The quadratic formula is a method that is used to find the roots of a quadratic equation. Several methods have been proposed for the solution of quadratic equations.
For a general quadratic equation of the form:
`ax^2 + bx + c = 0`
where a,b,c are constants (can be -ve) and where a !=0, the quadratic formula is given by
` x=(-b+-sqrt(b^2-4ac))/(2a)`
This formula is so important that you must learn it by heart for good
According to this solution, there are two roots of the quadratic equation. And they are given as:
`x=(-b+sqrt(b^2-4ac))/(2a) ` , ` x=(-b- sqrt(b^2-4ac))/(2a) `
i.e. one with the positive sign, while the other has a negative sign.
In the expression for the roots of a quadratic equation, the term present inside the square root `(b^2-4ac)` is termed as discriminant and the value of this expression is of great significance.
There are 3 different possibilities with regards to the value of discriminant and the consequences of each type of value have been discussed below:
If the discriminant has a positive value, the two roots of the quadratic equation are both distinct and real.
If the discriminant is equal to zero, the two roots are equal and real.
The third possibility arises when the discriminant is negative i.e. less than zero. In this case, the two roots are distinct and conjugate complex number.
In order to solve this equation using quadratic formula discussed above, we need to compare the coefficients of the given equation with the general form of quadratic equation:
`ax^2 + bx + c = 0`
Comparison shows that
a = 1
b = - 5
c = 3
Using the formula, we have
` x=(-b+-sqrt(b^2-4ac))/(2a) `
=>` x=(-(-5) +- sqrt((-5)^2-4(1)(3)))/(2(1))`
=> ` x=( +5 +- sqrt(25 - 12 ))/2`
=> ` x=( +5 +- sqrt(13 ))/2`
Where the two roots are real and distinct ( 13 > 0 )
Compare the given equation with the general form of quadratic equation
`ax^2 + bx + c = 0`
Comparison shows that
a = 4
b = - 6
c = 2
Use the quadratic formula to find x.
`x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-(-6)+-sqrt([(-6)]^2-4(4)(2)))/(2(4))`
`x=( 6 +-sqrt( 36-32 ))/8`
`x=( 6 +-sqrt( 4 ))/8`
`x=( 6 +-2 )/8`
=> `x=( 6 + 2 )/8 ` and `x=( 6 - 2 )/8`
=> x=1 and ` x=( 4 )/8 =( 1 )/2 `
Compare the given equation with the general form of quadratic equation
`ax^2 + bx + c = 0 `
But wait a minute, where is c? What is different in this quadratic equation? The right hand-side of this equation is non-zero.
We can subtract 11 from both sides of the equation to get
`x^2 + 4x - 11 = 11 - 11`
=> ` x^2 + 4x - 11 = 0`
Comparison shows that
a = 1
b = 4
c = - 11
Use the quadratic formula to find x.
`x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-4+-sqrt((4)^2-4(1)(-11)))/(2(1))`
`x=(-4+-sqrt( 16+44 ))/2`
`x=(-4+-sqrt( 60 ))/2`
`x=(-4+-2sqrt( 15 ))/2`
`x= -2+-sqrt( 15 ) `
Compare the given equation with the general form of quadratic equation
`ax^2 + bx + c = 0`
Comparison shows that
a = 1
b = 4
c = 4
Use the quadratic formula to find x.
`x=(-b+-sqrt(b^2-4ac))/2a`
`x=(-4+-sqrt((4)^2-4(1)(4)))/(2(1))`
`x=(-4+-sqrt( 16-16 ))/2`
`x=(-4+-0 )/2`
`x= -2,-2`
Hence the two roots are real and duplicate since `b^2 - 4ac = 0 `
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