Solve 2x² + 5x + 3 = 0 by splitting the middle term.

−1 and -3/2. ac = 6, split 2 and 3: 2x² + 2x + 3x + 3 = 2x(x+1) + 3(x+1) = (2x+3)(x+1) = 0 → x = -3/2 or −1.

Solve $2 x^{2} + 5 x + 3 = 0$ by splitting the middle term.

Class 10Quadratic EquationsAlgebra — Quadratics

ac = 6, split 2 and 3:

2 x^{2} + 2 x + 3 x + 3 = 2 x (x + 1) + 3 (x + 1) = (2 x + 3) (x + 1) = 0

→

x = - \frac{3}{2}

or −1.

Answer: A) −1 and

- \frac{3}{2}

Rewrite

a x^{2} + b x + c = 0

as a product

(p x + q) (r x + s) = 0

, then use the **zero-product property**: a product is zero iff one of its factors is zero.

**Steps for

x^{2} + b x + c = 0 * * (a = 1)

:
1. Find two numbers that MULTIPLY to

c

and ADD to

b

.
2. Use them as the constants in

(x + p) (x + q)

.

**Example**:

x^{2} - 7 x + 12 = 0

→ product 12, sum −7 → numbers −3 and −4 →

(x - 3) (x - 4) = 0

→

x = 3

or 4.

**For

a x^{2} + b x + c

with a ≠ 1**: split the middle term — find two numbers multiplying to

a \times c

and summing to

b

. Then group.

Example:

2 x^{2} + 7 x + 3 = 0. a \times c = 6

, sum = 7 → 6 and 1. Split:

2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + 1 (x + 3) = (2 x + 1) (x + 3) = 0

→

x = - 1/2

or −3.

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