Question

1)a^2-49= 2)(x-5)(x+6)=​

Answer 1

1) а² - 49 = (а - 7)(а + 7)

2) (x - 5)(x + 6) = x² + 6x - 5x - 30 = x² + x - 30

Answer 2

Ответ:

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Answer 3

Let's solve these two equations:

1. $a^2 - 49 = 0$

First, we can factor the left side of the equation:

$a^2 - 49 = (a - 7)(a + 7) = 0$

Now, we can use the zero-product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be equal to zero. So:

$a - 7 = 0$ or $a + 7 = 0$

Solving for $a$ in each equation:

For $a - 7 = 0$: $a = 7$

For $a + 7 = 0$: $a = -7$

So, there are two solutions to the equation $a^2 - 49 = 0$, which are $a = 7$ and $a = -7$.

2. $(x - 5)(x + 6) = 0$

Again, we can use the zero-product property here:

$x - 5 = 0$ or $x + 6 = 0$

Solving for $x$ in each equation:

For $x - 5 = 0$: $x = 5$

For $x + 6 = 0$: $x = -6$

So, there are two solutions to the equation $(x - 5)(x + 6) = 0$, which are $x = 5$ and $x = -6$.