Question

2/x+4 + 1 = 3/x²+8x+16​

Answer 1

Відповідь:

2/(x+4) + 1 = 3/(x²+8x+16 ) -----x nie= -4

2/(x+4) + 1 = 3/((x+4)^2) /*(x+4)^2

2(x+4)+ (x+4)^2 = 3

2x+8 + x^2+8x+16=3

x^2+10x+24-3=0

x^2+10x+21=0

D=b^2-4ac = 100-84 = 16

sqrt(D)=4

x1=(-10+2)/2=-8/2=-4

x1=(-10-2)/2=-12/2=-6

Пояснення:

Answer 2

To solve the equation $\frac{2}{x+4} + 1 = \frac{3}{x^2+8x+16}$, you'll first want to find a common denominator for the fractions on the left side of the equation. The common denominator here is $x^2+8x+16$, which can be factored as $(x+4)^2$. So, rewrite the equation with the common denominator:

$\frac{2}{x+4} + 1 = \frac{3}{(x+4)^2}$

Now, multiply both sides of the equation by $(x+4)^2$ to eliminate the fractions:

$2(x+4)^2 + (x+4)^2 = 3$

Now, you have a quadratic equation. Expand and simplify the left side:

$2(x^2 + 8x + 16) + (x^2 + 8x + 16) = 3$

Now, distribute and combine like terms:

$2x^2 + 16x + 32 + x^2 + 8x + 16 = 3$

Combine like terms again:

$3x^2 + 24x + 48 = 3$

Now, subtract 3 from both sides to set the equation to zero:

$3x^2 + 24x + 48 - 3 = 0$

$3x^2 + 24x + 45 = 0$

Now, you have a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 3$, $b = 24$, and $c = 45$. You can solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plug in the values for $a$, $b$, and $c$:

$x = \frac{-24 \pm \sqrt{24^2 - 4(3)(45)}}{2(3)}$

Now, calculate the discriminant ($b^2 - 4ac$):

$24^2 - 4(3)(45) = 576 - 540 = 36$

So, the discriminant is 36, which is a perfect square. You have two possible solutions:

1. $x = \frac{-24 + 6}{6} = \frac{-18}{6} = -3$

2. $x = \frac{-24 - 6}{6} = \frac{-30}{6} = -5$

So, the solutions to the equation are $x = -3$ and $x = -5$.