Question

8. 1) a^²+ 2a - 15; 3) x^2+ 15x + 54​

Answer 1

Д=4+60=64
х=3 х=-5
2)д=225- 216=9
х=12
х=-3

Answer 2

It looks like you've provided two quadratic expressions. To work with these expressions, we can factor them to find their roots (solutions). Here's how you can factor each of them:

1. a^2 + 2a - 15: To factor this expression, we need to find two numbers that multiply to -15 (the constant term) and add up to 2 (the coefficient of the linear term).

The two numbers that meet these criteria are +5 and -3 because 5 * (-3) = -15 and 5 + (-3) = 2.

So, we can factor the expression as follows: a^2 + 2a - 15 = (a + 5)(a - 3)

Now, you have factored the expression into two binomials: (a + 5) and (a - 3). To find the roots (solutions), set each binomial equal to zero:

a + 5 = 0 a - 3 = 0

Solving for 'a' in each equation:

For a + 5 = 0: a = -5

For a - 3 = 0: a = 3

So, the solutions to the equation a^2 + 2a - 15 = 0 are a = -5 and a = 3.

2. x^2 + 15x + 54: To factor this expression, we need to find two numbers that multiply to 54 (the constant term) and add up to 15 (the coefficient of the linear term).

The two numbers that meet these criteria are +18 and +3 because 18 * 3 = 54 and 18 + 3 = 21.

So, we can factor the expression as follows: x^2 + 15x + 54 = (x + 18)(x + 3)

Now, you have factored the expression into two binomials: (x + 18) and (x + 3). To find the roots (solutions), set each binomial equal to zero:

x + 18 = 0 x + 3 = 0

Solving for 'x' in each equation:

For x + 18 = 0: x = -18

For x + 3 = 0: x = -3

So, the solutions to the equation x^2 + 15x + 54 = 0 are x = -18 and x = -3.