SOLVING TRINOMIALS WITH LEADING COEFFICIENT OF 1

Solve the following quadratic equations.

Problem 1 :

x² + 7x + 6 = 0

Solution:

x² + 7x + 6 = 0

x² + x + 6x + 6 = 0

x(x + 1) + 6(x + 1) = 0

(x + 1)(x + 6) = 0

x + 1 = 0 or x + 6 = 0

x = -1 or x = -6

So, the solutions are x = -1 or x = -6.

Problem 2 :

x² - 8x + 12 = 0

Solution:

x² - 8x + 12 = 0

x² - 2x - 6x + 12 = 0

x(x - 2) - 6(x - 2) = 0

(x - 2)(x - 6) = 0

x - 2 = 0 or x - 6 = 0

x = 2 or x = 6

So, the solutions are x = 2 or x = 6.

Problem 3 :

x² + 5x - 24 = 0

Solution:

x² + 5x - 24 = 0

x² + 8x - 3x - 24 = 0

x(x + 8) - 3(x + 8) = 0

(x + 8)(x - 3) = 0

x + 8 = 0 or x - 3 = 0

x = -8 or x = 3

So, the solutions are x = -8 or x = 3.

Problem 4 :

x² - 3x + 1 = 0

Solution:

Comparing x² - 3x + 1 = 0 and ax² + bx + c = 0, we get

a = 1, b = -3 and c = 1

By using quadratic formula,

Problem 5 :

x² + 3x - 3 = 0

Solution:

Comparing x² + 3x - 3 = 0 and ax² + bx + c = 0, we get

a = 1, b = 3 and c = -3

By using quadratic formula,

Problem 6 :

x² + 4x + 2 = 0

Solution:

Comparing x² + 4x + 2 = 0 and ax² + bx + c = 0, we get

a = 1, b = 4 and c = 2

By using quadratic formula,

Problem 7 :

x² + 8x = 0

Solution:

x² + 8x = 0

x(x + 8) = 0

x = 0 or x + 8 = 0

x = 0 or x = -8

So, the solutions are x = 0 or x = -8.

Problem 8 :

x² - 10x = 0

Solution:

x² - 10x = 0

Since we have only two terms in the given polynomial, we can factor the common term to find factors.

x(x - 10) = 0

x = 0 or x - 10 = 0

x = 0 or x = 10

So, the solutions are x = 0 or x = 10. 

Problem 9 :

5x² - 10x = 0

Solution:

5x² - 10x = 0

5x(x - 2) = 0

5x = 0 or x - 2 = 0

x = 0 or x = 2

So, the solutions are x = 0 or x = 2. 

Problem 10 :

x² - 81 = 0

Solution:

x² - 81 = 0

Using square root property, we can solve it.

x² = 81

Taking square root on both sides, we get

x = √81

x = ± 9

So, the solutions are {9, -9}. 

Problem 11 :

x² - 121 = 0

Solution:

x² - 121 = 0

x² = 121

Taking square root on both sides, we get

x = √121

x = ± 11

So, the solutions are {11, -11}.

Problem 12 :

25 = 10x - x²

Solution:

x² - 10x + 25 = 0

Write equation in factored form.

(x - 5)² = 0

x - 5 = 0

x = 5

So, the solution is x = 5.

Problem 13 :

36 = 13x - x²

Solution:

x² - 13x + 36 = 0

x² - 9x - 4x + 36 = 0

x(x - 9) - 4(x - 9) = 0

(x - 4) (x - 9) = 0

x - 4 = 0 or x - 9 = 0

x = 4 or x = 9

So, the solutions are x = 4 or x = 9.