How do i caluclate the roots of the equation for:
x^2-3x-12=0
(x^2-4)=0 (x+3)=0
x^2=4 x+3+0
x=4 x=-3
Am i on the right path or just way off??
I am assuming there is only the one equation
x^2 - 3x - 12 = 0, because what follows makes no sense
First of all your quadratic does not factor, so you will have to either use the quadratic equation formula or complete the square.
let's use the formula:
x = (3 ± √(9-4(1)(-12))/2
= (3 ± √57)/2
Use your calculator if you need a decimal equivalent.
Ok i may be dense but i don't ge tthe formula cause when i work out
3 �} �ã(9-4(1)(-12))/2
I get 60 not 57, and i don't understand how you got 9 or 4 either
for any quadratic equation
ax^2 + bx + c = 0
x = (-b ± √(b^2 - 4ac))/(2a)
so in our case
a = 1, b = -3, and c = -12
look at the second example on this page
http://www.nipissingu.ca/calculus/tutorials/quadratics.html
You are on the right track! To find the roots of a quadratic equation, you need to solve for the values of x that make the equation true. Let's go through your steps:
1. x^2-3x-12=0: This is a quadratic equation in standard form. To solve it, you can use the factoring method, completing the square method, or quadratic formula. In this case, factoring is the simplest approach.
Factor the quadratic equation:
(x-4)(x+3) = 0
Now, set each factor equal to zero and solve for x:
x - 4 = 0 -> x = 4
x + 3 = 0 -> x = -3
So, the roots of the equation x^2-3x-12=0 are x = 4 and x = -3.
2. x^2 - 4 = 0: This equation is already factored as a difference of squares.
(x+2)(x-2) = 0
Set each factor equal to zero to find the roots:
x + 2 = 0 -> x = -2
x - 2 = 0 -> x = 2
So, the roots of the equation x^2 - 4 = 0 are x = -2 and x = 2.
Keep in mind that factoring is not always the easiest method to solve quadratic equations, especially for more complex equations. In those cases, you can use the quadratic formula, which is a general formula for finding the roots of any quadratic equation.