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.
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.
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
No because quadratic equations only have 2 X-Intercepts
If the solutions are p and q, then the quadratic is (x-p)(x-q) = 0 or x2 - (p+q)x + pq = 0 Hope this is what the question meant!
No. By definition, a quadratic equation can have at most two solutions. For a quadratic of the form ax^2 + bx + c, when the discriminant of a quadratic, b^2 - 4a*c is positive you have two distinct real solutions. As the discriminant becomes smaller, the two solutions move closer together. When the discriminant becomes zero, the two solutions coincide which may also be considered a quadratic with only one solution. When the discriminant is negative, there are no real solutions but there will be two complex solutions - that is those involving i = sqrt(-1).
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.
the maximum number of solutions to a quadratic equation is 2. However, usually there is only 1.
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
No because quadratic equations only have 2 X-Intercepts
12
If the solutions are p and q, then the quadratic is (x-p)(x-q) = 0 or x2 - (p+q)x + pq = 0 Hope this is what the question meant!
No. By definition, a quadratic equation can have at most two solutions. For a quadratic of the form ax^2 + bx + c, when the discriminant of a quadratic, b^2 - 4a*c is positive you have two distinct real solutions. As the discriminant becomes smaller, the two solutions move closer together. When the discriminant becomes zero, the two solutions coincide which may also be considered a quadratic with only one solution. When the discriminant is negative, there are no real solutions but there will be two complex solutions - that is those involving i = sqrt(-1).
If the discriminant of the quadratic equation is greater than zero then it will have two different solutions. If the discriminant is equal to zero then it will have two equal solutions. If the discriminant is less than zero then it will have no real solutions.
The solutions to a quadratic equation on a graph are the two points that cross the x-axis. NB A graphed quadratic equ'n produces a parabolic curve. If the curve crosses the x-axis in two different points it has two solution. If the quadratic curve just touches the x-axis , there is only ONE solution. It the quadratic curve does NOT touch the x-axis , then there are NO solutions. NNB In a quadratic equation, if the 'x^(2)' value is positive, then it produces a 'bowl' shaped curve. Conversely, if the 'x^(2)' value is negative, then it produces a 'umbrella' shaped curve.
A quadratic equation always has TWO (2) solutions. They may be different, the same, or non-existant as real numbers (ie they only exist as complex numbers).
I gotchu homie: It's The equation has x = 4 and x = -4 as its only solutions.
Solutions: x = 9 and x = 1 Factored: (x-9(x-1) = 0 Equation: x2-10x+9 = 0