How do you solve using the completing the square method $x^{2} + 13 x + 42 = 0$ $x ^2 + 13 x + 42 = 0$ ?

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How do you solve using the completing the square method $x^{2} + 13 x + 42 = 0$ $x ^2 + 13 x + 42 = 0$ ?

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Answer 1 · 2016-03-20T22:23:53

For a very detailed example of method see
http://socratic.org/s/asVauX76

$Vertex \to (x, y) = (- \frac{13}{2} - \frac{1}{4}) = (- 6 \frac{1}{2}, - \frac{1}{4})$ $"Vertex" ->(x,y)=(-13/2 -1/4) = (-6 1/2,-1/4)$
I will let you solve the other bits!

Explanation:

Given: $x^{2} + 13 x + 42 = 0$ $" "x^2+13x+42=0$

The process applied introduces an error that is removed by adding the correction factor of $k$ $k$

Write as: $(x^{2} + 13 x) + 42 = 0$ $" "(x^2+13x)+42=0$

Add the correction factor of $k$ $k$

$(x^{2} + 13 x) + 42 + k = 0$ $" "(x^2+13x)+42+k=0$

Take the power to outside the brackets

${(x + 13 x)}^{2} + 42 + k = 0$ $" "(x+13x)^2+42+k=0$

Halve the coefficient of $13 x$ $13x$

${(x + \frac{13}{2} x)}^{2} + 42 + k = 0$ $" "(x+13/2 x)^2+42+k=0$

Remove the $x$ $x$ from $\frac{13}{2} x$ $13/2 x$

${(x + \frac{13}{2})}^{2} + 42 + k = 0$ $" "(x+13/2)^2+42+k=0$

The error comes from ${(+ \frac{13}{2})}^{2} = + \frac{169}{4}$ $(+13/2)^2 =+ 169/4$

So $k = - \frac{169}{4}$ $k=-169/4$ giving:

${(x + \frac{13}{2})}^{2} + 42 - \frac{169}{4} = 0$ $" "(x+13/2)^2+42-169/4=0$

${(x + \frac{13}{2})}^{2} - \frac{1}{4} = 0$ $" "(x+13/2)^2 -1/4=0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Vertex \to (x, y) = (- \frac{13}{2} - \frac{1}{4}) = (- 6 \frac{1}{2}, - \frac{1}{4})$ $"Vertex" ->(x,y)=(-13/2 -1/4) = (-6 1/2,-1/4)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Find x-intercepts by substituting 0 for y and solve in the normal way
y-intercept is the constant in the original equation = +42

I will let you solve for the x-intercepts

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How do you solve using the completing the square method $x^{2} + 13 x + 42 = 0$ $x ^2 + 13 x + 42 = 0$ ?

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How do you solve using the completing the square method x2+13x+42=0x ^2 + 13 x + 42 = 0?

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How do you solve using the completing the square method $x^{2} + 13 x + 42 = 0$ $x ^2 + 13 x + 42 = 0$ ?