5y2 : the only factors are 5y and y
Similarly the only factors of 5 are 5 and 1
these figures must be arranged so that their indiviual multiples total 26y:
This can only be (5y + 1)(y + 5)
So x = -1/5 or -5
To find the solution to this equation, you need to rearrange the terms and solve for the variable. 4 = 2b + b^2 can be rewritten as b^2 + 2b - 4 = 0. You can then solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
Sum
Whether or not that there is a solution to a quadratic equation,
Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.
The solution to a math problem involving a quadratic equation is the values of the variable that make the equation true, typically found using the quadratic formula or factoring.
If you have a quadratic equation and there is no x to put into the equation to get zero. The graph is like a U that is above the x axis or a cap that is below the x axis.
To find the solution to this equation, you need to rearrange the terms and solve for the variable. 4 = 2b + b^2 can be rewritten as b^2 + 2b - 4 = 0. You can then solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
Sum
x2
Is it possible for a quadratic equation to have no real solution? please give an example and explain. Thank you
If the discriminant of a quadratic equation is less than zero then it has no solutions.
Whether or not that there is a solution to a quadratic equation,
It has one real solution.
1,2.5
Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.