Question

(х^2+2)^2-5(х^2+2)-6=0

Answer 1

Пусть (х²+2)=а, а∈R

a²-5a-6=0

D=(-(-5))²-4×1×(-6)=25+24=49

a1=(-(-5)-√49)/2×1=(5-7)/2=-2/2=-1

a2=(-(-5)+√49)/2×1=(5+7)/2=12/2=6

x²+2=a1

x²+2=-1

x²=-1-2

x²=-3

x1=-√3i=-1,73i

x2=√3i=1,73i

x²+2=a2

x²+2=6

x²=6-2

x²=4

x1=-√4

x1=-2

x2=√4

x2=2

Answer 2

To solve the quadratic equation (х^2+2)^2 - 5(х^2+2) - 6 = 0, we can use a substitution to make the equation easier to handle. Let's make a substitution: let y = (х^2 + 2). Now, we have the equation:

y^2 - 5y - 6 = 0

This is a simple quadratic equation in y. To solve for y, we can factor the equation or use the quadratic formula.

Factorization method: (y - 6)(y + 1) = 0

Now set each factor to zero and solve for y:

1. y - 6 = 0 → y = 6
2. y + 1 = 0 → y = -1

Since we substituted y = (х^2 + 2), we have two possibilities for х^2 + 2:

1. х^2 + 2 = 6 → х^2 = 6 - 2 → х^2 = 4 → х = ±√4 → х = ±2
2. х^2 + 2 = -1 → х^2 = -1 - 2 → х^2 = -3

However, the second solution х^2 = -3 leads to complex numbers, which are not real solutions in this context. So, we discard that solution.

Therefore, the real solutions for the original equation (х^2+2)^2 - 5(х^2+2) - 6 = 0 are х = 2 and х = -2.