Finding the X-Axis's?

Question

We are doing quadratic equations in class and I need help in trying to find the TWO x-axis's.

I remember that to find the x-axis you make y equal to 0. But for example take this equation:

x (squared) + 5x + 4 = 0
x (squared) + 5x = -4
Then what am I supposed to do with the squared, do I square root it?

I also have another question..

In this equation:

x = -b + (square root of) (b(squared) - 4ac)
----------------------------------------------
2 (a)

is that how I find the two x-intercepts! please help! thank you!

Also if the 2nd equation I wrote is not used for x-intercepts, what is it used for?

Answer 1

Hi Dear Jazz

x ^2 + 5x +4 = 0

∆ = b ^2 - 4ac
∆ = (5^2) - 4(1)(4) = 25 - 16
∆ = 9 & ∆ >= 0

x 1 = - b + √ ∆ / 2a = -5 + √9 / 2 = -5 + 3 / 2 = -2/2 = - 1
x 2 = - b - √ ∆ / 2a = -5 - √9 / 2 = -5 - 3 / 2 = -8 /2 = - 4

and x-intercepts is where y = 0 (The point at which a curve or function crosses the x-axis ( when y = 0 in two dimensions))
so the roots ( x1 & x2) are x-intercepts as well .
x1 = -1
x2 = -4
x-intercepts are ( -1 , 0 ) , ( -4 , 0)

Good Luck Sweetheart.

Answer 2

Oh dear!! There are *not* two x-axis. Only one. But there *are* two x intercepts (points where the curve crosses over the x-axis)

Let's look at your first equation x² + 5x + 4 = 0. This equation can be factored (You *do* remember factoring? This is a big piece of the reason that you learned it. And, if you *didn't* learn it, you better backtrack and learn it *real* quick ☺) into
(x+4)*(x+1) = 0
Now, what this means is that one of the two factors
(x+4 or x+1) has to be zero. So there are *two* ways that can happen. Either x = -4 or x = -1. And what happens if we substitute either -4 or -1 for x?
(-4)² + 5*(-4) + 4 = 16 - 20 + 4 = 0 and
(-1)² + 5*(-1) + 4 = 1 - 5 + 4 = 0
So x = -4 and x = -1 are the two x intercepts (also called 'roots') that you are looking for.

The equation x = (-b ±√(b² - 4ac))/(2a) is called the 'quadratic formula. And notice that there is a plus *and* a minus in front of the square root sign. Not just the plus sign you put in. This equation is used with what is called the 'standard form' of the quadratic equation
ax² + bx + c = 0
In the equation you have, a = 1, b = 5, and c = 4.
So the first x intercept is at
x = (-5 + √(5²-4*1*4))/(2*1) = -1 and the second is at
x = (-5 - √(5²-4*1*4))/(2*1) = -4
Note that you use the + for one root and the - for the other root in the equation.

Why have two ways to solve a quadratic? Many quadratic polynomials are just very *very* difficult to factor. You could spend all day trying to factor something like 2x² + 227x - 10950 to get
(x+150)*(2x-73), but with the quadratic equation all you have to do is plug in the numbers ☺


Doug

Answer 3

x^2 + 5x + 4 = 0 (factor)
(x+4)(x+1) = 0
=> (x+4) = 0 or (x+1) = 0
Therefore x =-4 or x=-1

these are your x-intercepts, or x axis's as you seem to like calling them.

you use the quadratic formula to solve for x only if you cannot factor easily. and it should be a +/- in front of the sqrt(b^2 - 4ac) as you must take the negative root into consideration.

Answer 4

first try to make factor s of the given equation
like the given equation u have
x^2+5x+4.1=0
for factorising multiply with the coeficient of x^2 and coefficient of x^0 i.e 1
in this example u get 4 and this 4 equals to =4*1 and 4+1=5
so it make factorising easy
x^2+4x+x+4=0
x(x+4)+1(x+4)=0
(x+4)(x+1)=0
x=-4 or -1
and the formula u mentioned is used when it is not factorized by any means and u get answer in points

Answer 5

Try using the FOIL method for factoring the equation. Then plug in the zero to find the axis. When that doesn't work, read the book like you should have done to begin with, instead of getting us to answer it for you.