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What is the sum of all possible values of x such that 2x(x + 10) = -50?

 Jan 24, 2022
 #1
avatar+675 
+1

 

 

2x ( x + 10 ) = -50  | :2

x ( x + 10 ) = -25 

x^2 + 10x = -25 | + 25

x^2 + 10x + 25 = 0

 

Use quadratic formula:

 

-b ± sqrt ( b^2 - 4ac )               -10 ± sqrt ( 10^2 - 4 * 1 * 25 )       -10 ± sqrt ( 100 - 100 )     -10 ± sqrt ( 0 )          -10

------------------------------     =     --------------------------------------- =  ------------------------------  =  --------------------- =  ----------- = -5 ± 0, so:

            2a                                                 2                                             2                                       2                       2

 

x_1 = -5

x_2 = 5

 

( - 5 ) + 5 = 0.

 

The answer is 0.

 Jan 24, 2022
 #2
avatar+36417 
+1

2x(x + 10) = -50     expand and simplify to

2x^2 + 20x + 50 = 0   or

x^2 + 10x + 25 =0

(x+5)(x+5) = 0     shows only one root x = - 5    (a 'double root')

     sum = -5     

 Jan 24, 2022
 #3
avatar+1209 
+2

2x(x + 10) = -50

2x^2 + 20x = -50

2x^2 + 20x + 50 = 0

Using VIeta's formulas, the sum of roots of the quadratic ax^2 + bx + c = 0 is -b/a.

 

-b/a = -20/2 = -10. 

 

(As ElectricPavlov has solved this question, the two roots are -5 and -5, which sum to -10 as expected.)

 Jan 24, 2022
 #4
avatar+360 
+1

\(2x(x + 10) = -50\)

\(2x^2+20x=-50\)

\(2x^2+20x+50=-50+50\)

\(2x^2+20x+50=0\)

\(x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\cdot \:2\cdot \:50}}{2\cdot \:2}\)

\(x_{1,\:2}=\frac{-20\pm \sqrt{0}}{2\cdot \:2}\)

\(x=\frac{-20}{2\cdot \:2}\)

\(x=-5\)

so the value is -5 

so -5!

hope this helps!,

      XxmathguyxX

 Jan 25, 2022

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