## anonymous one year ago 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

1. anonymous

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

2. 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

3. UsukiDoll

is your equation $-x^2-10x-3$ ?

4. 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.

5. MrNood

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

6. UsukiDoll

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

7. MrNood

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

8. 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

9. MrNood

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

10. 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

11. 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

12. UsukiDoll

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

13. MrNood

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

14. 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)

15. 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|

16. 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

17. UsukiDoll

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

18. UsukiDoll

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

19. anonymous

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

20. UsukiDoll

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

21. 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 $\large (x-5+\sqrt{22})(x-5-\sqrt{22})$

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 :/

23. 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

24. 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 ....

25. 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 $\Large{\color{green}{x=\frac{ -10 \pm \sqrt{16 \times 7} }{ 2 }}}$ $\Large{\color{green}{x=\frac{ -10 \pm4 \sqrt{7} }{ 2 }}}$

26. UsukiDoll

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

27. UsukiDoll

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

28. anonymous

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

29. anonymous

Amazing Job @UsukiDoll

30. TheViper

XD @waterineyes

31. anonymous

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

32. 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=

33. 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