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Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a
Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a
Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a
Use the quadratic formula. In x2 - 4x + 29 = 0, a is 1, b is -4, c = 29. The quadratic formula is: x = (-b plusminus squareroot(b2 - 4ac)) / 2a
What else can I help you with?
Find three different ordered pairs that are solutions of the equation y equals 2x-1?
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What must be true of an equation before you can use the quadratic formula to find the solutions?
the formula you are going to use to answer the equation
How do you find all real solutions to the system of equations 4x 2 plus y 2 equals 100 and 4x 2 minus y 2 equals 62?
Add the two equations together. This will give you a single equation in one variable. Solve this - it should give you two solutions. Then replace the corresponding variable for each of the solutions in any of the original equations.
Find all solutions in positive integers for the equation x minus y to the power of 4 equals the LCM of x and y?
Dude, stop trying to cheat on your own maths enrichment task loser.
Find the inverse variation equation if x equals 5 when y equals 9?
The equation is xy = 5*9 = 45 Alternatively, y = 45/x
How do you factor x2-4x-80 equals 0?
Use the quadratic equation formula to find the solutions to this equation.
How can you find the complex solutions of any quadratic equation?
I suggest you use the quadratic formula.
What statement must be true of an equation before you can use the quadratic formula to find the solutions?
The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.
How many solutions are there in the equation 2x equals 10?
There is one solution. To find it, divide both sides of the equation by 2. This leaves you with x=5, where 5 is your solution.
How can someone find how many solutions there are in an equation?
you can find it by counting how many numbers they are in the equation
What is the answer for x x -6x-8 equals -56?
x² - 6x -8 = -56 <=> x² -6x + 48 = 0 We now calculate the discriminant (which equals b²-4ac for an equation of the form ax² + bx + c-: D = b²-4ac = (-6)² - 4*1*48 = 36 - 192 = -156 I don't know what kind of course in calculus you are taking, but if you only want the real answers to this equation, you can stop here because the discriminant is negative, meaning there are no real solutions. However, there are complex solutions to this equation The complex roots of D are sqrt(156)*i=12,5*i and -sqrt(156)*i=-12,5*i There are two solutions to a quadratic equation, namely: x1 = (-b + sqrt(D))/(2*a) and x2 = (-b - sqrt(D))/(2*a) so the two solutions we find for this equation are: x1 = (-b + sqrt(D))/(2*a) = (6+12,5i)/2 = 3+6,25i and x2 = (-b - sqrt(D))/(2*a) = (6-12,5i)/2 = 3-6,25i x1 and x2 are complex solutions to this quadratic equation.
Find three different ordered pairs that are solutions of the equation y equals 2x-1?
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Why can a ghost solve the equation x square root equals negative four but not the equation x square root equals four?
A ghost can "solve" the equation ( x \sqrt{} = -4 ) in a figurative sense because the equation implies the presence of complex numbers, where ( x ) can take on imaginary values. Specifically, if we interpret ( x \sqrt{} ) as ( x ), then we can find a solution using ( x = -2i ). However, in the case of ( x \sqrt{} = 4 ), the solutions are real and positive, which might be outside the ghost's ethereal realm, suggesting that the ghost prefers the complexity of imaginary solutions over straightforward real ones.
Is this true according to the zero product rule the solutions to the simpler equations are also the solutions to the original equation?
Yes, according to the zero product rule, if a product of factors equals zero, then at least one of the factors must be zero. Therefore, if you simplify an equation into factors and find the solutions for those simpler equations, those solutions will also hold true for the original equation, provided that the simplification was valid and no extraneous solutions were introduced.
How do you find the number of solutions in a quadratic equation?
Factoring by the AC method, difference of squares, perfect square trinomial. If not factorable by those ways, you can use the quadratic formula. You can also find zeros by synthetic division. If there are not any real solutions, then the solutions are said to be complex, they do not cross the x axis.
Can you solve x plus 2y equals 2?
No. There is not enough information in the equation x + 2y = 2, by itself, to solve it. There are an infinite number of solutions. A second equation, or information to allow a second equation to be derived, must be given to find a solution.
How do you find quadratic equation with only 2 solutions?
If the discriminant of the quadratic equation is equal or greater than zero it will have 2 solutions if it is less than zero then there are no solutions.
