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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

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  1. anonymous
    • one year ago
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    \[-x^{2}-10x-3 \rightarrow x=5\pm \sqrt{25-3}\] \[f(x)= (x-5-\sqrt{22})(x-5+\sqrt{22})\]

  2. anonymous
    • one year ago
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    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
    • one year ago
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    is your equation \[-x^2-10x-3 \] ?

  4. anonymous
    • one year ago
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    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
    • one year ago
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    there is a difference between \[-x ^{2} and (-x)^{2}\] which do you mean?

  6. UsukiDoll
    • one year ago
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    Maybe let's assume that it is -x^2 ? since the original poster did say that they couldn't factor ?!

  7. MrNood
    • one year ago
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    y= -x^2-10x-3 has 2 real roots, but is not factorisable

  8. UsukiDoll
    • one year ago
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    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
    • one year ago
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    same is true for y= (-x)^2 - 10x -3

  10. MrNood
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    only part of his version is correct. There are sign errors

  13. MrNood
    • one year ago
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    using the quadratic formula will get the 2 solutions. This is not 'factorisation' in my understanding of the term

  14. UsukiDoll
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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
    • one year ago
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    on top of that we can't neglect the negative sign for -10x either so b = -10

  18. UsukiDoll
    • one year ago
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    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
    • one year ago
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    theposter has used (-x)^2 not -x^2

  20. UsukiDoll
    • one year ago
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    We wouldn't know until the original poster clarified it.

  21. UsukiDoll
    • one year ago
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    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
    • one year ago
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    @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
    • one year ago
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    \[(-x)^2-10x-3=x^2-10x-3\] \[\Large{\color{green}{x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }}}\] 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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \[\large \frac{-10}{2}+\frac{4 \sqrt{7}}{2} \rightarrow -5+2 \sqrt{7}\] \[\large \frac{-10}{2}-\frac{4 \sqrt{7}}{2} \rightarrow -5-2 \sqrt{7}\]

  27. UsukiDoll
    • one year ago
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    should have something like \[\large (x-5+2 \sqrt{7})(x-5-2\sqrt{7})\]

  28. anonymous
    • one year ago
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    Waah Shri Dharacharya..!!! @TheViper ... :P

  29. anonymous
    • one year ago
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    Amazing Job @UsukiDoll

  30. TheViper
    • one year ago
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    XD @waterineyes

  31. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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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