Question

1. 64х²+х=0 2.у³+2у²+у=0 1.а²+2аb+b²-16= 2.а²-х²-2ху-у²=

Answer 1

4585-2158915-1051-0659743-25

Answer 2

1. To solve the equation 64x² + x = 0, we can factor out x:

x(64x + 1) = 0

This equation will be satisfied if either x = 0 or (64x + 1) = 0.

For x = 0, the equation is satisfied.

For (64x + 1) = 0, we solve for x:

64x + 1 = 0 64x = -1 x = -1/64

So, the solutions to the equation 64x² + x = 0 are x = 0 and x = -1/64.

2. To solve the equation у³ + 2у² + у = 0, we can factor out у:

у(у² + 2у + 1) = 0

This equation will be satisfied if either у = 0 or (у² + 2у + 1) = 0.

For у = 0, the equation is satisfied.

For (у² + 2у + 1) = 0, we can factor it further:

(у + 1)² = 0

Taking the square root of both sides:

у + 1 = 0 у = -1

So, the solutions to the equation у³ + 2у² + у = 0 are у = 0 and у = -1.

3. The equation а² + 2аb + b² - 16 = 0 represents a quadratic equation in two variables, a and b. To solve it, we can use the quadratic formula:

a = (-2b ± √(4b² - 4(b² - 16))) / 2

Simplifying the expression inside the square root:

a = (-2b ± √(4b² - 4b² + 64)) / 2 a = (-2b ± √(64)) / 2 a = (-2b ± 8) / 2

Simplifying further:

a = -b ± 4

So, the solutions to the equation а² + 2аb + b² - 16 = 0 are a = -b + 4 and a = -b - 4.

4. The equation а² - х² - 2ху - у² = 0 is a quadratic equation in two variables, a and y. To solve it, we can try to factorize it or complete the square, but in this case, it seems difficult to find an exact factorization.

However, we can rewrite the equation as follows:

(a - y)² - (x + y)² = 0

This equation represents a difference of squares. We can factorize it using the identity:

a² - b² = (a + b)(a - b)

Applying the identity to our equation:

((a - y) + (x + y))((a - y) - (x + y)) = 0

Simplifying further:

(a + x)(a - 2y - x) = 0

So, the solutions to the equation а² - х² - 2ху - у² = 0 are given by the cases a + x = 0 and a - 2y - x = 0.