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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
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Find three different ordered pairs that are solutions of the equation y equals 2x-1?
3
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 can you find the complex solutions of any quadratic equation?
I suggest you use the quadratic formula.
How do you factor x2-4x-80 equals 0?
Use the quadratic equation formula to find the solutions to this equation.
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?
3
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.
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.
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 you find the solution of a quadratic equation by graphing its quadratic equation?
When you graph the quadratic equation, you have three possibilities... 1. The graph touches x-axis once. Then that quadratic equation only has one solution and you find it by finding the x-intercept. 2. The graph touches x-axis twice. Then that quadratic equation has two solutions and you also find it by finding the x-intercept 3. The graph doesn't touch the x-axis at all. Then that quadratic equation has no solutions. If you really want to find the solutions, you'll have to go to imaginary solutions, where the solutions include negative square roots.
Solutions to the quadratic equation x2 - 5x plus 4 equals 0 are?
Oh, dude, for this quadratic equation x^2 - 5x + 4 = 0, the solutions are just the roots of the equation. You can find them by either factoring the quadratic or using the quadratic formula. So, like, the solutions are x = 1 and x = 4. Easy peasy lemon squeezy!
