x - 2 = x^2 + x - 2. ✓

3. Apply the zero-product property.

If a product of two factors equals zero, at least one factor must be zero:

x+2=0orx−1=0x + 2 = 0 \quad \text{or} \quad x - 1 = 0

x=−2orx=1x = -2 \quad \text{or} \quad x = 1

4. Collect the roots into roster form.

The solution set is the collection of every value of xx that satisfies the equation — here, exactly two values: −2-2 and 11. Written in roster form:

{−2,1}\{-2, 1\}

Tip

Order doesn't matter in a set — {−2,1}\{-2, 1\} and {1,−2}\{1, -2\} describe the same set. Roster form just needs every element listed once, in any order.

✓Final answer

The solution set of x2+x−2=0x^2 + x - 2 = 0, in roster form, is {−2,1}\{-2, 1\}.

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