How do you solve using the completing the square method 2x^2 + 8x - 25 = 0?

1 Answer
Aug 24, 2016

=> x~~-6.06 " and " 2.06 to 2 decimal places

Explanation:

Given:" "2x^2+8x-25=0 ....................Equation(1)

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color(blue)("Completing the square")

write as y=2(x^2+4x)-25=0

Introduce the correction k to compensate for the changes we are going to make. They introduce an error.

2(x^2+4x)-25+k=0 larr" at this stage "k=0

Take the power of 2 outside the brackets

2(x+4x)^2-25+k=0

Remove the x from 4x

2(x+4)^2-25+k=0

Halve the 4

2(x+2)^2-25+k=0............................. Equation(2)

From the part: color(red)(2)(xcolor(blue)(+2))^2 you get the constant color(red)(2)color(blue)(xx2)^2=8 which is the error.

so 8+k=0=> k=-8 Substituting into equation(2) gives

2(x+2)^2-25-8=0............................. Equation(2_a)

2(x+2)^2-33=0
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color(blue)("Determine "x_("intercepts"))

Write as(x+2)^2=33/2

Square root both sides

x+2=sqrt(33/2)

=>x=-2+-sqrt(33/2)

=> x~~-6.06 " and " 2.06 to 2 decimal places

Tony B