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Charra [1.4K]
3 years ago
12

Solve the equation for all values of x by completing the square. 3x^2-60x+255=0

Mathematics
1 answer:
erik [133]3 years ago
4 0

Answer:

x=5 and x=15

Step-by-step explanation:

3x^2-60x+225=0

3x^2-60x+225/3=0/3

x^2-20x+75=0

x^2-5x-15x+75=0

x(x-5)-15(x-5)=0

(x-5)(x-15)=0

x-5=0 x-15=0

x=5 x=15

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2 years ago
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Diano4ka-milaya [45]

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2 years ago
Given the function how do I solve
BARSIC [14]

Answer: (a) -7      (b) \bold{\dfrac{4}{25}}      (c) \bold{-\dfrac{5}{2}}    (d) 4

Step-by-step explanation:

A) f(x) = 2x - 3      when x is between -5 and -2 (including -5 and -2)

B) f(x) = x²            when x is between -2 and 2 (including 2)

C) f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}             when x is between 2 and 5 (including 5)

a) Equation A includes x = -2

   f(x) = 2x - 3

   f(-2) = 2(-2) - 3

           =  -4    - 3

           =       -7

b) Equation B includes x = -\dfrac{2}{5}

   f(x) = x²

   f\bigg(-\dfrac{2}{5}\bigg) = \bigg(-\dfrac{2}{5}\bigg)^2

        =\dfrac{4}{25}

c) Equation C includes x = 4

   f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}

   f(4)=-\dfrac{3}{2}(4)+\dfrac{7}{2}

       =-\dfrac{12}{2}+\dfrac{7}{2}

       =-\dfrac{5}{2}

d) Try each equation to see if x falls within the values given.

A) f(x) = 2x - 3

   -2.5 = 2x - 3

    0.5 = 2x

  0.25 = x    NOT VALID since x should be between -5 and -2

B) f(x) = x²

   -2.5 = x²

   \sqrt{-2.5}=x  NOT VALID since x cannot be an imaginary number

C) f(x)=-\dfrac{3}{2}x+\dfrac{7}{2}

    -\dfrac{5}{2}=-\dfrac{3}{2}x+\dfrac{7}{2}

    -\dfrac{12}{2}=-\dfrac{3}{2}x

    -\dfrac{12}{2}\bigg(\dfrac{2}{3}\bigg)=x

    4 = x        VALID since x is between 2 and 5

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Yes ! Hi I’m here lol
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