Solve the equation 2x^2 + 3 – 41 = –15 to the nearest tenth.

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 x = {-4.4, +2.9}

Step-by-step explanation:

We assume you want to solve ...

  2x^2 +3x -41 = -15

Adding 41 and factoring out the leading coefficient gives ...

  2(x^2 +3/2x) = 26

Dividing by 2 makes it ...

  x^2 +3/2x = 13

We can add the square of half the x-coefficient to "complete the square."

  x^2 +3/2x +(3/4)^2 = 13 +(3/4)^2

  (x +3/4)^2 = 13.5625 . . . . write the left side as a square

  x +3/4 = ±√13.5625 . . . . . take the square root

  x = -0.75 ±3.683 = {-4.433, +2.933} . . . . subtract 3/4 and evaluate

The solutions are approximately x = -4.4 and x = 2.9.

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

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