How do I find zeroes of quadratic functions?
Saying that we are "finding the zeros of quadratic functions" is another way of saying that we are finding its x-intercepts.
To do that, you state the quadratic equation as
y = ax^2 + bx + c
From there, all you are doing is solving for x when y = 0.
There are four ways to do this:
The first is by factoring. The second method is by using square roots. The third is by completing the square. The last of the methods is by using the quadratic formula.
There is not room to explain all of these here, but the link I am including explains each of these in detail.
you can find the zeros of the quadratic funtion by
* factorizing example (x+2)(x-2)=0
* squreroot example -25x^2=o
root x^2=root 25
To find the zeros of the quadratic equation, you have to find out the values of the unknwons for which the quadratic expression becomes zero. The quadratic equation is an equation of second degree equation of an unknown (or variable ).
1) x^2 = 9. Solution: x = + (9)^(1/2) = 3 or x = -(9)^(1/2) =-3
Second way: x^2=9 . Subtract 9 from both sides. Then x^2-9= 0 . Factorising the left you get:(x-3)(x+3) = 0. To get zero of the product, one of the products, x-3 =0 or x+3 = 0 which gives you the zeros of the equation, x^2 = 9 or x^2 - 9 = 0 to be x=3 or x = - 3.
2) 5x^2 = 14: Solution x^2 = 14/5. So x= -or+ [sqrt(14/5)]
3) Now you can generalise the procedure. But I give you a numerical type example of the type ax^2+bx+c = 0 form.
x^2-3x+2=0 could be re written like:
x^2-3x+(3/2)^2-(3/2)^2+2 =0, adding subtracting (3/2)^2 to make it perfect square and number .
(x-3/2)^2 -(3/2)^2+2 = 0
(x-3/2)^2 = (3/2)^2- 2 = 9/2 - 2 =1/4
Taking square root,
x-3/2 = + or-1 sqrt(1/4) = +or (1/2)
x = 3/2+1/2 = 2 or x= 3/2-1/2 =1.