How do you solve $x^{2} - 5 x - 6 = 0$ $x^2-5x-6=0$ by completing the square?

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How do you solve $x^{2} - 5 x - 6 = 0$ $x^2-5x-6=0$ by completing the square?

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Answer 1 · 2016-09-07T19:40:34

$x = - 1 and x = 6$ $x=-1" and "x=6$

Please do not change this solution. It is a reference solution.

Explanation:

$End objective$ $color(blue)("End objective")$

Given the standard form $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

We end up with form $y = a {(x + \frac{b}{2 a})}^{2} + c + k$ $y=a(x+ b/(2a))^2+c+k$

The problem is that changing the original equation to this form introduces an error. So $k$ $k$ is a necessary correction. The value of which is determined by:

$a {(\frac{b}{2 a})}^{2} + k = 0$ $a(b/(2a))^2+k=0$

$For this question a = 1$ $color(magenta)("For this question "a=1)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Completing the square$ $color(blue)("Completing the square")$

Given: $y = 0 = x^{2} - 5 x - 6$ $" "y=0=x^2-5x-6$ .......................Equation(1)

$Step 1$ $color(brown)("Step 1")$

Write as: $(x^{2} - 5 x) - 6 + k \leftarrow At this stage k has no value$ $(x^2-5x)-6+k larr" At this stage "k" has no value"$

'...................................................................................................
$Step 2$ $color(brown)("Step 2")$
Move the power from $x^{2}$ $x^2$ to outside the brackets

${(x - 5 x)}^{2} - 6 + k$ $(x-5x )^(color(magenta)(2))-6+k$

$Now k starts to take on values that change at each step$ $color(purple)(" Now "k" starts to take on values that change at each step")$

'...................................................................................................
$Step 3$ $color(brown)("Step 3")$
Discard the $x$ $x$ from $5 x$ $5x$ inside the brackets

${(x - 5)}^{2} - 6 + k$ $(x-5)^2-6+k$

'...................................................................................................
$Step 4$ $color(brown)("Step 4")$

Halve the 5 inside the brackets

${(x - \frac{5}{2})}^{2} - 6 + k$ $(x-5/2)^2-6+k$

'...................................................................................................
$Step 5$ $color(brown)("Step 5")$

Determine the value of $k$ $k$
${(x - \frac{5}{2})}^{2} - 6 + k$ $(xcolor(magenta)(-5/2))^(color(magenta)(2))-6+color(magenta)(k)$ ......................Equation(2)

Let ${(- \frac{5}{2})}^{2} + k = 0 \to Note that this is the bit:$ $(color(magenta)(-5/2))^(color(magenta)(2))+ color(magenta)(k)=0" "rarr" Note that this is the bit:"$
$222222222222222222222222 a {(\frac{b}{2 a})}^{2} + k = 0$ $color(white)("222222222222222222222222") "a(color(magenta)(b/(2a)))^(color(magenta)(2))+color(magenta)(k)=0$

$ddddddddddddddddddddd$ $color(white)("ddddddddddddddddddddd")$ where in this case $a = 1$ $a=1$

$\Rightarrow \frac{25}{4} + k = 0 \Rightarrow k = - \frac{25}{4}$ $=>25/4+k=0" " =>" " k= -25/4$

So by substitution equation(2) becomes:

$y = 0 = {(x - \frac{5}{2})}^{2} - 6 - \frac{25}{4}$ $y=0=(x-5/2)^2-6-25/4$

$color(blue)(y=0=(x-5/2)^2-49/4)color(white)(.)......Equation(2_a)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(brown)("Step 6 - Determine the vertex")$

Using $Eqn(2_a)$
Vertex $->(x,y)=[(-1)xx(-5/2)color(white)("ddd"),color(white)("d")-49/4]$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Solving for "y=0)$

Add $49/4$ to both sides

$(x-5/2)^2=49/4$

Square root both sides

$x-5/2=+-sqrt(49/4) = +-7/2$

Add $5/2$ to both sides

$x=5/2+-7/2$

$x=-1" and "x=6$

Tony B

Answer 2 · 2017-12-05T21:56:00

As Tony B. states the answer is $x=6$ or $x=-1$ . The explanation here uses a different method to complete the square.

Explanation:

In this method a different approach is used to complete the square. The idea is to show more clearly how the coefficient $5$ gets "halved".

Start with the expression

$x^2-5x=x(x-5)$

These expression is the same as the first two terms of the quadratic expression originally given. We render this as a difference of squares:

$x(x-5)=u^2-v^2=(u+v)(u-v)$

And then we match the two sets of factors:

$x=u+v$

$x-5=u-v$

Now add these two equations together and note that in eliminating $v$ we make a coefficient of $2$ on $u$ . This is how we end up "halving the $5$ :

$2x-5=2u$ so $u=(x-5/2)$

Then take the difference of the original two equations to get $v$ which is just a constant, since $u$ and $x$ cancel at the same time:

$5=2v$ so $v=5/2$ ; again half of $5$ appears.

This works because the original quadratic expression had a coefficient of $1$ on $x^2$ . If the original quadratic expression has a different coefficient, like $2$ in $2x^2-10x-12=0$ , we would have divided by that coefficient on $x^2$ at the beginnjng to get (in this case) $x^2-5x-6=0$ with a coefficient of $1$ on $x^2$ , and then we could proceed as above.

Then we have:

$x^2-5x=u^2-v^2=(x+5/2)^2-({25}/4)$

And the original equation becomes:

$x^2-5x-6=0$

$(x-5/2)^2-({25}/4)-6=0$

$(x-5/2)^2=({49}/4)$

So we take both possible square roots noting that ${49}/4$ happens to be a perfect square:

$(x-5/2)=+(7/2)$ so $x=6$ or

$(x-5/2)=-(7/2)$ so $x=-1$

If we did not have a perfect square the final answer would contain square roots.

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How do you solve $x^{2} - 5 x - 6 = 0$ $x^2-5x-6=0$ by completing the square?

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