Can someone please help me ? Im trying to factor this equation.
F(x)= (-x)^2 - 10x -3
Dont just give me the answer, i need explanation please. I'm doing a COMPASS test on weds

- anonymous

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- anonymous

\[-x^{2}-10x-3 \rightarrow x=5\pm \sqrt{25-3}\]
\[f(x)= (x-5-\sqrt{22})(x-5+\sqrt{22})\]

- anonymous

how do you get the square root of 22 ? Can you make it an easier explanation for me because i first set it up as (x )(x ) and i couldnt figure out the rest because no factors of -3 were equal to the sum -10

- UsukiDoll

is your equation
\[-x^2-10x-3 \] ?

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## More answers

- anonymous

there exists the formula: \[\Delta= \frac{ -\frac{ b }{ 2 }\pm \sqrt{(\frac{ b }{ 2 }})^{2}-ac }{ a }\]
where a,b and c are the numbers in front of x^2, x, and the last term, respectively.

- MrNood

there is a difference between
\[-x ^{2} and (-x)^{2}\]
which do you mean?

- UsukiDoll

Maybe let's assume that it is -x^2 ? since the original poster did say that they couldn't factor ?!

- MrNood

y= -x^2-10x-3
has 2 real roots, but is not factorisable

- UsukiDoll

true @MrNood because when we used the discriminant formula \[b^2-4ac \] letting a = -1, b = -10, and c = -3 we don't have a perfect square

- MrNood

same is true for
y= (-x)^2 - 10x -3

- MrNood

of course we can use the quadratic formula ot get the solution
BUT the OP asks for factorisation
@francemaz
I think your version of the equation is not correct

- UsukiDoll

We have no choice but to use the quadratic formula to get the roots so we can do factorization. Otherwise, what's the alternative? @MrNood

- UsukiDoll

only part of his version is correct. There are sign errors

- MrNood

using the quadratic formula will get the 2 solutions.
This is not 'factorisation' in my understanding of the term

- UsukiDoll

then just leave the problem alone afterwards? are you serious? By the discriminant formula, we either have a perfect square (yes we can factor) or not a perfect square (we have to use the quadratic equation)

- anonymous

given eq can be written as (-x)^2+ 10(-x) -3
now use quadratic formula using -x instead of x and a=1, b= 10 , c=-3,|dw:1436865132787:dw|

- UsukiDoll

\(\color{blue}{\text{Originally Posted by}}\) @eninone
given eq can be written as (-x)^2+ 10(-x) -3
now use quadratic formula using -x instead of x and a=1, b= 10 , c=-3,Created with Raphaël-x = -b 4acReply Using Drawing
\(\color{blue}{\text{End of Quote}}\)
huh I thought if we have -x^2 supposedly our a = -1?
because for ax^2+bx+c
-x^2 means that a is -1

- UsukiDoll

on top of that we can't neglect the negative sign for -10x either so b = -10

- UsukiDoll

the equation we are given is either
\[f(x) = -x^2-10x-3 \]
or
\[f(x) =(-x)^2-10x-3 \rightarrow x^2-10x-3\]

- anonymous

theposter has used (-x)^2 not -x^2

- UsukiDoll

We wouldn't know until the original poster clarified it.

- UsukiDoll

Anyway, assuming that we have what @francemaz ended up with
We are given the equation \[-x^2-10x-3\]
The standard quadratic equation is in the form of \[ax^2+bx+c\]
to figure out if our equation can factor or not we use the discriminant formula which is \[b^2-4ac\]
Since we are given \[-x^2-10x-3 \] and it is in the form of \[ax^2+bx+c\]
we have a = -1, b =-10, and c=-3.
So now we plug these values into the discriminant formula
If we have a perfect square like 9, 16, 25... then it is possible to factor this equation. However, if we don't have a perfect square like 12 or 78, then we need to use the quadratic formula which is
\[\LARGE \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \\ \end{array}\]
So back to the discriminant formula,
\[b^2-4ac\]
we plug in
a = -1, b =-10, and c=-3.
\[(-10)^2-4(-1)(-3)\]
\[100-12=88\]
Ok. Not a perfect square, so we need to use the quadratic formula
\[\LARGE \begin{array}{*{20}c} {x = \frac{{ 10 \pm \sqrt {88} }}{{-2}}} \\ \end{array}\]
so now we need to simplify \[\sqrt{88}\]
We need to find the smallest perfect square number so we can simplify \[\sqrt{88}\].
The only smallest perfect square number that can work is 4
because 4 x 22 = 88. Therefore
\[\LARGE \begin{array}{*{20}c} {x = \frac{{ 10 \pm \sqrt {4 \cdot 22} }}{{-2}}} \\ \end{array}\]
The square root of 4 is 2
\[\LARGE \begin{array}{*{20}c} {x = \frac{{ 10 \pm 2\sqrt { 22} }}{{-2}}} \\ \end{array}\]
so now we split this fraction up into two cases: one with the addition sign and one with the subtraction sign
\[\LARGE \begin{array}{*{20}c} {x = \frac{{ 10 + 2\sqrt { 22} }}{{-2}}} \\ \end{array}\]
\[\LARGE \begin{array}{*{20}c} {x = \frac{{ 10 - 2\sqrt { 22} }}{{-2}}} \\ \end{array}\]
Now we can just simplify by splitting the fractions
\[\large x = \frac{10}{-2}+\frac{2\sqrt{22}}{-2} \]
\[\large x = -5-\sqrt{22}\]
\[\large x = -5+\sqrt{22}\]
so our factorization looks like this
\[\large (x-5+\sqrt{22})(x-5-\sqrt{22})\]

- UsukiDoll

@Coca_Lola4 here is the in-depth extension of an explanation on what @francemaz wrote. Sorry for the dramasode that happened after his post :/

- TheViper

\[(-x)^2-10x-3=x^2-10x-3\]
\[\Large{\color{green}{x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }}}\]
The method is Sri Dharacharya Method or u can say Quadratic Equation.
Here a=coefficient of \(x^2\) b=coefficient of x
c=constant term

- sohailiftikhar

every equation can be solve by quadratic formula so If you are facing fromlem in factorizing a equation or can't solve it by completing aquar then just use quadratic formula then you will get your ans ..@UsukiDoll clarify it nicely ....

- UsukiDoll

Ok. Let's do the other version...just in case
so suppose we have the equation \[x^2-10x-3 \]
then a = 1, b = -10, and c =-3
Using the discriminant formula we have
\[b^2-4ac = (-10)^2-4(1)(-3)\]
\[100-4(-3) = 100+12 = 112 \]
also not a perfect square
\[\Large{\color{green}{x=\frac{ -10 \pm \sqrt{112} }{ 2 }}}\]
so we need the smallest perfect square which in this case 112 = 7 x 16
\[\Large{\color{green}{x=\frac{ -10 \pm \sqrt{16 \times 7} }{ 2 }}}\]
\[\Large{\color{green}{x=\frac{ -10 \pm4 \sqrt{7} }{ 2 }}}\]

- UsukiDoll

\[\large \frac{-10}{2}+\frac{4 \sqrt{7}}{2} \rightarrow -5+2 \sqrt{7}\]
\[\large \frac{-10}{2}-\frac{4 \sqrt{7}}{2} \rightarrow -5-2 \sqrt{7}\]

- UsukiDoll

should have something like
\[\large (x-5+2 \sqrt{7})(x-5-2\sqrt{7})\]

- anonymous

Waah Shri Dharacharya..!!! @TheViper ... :P

- anonymous

Amazing Job @UsukiDoll

- TheViper

XD @waterineyes

- anonymous

Yeah, it was -x^2 but it looked more confusing if i didnt put the ( ) sorry. I forgot about the quadratic formula, thanks

- anonymous

Im still soooo confused. Heres the original question
Consider the parabola given by the equation:
f(x)=−x2−10x−3
Find the two values of x that make f(x)=0. Round your answers to two decimal places. Write the values as a list, separated by commas:
x=

- UsukiDoll

Ok so... we're dealing with the same equation, except we have the find the equilibrium that makes the function 0
so we have to find all x values that makes f(x) = 0
so we set the equation to 0
0 = -x^2-10x-3 and practically do the same thing as I've done above because it's not factorable and the discriminant formula will produce 88 which is not a perfect square.
you will have the same answer as factorization.. only it's going to be the roots that we need after using the quadratic formula.
\[\large x = -5-\sqrt{22} \]
\[\large x = -5+\sqrt{22} \]
so these are your x values that will make your function 0.
The process of factorization and finding all x values that makes f(x) = 0 is the same only for factorization we take it a step further and write
\[\large (x-5+\sqrt{22})(x-5-\sqrt{22}) \]
Suppose we want to check and see if these roots do make the function go to 0
Then we plug in one of the values and simplify to see if we have 0=0
\[\large 0 = -x^2-10x-3 \]
For \[\large x = -5-\sqrt{22} \]
\[\large 0 = -(-5-\sqrt{22})^2-10(-5-\sqrt{22})-3 \]
\[\large 0 = -(-5-\sqrt{22})^2+50+10 \sqrt{22} -3 \]
\[[\large -(-5-\sqrt{22})^2 \rightarrow -47-10\sqrt{22}]\]
\[\large 0 = -47-10 \sqrt{22}+50+10 \sqrt{22} -3 \]
\[\large 0 = -47-3+50-10 \sqrt{22}+10 \sqrt{22} \] (rearrange terms)
\[\large 0 = -50+50-10 \sqrt{22}+10 \sqrt{22} \] (combine like terms)
\[\large 0 = 0-10 \sqrt{22}+10 \sqrt{22} \]
\[\large 0 = 0-0 \]
\[\large 0 = 0\]
Similarly for
\[\large x = -5+\sqrt{22} \]
\[\large 0 = -( -5+\sqrt{22} )^2-10( -5+\sqrt{22} )-3 \]
\[\large 0 = -( -5+\sqrt{22} )^2+50-10\sqrt{22}-3 \]
Expand
\[\large -( -5+\sqrt{22} )^2 \rightarrow 10 \sqrt{22}-47\]
\[\large 0 =10 \sqrt{22}-47+50-10\sqrt{22}-3 \] (rearrange terms)
\[\large 0 =10 \sqrt{22}-10\sqrt{22}-47+50-3 \] (combine like terms)
\[\large 0 =10 \sqrt{22}-10\sqrt{22}+3-3 \]
\[\large 0 =10 \sqrt{22}-10\sqrt{22}+0 \]
\[\large 0 =0+0 \]
\[\large 0 =0 \]
We have successfully found our x-values that will make our function f(x) = 0

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