Question

2(x^2+2x)^2-7(x^2+2x)+3=0

Answer 1

X1=1
x2=-3 вот и ответ но решения нету

Answer 2

To solve the quadratic equation 2(x^2+2x)^2 - 7(x^2+2x) + 3 = 0, let's first simplify the equation by setting a variable substitution. Let's set y = x^2 + 2x. Now, the equation becomes:

2y^2 - 7y + 3 = 0

Now, we can solve this quadratic equation for y using factoring or the quadratic formula. Let's try factoring:

2y^2 - 7y + 3 = 0

To factor this, we look for two numbers whose product is (2 * 3 = 6) and whose sum is -7.

The numbers are -6 and -1, since (-6) * (-1) = 6, and (-6) + (-1) = -7.

Now, rewrite the middle term (-7y) using -6y and -1y:

2y^2 - 6y - 1y + 3 = 0

Now, factor by grouping:

(2y^2 - 6y) + (-1y + 3) = 0

Factor out the common terms:

2y(y - 3) - 1(y - 3) = 0

Now, notice that (y - 3) is a common factor:

(y - 3)(2y - 1) = 0

Now, set each factor equal to zero and solve for y:

1. y - 3 = 0 y = 3

2. 2y - 1 = 0 2y = 1 y = 1/2

Now that we have the values of y, we can revert to the substitution:

1. y = x^2 + 2x x^2 + 2x = 3 x^2 + 2x - 3 = 0

2. y = x^2 + 2x x^2 + 2x = 1/2 x^2 + 2x - 1/2 = 0

Now, we have two separate quadratic equations for x. Let's solve them:

1. x^2 + 2x - 3 = 0

This equation can be factored as follows:

(x + 3)(x - 1) = 0

Setting each factor to zero:

x + 3 = 0 x = -3

x - 1 = 0 x = 1

So, for this case, we have two solutions: x = -3 and x = 1.

2. x^2 + 2x - 1/2 = 0

This equation is not easily factorable, so let's use the quadratic formula:

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = 1, b = 2, and c = -1/2:

x = (-(2) ± √((2)^2 - 4(1)(-1/2))) / 2(1) x = (-2 ± √(4 + 2)) / 2 x = (-2 ± √6) / 2

Now, we have two solutions for this case as well:

x = (-2 + √6) / 2 ≈ 0.82 x = (-2 - √6) / 2 ≈ -2.82

So, the solutions to the original equation are x = -3, x = 1, x ≈ 0.82, and x ≈ -2.82.