Question

1)x^4-5x^2-6=0 2)4x^4-5x^2+1=0 3)x^4-9x^2+20=0 4)16y^4+145y^2+9=0 5)12y4-7y^2+1=0

Answer 1

1) x⁴-5x²-6=0

t²-5t-6=0

t= -1

t=6

x²= -1

x²=6

x€/R

x= -√6

x=√6

x1= -6, x2=√6.

2)4x⁴-5x²+1=0

4t²-5t+1=0

t=1/4

t=1

x²=1/4

x²=1

x= -1/2

x=1/2

x= -1

x=1

x1= -1, x2= -1/2, x3=1/2, x4=1.

3) x⁴-9x+20=0

t²-9t+20=0

t=4

t=5

x²=4

x²=5

x= -2

x=2

x= -√5

x=√5

x1= -√5, x2= -2, x3=2, x4=√5.

4)16y⁴+145y²+9=0

y= €/R.

5) 12y4-7y²+1=0

48y-7y²+1=0

-7y²+48y+1=0

7y²-48y-1=0

y= -(-48)±√(-48)²-4×7×(-1)/2×7

y=48±√2304+28/14

y=48±2332/14

y=48+2√583/14

y=48-2√583/14

y=24+√583/7

y=24-√583/7

y1=24-√583/7, x2=24+√583/7.

Answer 2

To solve these equations, we'll use substitution to simplify them. Let's tackle each equation one by one:

1. x^4 - 5x^2 - 6 = 0

Let's use substitution. Let y = x^2. Then the equation becomes:

y^2 - 5y - 6 = 0

Now, we can factor this quadratic equation:

(y - 6)(y + 1) = 0

So, either y - 6 = 0 or y + 1 = 0:

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

Now, we need to substitute back for y:

1. x^2 = 6 => x = ±√6
2. x^2 = -1 (No real solutions, as the square of a real number cannot be -1)

Therefore, the real solutions to the equation are x = √6 and x = -√6.

2. 4x^4 - 5x^2 + 1 = 0

Let's use substitution again. Let y = x^2. Then the equation becomes:

4y^2 - 5y + 1 = 0

Now, we can factor this quadratic equation:

(4y - 1)(y - 1) = 0

So, either 4y - 1 = 0 or y - 1 = 0:

1. 4y - 1 = 0 => 4y = 1 => y = 1/4
2. y - 1 = 0 => y = 1

Now, we need to substitute back for y:

1. x^2 = 1/4 => x = ±(1/2)
2. x^2 = 1 => x = ±1

Therefore, the real solutions to the equation are x = 1/2, x = -1/2, x = 1, and x = -1.

3. x^4 - 9x^2 + 20 = 0

Let's use substitution again. Let y = x^2. Then the equation becomes:

y^2 - 9y + 20 = 0

Now, we can factor this quadratic equation:

(y - 5)(y - 4) = 0

So, either y - 5 = 0 or y - 4 = 0:

1. y - 5 = 0 => y = 5
2. y - 4 = 0 => y = 4

Now, we need to substitute back for y:

1. x^2 = 5 => x = ±√5
2. x^2 = 4 => x = ±2

Therefore, the real solutions to the equation are x = √5, x = -√5, x = 2, and x = -2.

I will now solve the remaining two equations:

4. 16y^4 + 145y^2 + 9 = 0

Let's use substitution again. Let z = y^2. Then the equation becomes:

16z^2 + 145z + 9 = 0

Now, we can factor this quadratic equation:

(4z + 1)(4z + 9) = 0

So, either 4z + 1 = 0 or 4z + 9 = 0:

1. 4z + 1 = 0 => 4z = -1 => z = -1/4
2. 4z + 9 = 0 => 4z = -9 => z = -9/4

Now, we need to substitute back for z:

1. y^2 = -1/4 (No real solutions, as the square of a real number cannot be negative)
2. y^2 = -9/4 (No real solutions, as the square of a real number cannot be negative)

Therefore, there are no real solutions to the equation.

5. 12y^4 - 7y^2 + 1 = 0

Let's use substitution again. Let z = y^2. Then the equation becomes:

12z^2 - 7z + 1 = 0

Now, we can factor this quadratic equation:

(3z - 1)(4z - 1) = 0

So, either 3z - 1 = 0 or 4z - 1 = 0:

1. 3z - 1 = 0 => 3z = 1 => z = 1/3
2. 4z - 1 = 0 => 4z = 1 => z = 1/4

Now, we need to substitute back for z:

1. y^2 = 1/3 => y = ±√(1/3)
2. y^2 = 1/4 => y = ±1/2

Therefore, the real solutions to the equation are y = √(1/3), y = -√(1/3), y = 1/2, and y = -1/2.