Quadratic equations are equations of degree 2. The degree is the highest exponent of the equation. For example, x^2+2x+1=0 is a quadratic equation because the highest exponent (degree) is 2. Quadratic equations are solved using the quadratic equation if factoring is not possible. This equation can be factored: 3x^2+4x+1=0. It is factored (3x+1)*(x+1)=0. The roots are x=-1/3 and x=-1. They are found by setting either of the two expressions in the parentheses equal to zero, then solving the equation. The first equation is 3x+1=0, subtracting 1 from both sides and dividing both sides by 3 arrives at the answer, x=-1/3. The second equation is x+1=0, subtracting 1 from both sides obtains the answer, x=-1. Quadratic equations always have two roots according to the Fundamental Theorem of Algebra. The roots are either both real or both complex.
The Quadratic Formula is -b/(2*a)+-((b^2-4*a*c)^1/2)/2*a, where a is the coefficient of the x^2 term, b is the coefficient of the x term, and c is the coefficient of the whole number term. The equation 10x^2-5x-1 is solved with a=10, b=-5, c=-1 as follows: -(-5)/(2*10)+-(((-5)^2-4*10*(-1))^1/2)/2*10=5/20+-(25+40)^1/2=(1/4+(65)^1/2)/20, (1/4-(65)^1/2)/20.
There can also be imaginary roots if the discriminant (b^2-4ac) is less than 0. The imaginary roots occur because if (b^2-4ac)^1/2 (called the “discriminant”) is a negative value, then it has complex roots because the square root of a negative value is an imaginary number. The imaginary numbers are represented as i=(-1)^1/2. The roots will be complex numbers (a+bi), (a-bi). The equation x^2+2x+2 has imaginary roots. Using the quadratic formula, a=1, b=2, c=2. Plugging in the values:
-2/2+-[(2^2-4*1*2)^1/2]/2 =-1+-[(-4)^1/2]/2=-1+-(2i)/2 =-1+-(i)=-1+i, -1-i. Now we have two complex numbers with a=-1 and b=+1, -1. The solutions can be used to check the validity of the answer: -1+i is as follows: (-1+i)^2+2(-1+i)+2=1-2i+i^2-2+2i+2=1+i^2=1-1=0.
Note that i^2=-1. This is because i=(-1)^1/2 so that i^2=((-1)^1/2)^2=-1. Imaginary roots always occur in multiples of 2. This is because if a complex number is a root, then so is its complement. The complement is the complex number with the same real coefficient but the opposite sign of the imaginary coefficient. The terminology is a+bi, a-bi, where a is a real number and i=(-1)^1/2. In the example the two complex roots are -1+i and -1-i.
If the value of the discriminant is greater than zero, then the two roots are real because the square root of a positive number is a real number. An example is x^2+3x+1=0. The discriminant is (b^2-4*a*c)^1/2, so we have (3^2-4*1*1)^1/2=(9-4)^1/2=(5)^ because a=1, b=3, c=1. The rest of the quadratic formula is -3/2*1+-((5)^1/2)/2*1=-3/2+-((5)^1/2)/2. The two roots are x=(-3+5^1/2)/2 and x=(-3-5^1/2)/2.
Sometimes a quadratic equation has a double root. This happens if the value under the square root is zero (b^2-4*a*c=0). An example is x^2+2x+1=0. Using the quadratic formula, a=1, b=2, c=1. Plugging in the values:
-2/2+-((2^2-4*1*1)^1/2)/2=-1+-0x=-1 The double root is x=-1. This can be checked by using the product x^2+2x+1=(x+1)(x+1)=0, solving for x gets x=-1, x=-1. The quadratic equation can be used to solve any even degree equations by factoring it into a product of two expressions that are half of the original equation. Examples are (x+4)^8=(x+4)^4(x+4)^4 =((x+4)^4)^2 and (x+9)^10=(x+9)^5(x+9)^5=((x+9) ^5)^2. The graph of a quadratic equation set to f(x) is a parabola.
