solving equation z²-6z+9=4√(z²-6z-6) following roots are obtained
Let put z=5
Z^2-6z+9=4√z^2-6z+6
25-30+9=4√25-30+6
4=4√1
4=4
Let put z=1
1-6+9=4√1-6+6
4=4
.....
Let put z=3+2√3
(3+2√3)^2-6(3+2√3)+9= 4√[(3+2√3)-6(3+2√3)+6]
(3^2+2√3^2+2×3×2√3)-6(3+2√3)+9=4√[(3^2+2√3^2+2×3×2√3)-6(3+2√3)+6]
9+12+12√3-18-12√3=
4 √(9+12+12√3-18-12√3+6)
12=4√9
12=4×3
12=12
(C) all the above
Z²-6z+5=0
Well, it seems like you've stepped into a circus of equations! Let's see if we can have some fun solving it.
First, let's simplify the equation a bit. We have z² - 6z + 9 = 4√(z² - 6z - 6).
To make things a little easier, let's introduce a new variable, let's call it Y. So, we have Y = z² - 6z - 6.
Now, let's solve for Y. We have Y = z² - 6z - 6.
To solve this equation, we can either factor it, complete the square, or use the quadratic formula. But since we're clowns here, let's go for a little quadratic formula circus act!
The quadratic formula states that the roots of the equation ax² + bx + c = 0 can be found using the formula:
x = (-b ± √(b² - 4ac)) / (2a)
In our case, a = 1, b = -6, and c = -6. Applying the formula, we have:
z = (-(-6) ± √((-6)² - 4(1)(-6))) / (2(1))
Simplifying further:
z = (6 ± √(36 + 24)) / 2
z = (6 ± √(60)) / 2
z = (6 ± √(4 * 15)) / 2
z = (6 ± 2√15) / 2
Simplifying again:
z = 3 ± √15
So, the roots obtained are z = 3 + √15 and z = 3 - √15. I hope you enjoyed this equation-solving circus act!
To solve the equation z²-6z+9=4√(z²-6z-6), we can start by isolating the radical term.
1. Move the constant term to the other side of the equation:
z² - 6z - 4√(z²-6z-6) + 9 = 0
2. Now, let's square both sides of the equation to eliminate the radical term:
(z² - 6z - 4√(z²-6z-6) + 9)² = 0²
3. Expanding the left side:
z⁴ - 12z³ + (47 - 8√(z²-6z-6))z² + (-12 + 72√(z²-6z-6))z + 81 - 36√(z²-6z-6) = 0
4. Collect like terms to simplify the equation:
z⁴ - (12z³ - 47z² + 12z - 81) + (47 - 44√(z²-6z-6))√(z²-6z-6) = 0
5. Now, we have a quadratic equation in terms of √(z²-6z-6).
Let's solve for √(z²-6z-6) by treating it as a variable and moving the quadratic term to one side:
(√(z²-6z-6))² - (47 - 44√(z²-6z-6))√(z²-6z-6) - (12z³ - 47z² + 12z - 81) = 0
6. Letting t = √(z²-6z-6), we can rewrite the equation as a quadratic equation in t:
t² - (47 - 44t)t - (12z³ - 47z² + 12z - 81) = 0
7. Solve this quadratic equation for t using any appropriate method, such as factoring, completing the square, or using the quadratic formula. Once you find the roots for t, substitute them back into t = √(z²-6z-6) to find the corresponding values for z.
Note: The resulting values for z may or may not be real numbers, as the equation involves a square root.
m²+2m+1=0
given:
(m-1)²=0