Question

(5x2-4)2 + 6(5x2- 4) - 7 = 0

Answer 1

Пошаговое объяснение:

1) 5х2-4=6

25х2-4=6

3)6х2=12

4)6х6=36

5)36+12=48

6)48-7=41

Answer 2

10x-8+60x-24-7=0

10x+60x=8+24+7

70x=39

x=39/70

Answer 3

To solve the equation (5x^2 - 4)^2 + 6(5x^2 - 4) - 7 = 0, you can use substitution to simplify the equation. Let's make a substitution:

Let y = 5x^2 - 4.

Now, rewrite the equation in terms of y:

y^2 + 6y - 7 = 0

This is a quadratic equation in terms of y. To solve it, you can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = 6, and c = -7. Plug these values into the quadratic formula:

y = (-6 ± √(6^2 - 4(1)(-7))) / (2(1))

y = (-6 ± √(36 + 28)) / 2

y = (-6 ± √64) / 2

y = (-6 ± 8) / 2

Now, you have two possible solutions for y:

1. y = (-6 + 8) / 2 = 2 / 2 = 1
2. y = (-6 - 8) / 2 = -14 / 2 = -7

Now, remember that we made a substitution at the beginning: y = 5x^2 - 4. So, you need to solve for x:

1. If y = 1: 5x^2 - 4 = 1 5x^2 = 5 x^2 = 1 x = ±1

2. If y = -7: 5x^2 - 4 = -7 5x^2 = -7 + 4 5x^2 = -3 x^2 = -3/5 (Note: This solution is imaginary because you can't take the square root of a negative number in the real number system.)

So, the solutions for x are: x = 1 x = -1 x = ±√(-3/5) (imaginary)

Therefore, there are two real solutions for x, which are x = 1 and x = -1, and two imaginary solutions for x.