What are Quadratic Equations
In the word Quadratic, ‘Quad’ means square. So we can define a Quadratic equation as a second degree polynomial equation in one variable namely, ‘x’. Here the variable gets squared, like x^2 and hence the name quadratic. In Algebra Quadratic Equations are the equations in a single variable and have the highest degree as 2.
It is also referred to as an ‘equation of degree two’ the highest degree being two. The standard form of a Quadratic Equation is given by, ax^2+bx+c=0 where ‘a’ is not equal to zero and a,b,c are coefficients which are the knowns, ‘x’ is an unknown and is a variable. For example, given 2x^2+4x +3=0 when we compare the terms with the standard equation we get, a=2, b=4 and c=3.
The solutions of these equations which are also called the roots can be found by using factorization method or using the special products. If there are no factors then completing the square method is used,
ax^2+bx+c=0
x^2+(b/a)x = -(c/a)
[x+(b/2a)]2 = -(c/a) + b^2/4a^2
[x+(b/2a)]2 = [b^2 – 4ac]/4a^2
x+(b/2a)= sqrt [(b^2 – 4ac)/4a^2]
x + (b/2a) = sqrt[(b^2 – 4ac)]/2a
x=[ -b sqrt(b^2 – 4ac)]/2a is the quadratic formula to find the roots.
While solving quadratic equations, first we calculate discriminant denoted by the letter‘d’. The formula to calculate discriminant is given by, d=b^2-4ac. Depending on the value of this discriminant the type of solution can be determined. The following are the possibilities,
If the discriminant is greater than zero, that is if b^2- 4ac > 0 then the equation has two solutions given by, x1= [-b+ sqrt(b^2- 4ac)]/2a and x^2 = [-b - sqrt(b^2- 4ac)]/2a
If the discriminant is equal to zero, that is if b^2- 4ac=0 then there is one solution given by,
x=-b/2a
If the discriminant is less than zero, that is if b^2-4ac < 0 then no solutions are defined for that equation
Having problem with Compound Interest Formula Monthly keep reading my upcoming posts, i will try to help you.
We shall now learn to solve some of the Quadratic Equations Problems
Solve x^2 + x – 4=0. In this problem factorization method cannot be used as the terms cannot be factored. So, we shall use the completing the squares method, that is quadratic formula, x=[ -b sqrt (b^2 – 4ac)]/2a.
Comparing the terms with the standard equation ax^2+bx+c=0 we get, a=1, b=1 and c=-4. Substituting these values we get,
x ={ -1sqrt[12- 4(1)(-4)]}/2(1)
=[-1sqrt(1+16)]/2
x =(-1)/2 is the solution
Solve x^2-16. The equation has two terms and it is difference of squares; [a^2-b^2=(a+b)(a-b)]
x^2-42=0 ; here a=x, b=4
(x+4)(x- 4)=0
x+4=0; x-4=0
x=4 is the solution
In the word Quadratic, ‘Quad’ means square. So we can define a Quadratic equation as a second degree polynomial equation in one variable namely, ‘x’. Here the variable gets squared, like x^2 and hence the name quadratic. In Algebra Quadratic Equations are the equations in a single variable and have the highest degree as 2.
It is also referred to as an ‘equation of degree two’ the highest degree being two. The standard form of a Quadratic Equation is given by, ax^2+bx+c=0 where ‘a’ is not equal to zero and a,b,c are coefficients which are the knowns, ‘x’ is an unknown and is a variable. For example, given 2x^2+4x +3=0 when we compare the terms with the standard equation we get, a=2, b=4 and c=3.
The solutions of these equations which are also called the roots can be found by using factorization method or using the special products. If there are no factors then completing the square method is used,
ax^2+bx+c=0
x^2+(b/a)x = -(c/a)
[x+(b/2a)]2 = -(c/a) + b^2/4a^2
[x+(b/2a)]2 = [b^2 – 4ac]/4a^2
x+(b/2a)= sqrt [(b^2 – 4ac)/4a^2]
x + (b/2a) = sqrt[(b^2 – 4ac)]/2a
x=[ -b sqrt(b^2 – 4ac)]/2a is the quadratic formula to find the roots.
While solving quadratic equations, first we calculate discriminant denoted by the letter‘d’. The formula to calculate discriminant is given by, d=b^2-4ac. Depending on the value of this discriminant the type of solution can be determined. The following are the possibilities,
If the discriminant is greater than zero, that is if b^2- 4ac > 0 then the equation has two solutions given by, x1= [-b+ sqrt(b^2- 4ac)]/2a and x^2 = [-b - sqrt(b^2- 4ac)]/2a
If the discriminant is equal to zero, that is if b^2- 4ac=0 then there is one solution given by,
x=-b/2a
If the discriminant is less than zero, that is if b^2-4ac < 0 then no solutions are defined for that equation
Having problem with Compound Interest Formula Monthly keep reading my upcoming posts, i will try to help you.
We shall now learn to solve some of the Quadratic Equations Problems
Solve x^2 + x – 4=0. In this problem factorization method cannot be used as the terms cannot be factored. So, we shall use the completing the squares method, that is quadratic formula, x=[ -b sqrt (b^2 – 4ac)]/2a.
Comparing the terms with the standard equation ax^2+bx+c=0 we get, a=1, b=1 and c=-4. Substituting these values we get,
x ={ -1sqrt[12- 4(1)(-4)]}/2(1)
=[-1sqrt(1+16)]/2
x =(-1)/2 is the solution
Solve x^2-16. The equation has two terms and it is difference of squares; [a^2-b^2=(a+b)(a-b)]
x^2-42=0 ; here a=x, b=4
(x+4)(x- 4)=0
x+4=0; x-4=0
x=4 is the solution
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