No because quadratic equations only have 2 X-Intercepts
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.
at first the first person to solve the quadratic equation is from the middle kingdom of Egypt. Greeks were also able to solve the quadratic equation but that was on the unproper way. Greeks were able to solve the quadratic equation by geometric method or equlid's method. equlid's method contains only three quadratic equation. dipohantus have also solved the quadratic equations but he have solved by giving only two roots any they both were only of positive signs.After that arbhatya also gave the two formulas for quadratic equation but the bentaguptahave only accepted only one of them after theat some of the Indian mathematican have also solved the quadratic equation who gave the proper definations and formula and in this way quadratic equation have been formed. Prabesh Regmi Kanjirowa National School
Solutions: x = 9 and x = 1 Factored: (x-9(x-1) = 0 Equation: x2-10x+9 = 0
If the discriminant - the part under the radical sign in the quadratic formula - is negative, then the result is complex, it is as simple as that. You can't convert a complex number to a real number. If a particular problem requires only real-number solutions, then - if the formula gives complex numbers - you can state that there is no solution.
There are several methods for solving quadratic equations, although some apply only to specific quadratic equations of specific forms. The methods include:Use of the quadratic formulaCompleting the SquareFactoringIterative methodsguessing
Quadratic curves only have two solutions when the discrimant is greater than or equal to zero.
No because quadratic equations only have 2 X-Intercepts
Graphing
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.
at first the first person to solve the quadratic equation is from the middle kingdom of Egypt. Greeks were also able to solve the quadratic equation but that was on the unproper way. Greeks were able to solve the quadratic equation by geometric method or equlid's method. equlid's method contains only three quadratic equation. dipohantus have also solved the quadratic equations but he have solved by giving only two roots any they both were only of positive signs.After that arbhatya also gave the two formulas for quadratic equation but the bentaguptahave only accepted only one of them after theat some of the Indian mathematican have also solved the quadratic equation who gave the proper definations and formula and in this way quadratic equation have been formed. Prabesh Regmi Kanjirowa National School
12
I may only be in 8th grade but I am absolutely positive that all quadratic equations have 2 solutions. No - They may have 0,1, or 2 answers For example, the problem x^2 + 8x +16 = 0 has only one solution -4. This is because the radical evaluates to 0 rendering the +/- sign irrelevant.
no only equations with x2 and lower powers can be considered quadratic. those with x3 cannot be considered quadratic, just as x2 cannot be considered linear
No, this is only true for equations of the quadratic form, actually it is the definition of a quadratic equation. for ex: those of form ax^3+bx^2+cx+d=0 then the highest power of the unknown is 3
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.
Solutions: x = 9 and x = 1 Factored: (x-9(x-1) = 0 Equation: x2-10x+9 = 0