Use the quadratic formula to slove the equation.
x^2-x=-7
This is what I have got so far and I do not think this is right:
x=1(-1)±√-1^2-4(1)(7)/2(1)
x=1±√1-4(7)/2
x=1±√-21/2
x=1±-21/2
and here I am lost!
Use rational exponents to simplify. √(5&x^10 )
Simplify by taking roots of the numerator and the denominator. Assume that all expressions under radicals represent positive numbers. ∛(〖125x〗^11/y^3 )
Multiply. (-2t)^2 (〖3t〗^7 )²
Divide and simplify. (3y-6)/14÷(y-2)/4y
change it to the standard form first
x^2 - x + 7 = 0
a=1, b=-1, c = 7
x = (1 ± √(1 - 4(1)(7))/2
= (1 ± √-27)/2
Your answer is an imaginary number or called a complex number, you could rewrite is as
x = (1 ± 3√3 i)/2
x^2-x+7=0
x= (1+-sqrt(1-28))/2=1/2 +- 1/2 sqrt-21
and sqrt-21= isqrt21
2) I don't understand the others.
Okay so do I write the answer as the solution set is:(1 �} 3�ã3 i)/2 , 0
To solve the equation x^2-x=-7 using the quadratic formula, you need to rearrange it in the form ax^2+bx+c=0. In this case, a=1, b=-1, and c=-7.
1. Start by substituting the values of a, b, and c into the quadratic formula:
x=(-b±√(b^2-4ac))/(2a)
2. Substitute the values into the formula:
x=(-(-1)±√((-1)^2-4(1)(-7)))/(2(1))
x=(1±√(1+28))/2
x=(1±√29)/2
So the solutions to the equation are x=(1+√29)/2 and x=(1-√29)/2.
Now let's move on to the second question:
To simplify √(5&x^10), we can rewrite it using rational exponents. The square root (√) can be expressed as a fractional exponent (1/2), and the exponent 10 can be written as (10/1):
√(5&x^10) = (5&x^10)^(1/2)
Now using the rule of exponentiation, we multiply the exponents:
(5&x^10)^(1/2) = 5^(1/2) * (x^10)^(1/2)
Simplifying further, we get:
√(5&x^10) = √5 * x^5
Next, let's simplify the expression ∛(125x^11/y^3).
To simplify by taking roots of the numerator and denominator, we can rewrite it using rational exponents. The cube root (∛) can be expressed as a fractional exponent (1/3), and the exponent 11 can be written as (11/1), and the exponent 3 can be written as (3/1):
∛(125x^11/y^3) = (125x^11/y^3)^(1/3)
Now using the rule of exponentiation, we multiply the exponents:
(125x^11/y^3)^(1/3) = (125)^(1/3) * (x^11)^(1/3) / (y^3)^(1/3)
Simplifying further, we get:
∛(125x^11/y^3) = 5x^(11/3) / y
Moving on to the next question, we need to multiply (-2t)^2 · (3t)^7.
(-2t)^2 = (-2)^2 · t^2 = 4t^2
(3t)^7 = 3^7 · t^7 = 2187t^7
Now multiplying both terms:
(-2t)^2 · (3t)^7 = (4t^2) · (2187t^7) = 8748t^9
For the last question, we need to divide and simplify (3y-6)/14 ÷ (y-2)/4y.
To divide fractions, we need to flip the second fraction and multiply:
(3y-6)/14 ÷ (y-2)/4y = (3y-6)/14 · (4y)/(y-2)
Now, let's simplify the expression:
(3y-6)/14 · (4y)/(y-2) = (3(y-2))/14 · (4y)/(y-2)
The (y-2) terms cancel out:
(3(y-2))/14 · (4y)/(y-2) = 3/14 · 4y = 12y/14 = 6y/7
So, the simplified expression is 6y/7.