You are finding the roots or solutions. These are the values of the variable such that the quadratic equation is true. In graphical form, they are the values of the x-coordinates where the graph intersects the x-axis.
A linear equation is that of a straight line. Any one of the infinitely many points on the line will be solutions. If the equation is in terms of the variables x and y, just pick any two values of x, solve for y and the results will be the coordinates of two solutions.
The equation is |x|2-3|x|+2=0 If x>0 then the equation becomes x2-3x+2=0 (x-2)(x-1)=0 x=1,2 We get two values for x. If x<0, then the equation is again x2-3x+2=0 We again get two values. Therefore, the total number of solutions=4.
A quadratic equation is one that can be written as y=Ax^2+Bx+C. The solutions are the values of x that make y=0. If an equation has solutions, say x=M and x=N, then Ax^2+Bx+C=(x-M)(x-N). For example: y=x^2-5x+6 So we want to find what values of x make the equation true: 0=x^2-5x+6 This happens at x=2, when y=(2)^2-5*(2)+6 =4-10+6 =0 and at x=3, when y=(3)^2-5*(3)+6 =9-15+6 =0 So the solutions are x=2 and x=3, and the equation can be written as y=(x-2)(x-3).
Select any three values of x in the domain of the equation. Solve the equation at these three points for the other variable, y. Then each (x, y) will be an ordered pair that is a solution of the equation.
Roots, zeroes, and x values are 3 other names for solutions of a quadratic equation.
You are finding the roots or solutions. These are the values of the variable such that the quadratic equation is true. In graphical form, they are the values of the x-coordinates where the graph intersects the x-axis.
Each value of x, when substituted in the equation, will give a true statement.
A linear equation is that of a straight line. Any one of the infinitely many points on the line will be solutions. If the equation is in terms of the variables x and y, just pick any two values of x, solve for y and the results will be the coordinates of two solutions.
The equation is |x|2-3|x|+2=0 If x>0 then the equation becomes x2-3x+2=0 (x-2)(x-1)=0 x=1,2 We get two values for x. If x<0, then the equation is again x2-3x+2=0 We again get two values. Therefore, the total number of solutions=4.
No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2No. The resulting equation has more solutions. For example, x = 2 has only one solution and that is x = 2.butx2= 4, the squared equation, has two solutions: x = +2 and x = -2
A quadratic equation is one that can be written as y=Ax^2+Bx+C. The solutions are the values of x that make y=0. If an equation has solutions, say x=M and x=N, then Ax^2+Bx+C=(x-M)(x-N). For example: y=x^2-5x+6 So we want to find what values of x make the equation true: 0=x^2-5x+6 This happens at x=2, when y=(2)^2-5*(2)+6 =4-10+6 =0 and at x=3, when y=(3)^2-5*(3)+6 =9-15+6 =0 So the solutions are x=2 and x=3, and the equation can be written as y=(x-2)(x-3).
Select any three values of x in the domain of the equation. Solve the equation at these three points for the other variable, y. Then each (x, y) will be an ordered pair that is a solution of the equation.
zeros values at which an equation equals zero are called roots,solutions, or simply zeros. an x-intercept occurs when y=o ex.) y=x squared - 4 0=(x-2)(x+2) (-infinity,-2)(-2,2) (2,infinity)
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1) When solving radical equations, it is often convenient to square both sides of the equation. 2) When doing this, extraneous solutions may be introduced - the new equation may have solutions that are not solutions of the original equation. Here is a simple example (without radicals): The equation x = 5 has exactly one solution (if you replace x with 5, the equation is true, for other values, it isn't). If you square both sides, you get: x2 = 25 which also has the solution x = 5. However, it also has the extraneous solution x = -5, which is not a solution to the original equation.
Yes and yes. eg x = y + 1 has an infinite number of solutions, and {sin(x) + cos(x) = 2} does not have a solution.