This is related to the Fundamental Theorem of Algebra; read about it for more information. Basically, this theorem states that any complex polynomial has at least one root; as a corrolary - in the complex number system - a polynomial of degree "n" can be divided into "n" linear factors. For example, x2 - 5x - 6 can be expressed as (x - 2) (x - 3). (The numbers may be complex for some polynomials.)
Therefore, the corresponding equation, x2 - 5x - 6 = 0, can be written as (x - 2) (x - 3) = 0. Since a product can only be zero if at least one of its factors is zero, this lets us split the equation into two parts: (x - 2) (x - 3) = 0 is equivalent to (x - 2) = 0 or (x - 3) = 0. Each linear equation has one solution; the equation thus has two solutions. (However, there may be repeated solutions, depending on the polynomial.)
Write the quadratic equation in the form ax2 + bx + c = 0 The roots are equal if and only if b2 - 4ac = 0. The expression, b2-4ac is called the [quadratic] discriminant.
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
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.
No. Well, it depends what you mean with "any quadratic equation". The quadratic formula can solve any equation that can be converted to the form: ax2 + bx + c = 0 Note that it involves only a single variable. There are other limitations as well; for example, no additional operations. If a variable, or the square of a variable, appears in the denominator (1/x, or 1/x2), then some might say that it is "quadratic", but it might no longer be possible to convert the equation into the standard form named above. Similarly, if you have additional operations such as square roots or higher roots, trigonometric functions, etc., it might not be possible to convert the equation into a form that can be solved by the quadratic formula.
A quadratic equation is univariate: it has only one variable. A quadratic equation cannot have two variables. So, if b and c are known then it is a quadratic equation in a; if a and b are known it is a quadratic in c.Another Answer:-The question given is Pythagoras' theorem formula for a right angle triangle
the maximum number of solutions to a quadratic equation is 2. However, usually there is only 1.
Write the quadratic equation in the form ax2 + bx + c = 0 The roots are equal if and only if b2 - 4ac = 0. The expression, b2-4ac is called the [quadratic] discriminant.
The equation ax^2 + bx + c = 0 where a, b and c are real and a is non-zero has discriminant D = b^2 – 4ac. Then,if D > 0 the equation has two real roots which are distinct;if D = 0 the equation has two real roots which are coincident;if D < 0 the equation has two roots which form a complex conjugate pair (advanced mathematics only).
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
No. The quadratic may have what's known as repeated roots, where it only has one root; for example, x2 + 2x + 1 = (x+1)(x+1) = 0 has only one root at x = -1. It always has roots, but can be imaginary roots, also, when no part of the graph intersects the X axis.
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.
No. Well, it depends what you mean with "any quadratic equation". The quadratic formula can solve any equation that can be converted to the form: ax2 + bx + c = 0 Note that it involves only a single variable. There are other limitations as well; for example, no additional operations. If a variable, or the square of a variable, appears in the denominator (1/x, or 1/x2), then some might say that it is "quadratic", but it might no longer be possible to convert the equation into the standard form named above. Similarly, if you have additional operations such as square roots or higher roots, trigonometric functions, etc., it might not be possible to convert the equation into a form that can be solved by the quadratic formula.
The quadratic formula can be used to solve an equation only if the highest degree in the equation is 2.
A quadratic equation is univariate: it has only one variable. A quadratic equation cannot have two variables. So, if b and c are known then it is a quadratic equation in a; if a and b are known it is a quadratic in c.Another Answer:-The question given is Pythagoras' theorem formula for a right angle triangle
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
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.
Only that the equation that you are trying to solve is a quadratic, that is to say, the powers of the variable are 2,1 and 0 (or any constant increment of these three numbers). Non-negativity of the discriminant is NOT a condition because you can still use the quadratic formula and get roots that are in the complex domain.