Question

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

Answer 1

(x+3)²+(x-4)²=2(4-x)(x+3)

(x+3)²+(x-4)² -2(4-x)(x+3)=0

(x+3)²+(x-4)² +2(х-4)(x+3)=0

(x+3)² +2(х-4)(x+3)+(x-4)²=0

(х+3+х-4)²=0

(2х-1)²=0

2х-1=0

х=0,5

Answer 2

Let's solve the given equation step by step:

First, expand the squared terms on both sides of the equation: (x+3)^2 + (x-4)^2 = 2(4-x)(x+3)

Expanding the squares: (x^2 + 6x + 9) + (x^2 - 8x + 16) = 2(4-x)(x+3)

Combining like terms: 2x^2 - 2x + 25 = 2(4-x)(x+3)

Next, distribute the 2 on the right side: 2x^2 - 2x + 25 = 8 - 2x + 12x - 6x^2

Rearranging the terms: 2x^2 - 2x + 25 = -6x^2 + 10x + 8

Moving all terms to one side of the equation: 8x^2 - 12x + 17 = 0

Now, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 8, b = -12, and c = 17.

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values: x = (-(-12) ± √((-12)^2 - 4 * 8 * 17)) / (2 * 8)

Simplifying: x = (12 ± √(144 - 544)) / 16 x = (12 ± √(-400)) / 16

The term inside the square root is negative, which means the quadratic equation has no real solutions. The equation does not intersect the x-axis, so there are no real values of x that satisfy the equation.

Therefore, the given equation has no solutions in the real number system.