Find the x-intercepts.x^2-x-10?

Answer 1

x^2 - x - 10 = 0

You solve by using the quadratic formula

x = -b +/- [sqrt(2) b^2 - 4ac] / 2a

x = -(-1) +/- [sqrt(2) (-1)^2 - 4(1)(-10)] / 2(1)

x = 1 +/- sqrt(2) 41 / 2

x = 3.7
x = -2.7

Answer 2

The "x-intercepts" are the points where the graph of the equation crosses the x axis. This is where the value of the equation is 0. The places where this occurs (the x-value where the function is worth zero) are called roots.

For a quadratic equation (a second degree polynomial in one variable), there are a few tricks.

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The quadratic solution (a.k.a. the quadratic formula):

If the function is written in this form:

f(x) = ax^2 + bx + c

then the roots are found with

x = [ -b +/- SQRT( b^2 - 4ac) ] / 2a

The +/- means "plus or minus" as minus numbers can be the square root of a positive number.
e.g., The square root of 4 is +2 and it is ALSO -2.

If the "determinant" (b^2 - 4ac) is negative, then the square root -- in real numbers -- does not exist and the equation has no real roots.
If the determinant is 0, then the square root is 0 and there is only one value of x (the curve simply touches the x-axis without crossing it). Some say that there is only one root, but most mathematicians will say that there are two identical roots (in advanced math, there is a subtle difference).

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Another way is to find the factors of the equation:

If you can write the equation in the form:

f(x) = (x-m)(x-n)
then +m and +n are the roots.
The demonstration is easy:
if x=m, then x-m=0 and 0 times whatever is 0.
Therefore, m is a root.

To find values for m and n:

(x-m)(x-n) = x^2 +(-m-n)x +mn

In your equation,
(-m-n) = -1
and
mn = -10

Sometimes it is easy, sometimes it is not.

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Another way is to complete a square, allowing you to have an equation that has squares on both side. Taking the square root reduces the equation to a first degree equation. Screaming monk has already shown how to do this.

PS
(in a previous version, I had a typo here, and I apologize)
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In weird cases (where a root is not a rational number), one can use an approximation method:

x^2 -x -10 = 0
x^2 = x + 10
x = SQRT(x+10)

And you do this iteratively, starting with a number close to a suspected root. The answer you find for x is used in the next calculation. If the curve is concave, the intermediate values of x will converge towards the real solution.
If they do not converge, flip it around:

0 = x^2 -x -10
x = x^2 - 10

This method is best done with a good calculator or with a spreadsheet (like Excel).

Using the square root option, and starting with 0, I get:
0
3.16227766
3.627985345
3.691610129
3.700217579
3.701380496
3.701537585
3.701558805
3.701561671
3.701562058
3.701562111
3.701562118
3.701562119

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Finally, if you stumble upon a root (let us say "m"), then you can find the other root by long division:

x^2 -x -10 divided by (x-m) will yield (x-n)

If m is an irrational number (as above), this could take a while. This trick is good if you try an easy value like an integer (x=1 or x=-2) and it gives you f(x)=0 right away. Having one root, the division may be a quick way to find the other root.

Answer 3

x^2-x-10 = 0
=> x = (1/2) [ 1 ± √41) ]

Answer 4

You need to set this equal to something for x intercepts to exist. To find the 'zeroes' or 'x' intercepts, set equal to zero. Then, you could complete the square, if you prefer:

0 = x^2 - x -10
10 = x^2 -x
Now take the single degree term's(-1x) coefficient (-1), halve it(-1/2), and then square it (1/4) Notice that squaring produces a positive number.
Place this value (1/4) on both sides of the equal sign to maintain integrity of the identity (keep left side equal to right side).
10 + 1/4 = x^2 -x + 1/4
10.25 = (x - 0.5)^2
sqrrt(10.25) = x-0.5
0.5 +/- 3.2 =x
so solution set for x is:
{3.7,-2.7} not an exact answer BTW

Answer 5

x^2 - x - 10 = 0,

x1 = 1/2+root(41)/2
x2 = 1/2-root(41)/2