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Explain why the roots of a quadratic equation are imaginary if the value of the discriminant is less than 0?
Wiki User
∙ 13y ago
The square of any real number is non-negative. So no real number can have a negative square. Consequently, a negative number cannot have a real square root. If the discriminant is less than zero, the quadratic equation requires the square root of that negative value, which cannot be real and so must be imaginary.
Wiki User
∙ 13y ago
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Which statement best explain why there is no real solution to the quadratic equation 2x2 plus x plus 7 equals 0?
The discriminant says; b^2 - 4ac 1^2 - 4(2)(7) = 1 - 56 < 1 So, less than 1 and no real roots
Explain a Quadratic equation and how to solve it?
A quadratic equation is any equation that can be expressed as ax2 + bx + c = 0.Note that the a, b and c are specified, x is the only unknown.Example:x2 - 10x - 24 = 0a, b, c are the coefficients of each term.Now x2 appears not to have a coefficient, but remember x2 is the same thing as 1x2 so the coefficient is 1. So a = 1.The second term has a coefficient of -10 because it has a minus, not plus sign in front of it so b = -10.Likewise for c, the third term. C = -24.So you have your terms.There are two popular ways of solving this.You can factorise the equation, or use the Quadratic Formula.I prefer to use the Quadratic Formula, as it is very straightforward, you just need to practise it.The quadratic formula is x = (-b±√(b2-4ac))/2a
Explain the conditions under which a quadratic equation can be solved using simple rearrangement versus using the quadratic formula?
The quadratic formula always works (as long as one considers complex numbers). "Simple rearrangement" may be quicker when the numbers look simple enough for you to decide (or rather guess) what the factors/ roots are by inspection (but the "rearrangement" method still works -- the numbers may just be more complicated). Probably the easiest quadratic is when the coefficient of x is zero (i.e. a polynomial of the form ax^2+b=0) or when there is no constant term (i.e. ax^2+bx=0) The quadratic formula cannot be used to solve an equation if a term in the equation has a degree higher than 2 (or if it can't be put in the form ax^2+bx+c=0). There are other more complex formulas for polynomials for degree 3 and 4.
Explain how you an determine if the number is a solution of an equation?
Substitute the number in the equation. If the resulting statement is true the number is a solution to the equation.
If a function is defined by an equation explain how to find its domain.?
The set of all values of x, for which the equation is true is the domain of the function defined by that equation.
Is it possible for a quadratic equation to have no real solution give examle ansd explain?
Is it possible for a quadratic equation to have no real solution? please give an example and explain. Thank you
Which statement best explain why there is no real solution to the quadratic equation 2x2 plus x plus 7 equals 0?
The discriminant says; b^2 - 4ac 1^2 - 4(2)(7) = 1 - 56 < 1 So, less than 1 and no real roots
Explain why x2 plus 4x plus 4 equals 0 has 2 x intercepts?
x2+4x+4 = 0 (x+2)(x+2) = 0 x = -2 and also x = -2 When the solutions to a quadratic equation are identical this means that both values are at the turning point or vertex of the parabola just on the x axis. d>=0,then roots of quadratic equation are real if D<0,then roots are imaginary
Explain the purpose of the discriminant?
the purpose...... it yet to be known
3 What is the objective of Discriminant Analysis. Explain the terms involved in Discriminant Analysis?
Discriminant Analysis is an analysis carried out to discriminate between the good debts and bad debts of a compan.
Explain how to find restrictions on the variable in a rational equation?
Substitute the variable in the denominator for a value that will make the denominator have a value of 0 or an imaginary number
Explain a Quadratic equation and how to solve it?
A quadratic equation is any equation that can be expressed as ax2 + bx + c = 0.Note that the a, b and c are specified, x is the only unknown.Example:x2 - 10x - 24 = 0a, b, c are the coefficients of each term.Now x2 appears not to have a coefficient, but remember x2 is the same thing as 1x2 so the coefficient is 1. So a = 1.The second term has a coefficient of -10 because it has a minus, not plus sign in front of it so b = -10.Likewise for c, the third term. C = -24.So you have your terms.There are two popular ways of solving this.You can factorise the equation, or use the Quadratic Formula.I prefer to use the Quadratic Formula, as it is very straightforward, you just need to practise it.The quadratic formula is x = (-b±√(b2-4ac))/2a
How many existing methods are there in solving quadratic equations?
There are 5 existing methods in solving quadratic equations. For the first 4 methods (quadratic formula, factoring, graphing, completing the square) you can easily find them in algebra books. I would like to explain here the new one, the Diagonal Sum Method, recently presented in book titled:"New methods for solving quadratic equations and inequalities" (Trafford 2009). It directly gives the 2 roots in the form of 2 fractions, without having to factor the equation. The innovative concept of the method is finding 2 fractions knowing their Sum (-b/a) and their Product (c/a). It is very fast, convenient and is applicable whenever the given quadratic equation is factorable. In general, it is hard to tell in advance if a given quadratic equation can be factored. However, if this new method fails to find the answer, then we can conclude that the equation can not be factored, and consequently, the quadratic formula must be used. This new method can replace the trial-and-error factoring method since it is faster, more convenient, with fewer permutations and fewer trials.
Explain the conditions under which a quadratic equation can be solved using simple rearrangement versus using the quadratic formula?
The quadratic formula always works (as long as one considers complex numbers). "Simple rearrangement" may be quicker when the numbers look simple enough for you to decide (or rather guess) what the factors/ roots are by inspection (but the "rearrangement" method still works -- the numbers may just be more complicated). Probably the easiest quadratic is when the coefficient of x is zero (i.e. a polynomial of the form ax^2+b=0) or when there is no constant term (i.e. ax^2+bx=0) The quadratic formula cannot be used to solve an equation if a term in the equation has a degree higher than 2 (or if it can't be put in the form ax^2+bx+c=0). There are other more complex formulas for polynomials for degree 3 and 4.
What algebra 2 text explains the quadratic formula best?
The use of ax^2 + bx + c=0. This is the formula and can be best used to explain it. A,b, c stand for different numerical coefficients and you use factoring to solve the equation.
Explain how you an determine if the number is a solution of an equation?
Substitute the number in the equation. If the resulting statement is true the number is a solution to the equation.
What equation for a straight line?
y=mx+b hopefully, i shouldn't have to explain to you each portion of the equation
