Question

find the zero x2-7x+12 and chech the relationship between zeros and cofficient​

Answer 1

Answer:

The zeros of x² - 7x + 12 are 3 and 4.

Step-by-step explanation:

Given,

• x² - 7x + 12 = 0

To Find,

• The zeros of polynomial.
• The relationship between zeros and coefficient.

Solution,

x² - 7x + 12 = 0

→ x² - 4x - 3x + 12 = 0

→ x (x - 4) - 3 (x - 4) = 0

→ (x - 3)(x - 4) = 0

x - 3 = 0

→ x = 3

→ α = 3

x - 4 = 0

→ x = 4

→ ß = 4

In Quadratic Polynomial,

p(x) = ax² + bx + c

p(x) = x² - 7x + 12

x² - 7x + 12 = ax² + bx + c

(1)x² + (-7)x + 12 = ax² + bx + c

a = 1

b = -7

c = 12

Checking the relationship between zeros and coefficient;

α + ß = -b/a

→ 3 + 4 = -(-7)/1

→ 7 = 7

→ LHS = RHS

αß = c/a

→ 3 × 4 = 12/1

→ 12 = 12

→ LHS = RHS

Required Answer,

• The zeros of x² - 7x + 12 are 3 and 4.