1) Solve for x and check
Radical (x^2-10x)=3i
2) Simplify
(5+ Radical 2)/(5-Radical 2)
3)describe the nature of the roots
x^2-x-6=0
x^2+4x+29=0
4) Solve for x and check
Radical (x^2 +4x +44) +3=2x
I got 7 and -5/3 as answers, was not sure though
2) Use the fact that (a + b)(a - b) = a^2 - b^2
3) Let's say your equation in the format ax^2 + bx + c = 0
In your two cases, a = 1.
Calculate the quantity b^2 - 4ac for each equation. If it is positive there are two real roots. If it is zero there is one. If it is negative, there are two complex roots
1) square both sides
x^2 - 10x = 9i^2
x^2 - 10x = -9
x^2 - 10x +9 = 0
(x-1)(x-9) = 0
x = 1 or x=9
check (since we squared)
if x=1
LS = √(1-10) = √-9 = 3i = RS
if x=9
LS = √(81-90) = √-9 = 3i = RS
so x = 1 or x=9
4) since you clearly squared both sides to solve, each answer you obtained has to be verified in the original equation
if x=7 it works
if x= -5/3 it does not work
so x = 7 is the only solution
1) To solve for x in the equation Radical (x^2-10x)=3i, we first need to isolate the radical term.
Start by squaring both sides of the equation to eliminate the radical:
(x^2-10x)^2 = (3i)^2
x^4 - 20x^3 + 100x^2 = -9
Rearrange the equation to the standard form (ax^2+bx+c=0):
x^4 - 20x^3 + 100x^2 + 9 = 0
Now, you can use a numerical method or a graphing calculator to find the solutions. Once you find the values of x, substitute them back into the original equation to check:
sqrt(x^2-10x) = 3i
sqrt((-5)^2-10(-5)) = 3i
sqrt(25+50) = 3i
sqrt(75) = 3i
Since the square root of a positive number cannot be imaginary, there are no real solutions for x in this equation.
2) To simplify the expression (5+ Radical 2)/(5-Radical 2), we can use a technique called rationalizing the denominator.
The rationalization process involves multiplying the numerator and denominator by the conjugate of the denominator, which in this case is (5+Radical 2).
(5+ Radical 2)/(5-Radical 2) * (5+ Radical 2)/(5+Radical 2)
= (25 + 10√2 + 5√2 + 2) / (25 - 2)
Simplifying the numerator gives:
= (27 + 15√2) / 23
So the simplified form of (5+ Radical 2)/(5-Radical 2) is (27 + 15√2) / 23.
3) The nature of the roots of a quadratic equation can be determined by looking at the discriminant (b^2 - 4ac).
a) For the equation x^2-x-6=0, the coefficients are a=1, b=-1, and c=-6. The discriminant is given by:
(-1)^2 - 4(1)(-6) = 1 + 24 = 25
Since the discriminant is positive, the equation has two distinct real roots.
b) For the equation x^2+4x+29=0, the coefficients are a=1, b=4, and c=29. The discriminant is given by:
(4)^2 - 4(1)(29) = 16 - 116 = -100
Since the discriminant is negative, the equation has two complex roots.
4) To solve for x in the equation Radical (x^2 + 4x + 44) + 3 = 2x, we can start by isolating the radical term and then square both sides to eliminate it.
Start by subtracting 3 from both sides of the equation:
Radical (x^2 + 4x + 44) = 2x - 3
Now, square both sides of the equation to eliminate the radical:
(x^2 + 4x + 44) = (2x - 3)^2
(x^2 + 4x + 44) = 4x^2 - 12x + 9
Rearrange the equation to the standard form (ax^2+bx+c=0):
3x^2 - 16x + 35 = 0
To solve this quadratic equation, you can use factoring, completing the square, or the quadratic formula. Once you find the values of x, substitute them back into the original equation to check.
It appears that you have already obtained the solutions of x = 7 and x = -5/3. To check if these are indeed the solutions, substitute them back into the original equation:
For x = 7:
Radical ((7)^2 + 4(7) + 44) + 3 = 2(7)
Radical (49 + 28 + 44) + 3 = 14 + 3
Radical (121) + 3 = 17
11 + 3 = 17
14 = 17 (This is not true, so x = 7 is not a valid solution.)
For x = -5/3:
Radical ((-5/3)^2 + 4(-5/3) + 44) + 3 = 2(-5/3)
Radical (25/9 - 20/3 + 44) + 3 = -10/3
Radical ((25 - 60 + 396)/9) + 3 = -10/3
Radical (361/9) + 3 = -10/3
19/3 + 3 = -10/3
28/3 = -10/3 (This is not true, so x = -5/3 is not a valid solution.)
Therefore, there might be a mistake in obtaining the solutions, and x = 7 and x = -5/3 are not valid solutions to the equation.