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

3 Answers

Jul 15, 2016

$x = 5$ $color(blue)(x=5)$ or $x = 1$ $color(blue)(x=1)$

Explanation:

Remember in squaring a general binomial
$XXX {(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $color(white)("XXX")(a+b)^2=a^2+2ab+b^2$

Given: $x^{2} - 6 x + 5 = 0$ $x^2-6x+5=0$

If $x^{2}$ $x^2$ and $- 6 x$ $-6x$ are the first two terms of the expansion of a squared binomial:
$XXX a = 1$ $color(white)("XXX")a=1$ , and
$XXX b = - 3$ $color(white)("XXX")b=-3$

So the third term must be $b^{2} = {(- 3)}^{2} = 9$ $b^2=(-3)^2=9$

To complete the square we add a $9$ $9$ to the first two terms (but to keep everything balanced we will need to subtract it again:
$XXX x^{2} - 6 x + 9 + 5 - 9 = 0$ $color(white)("XXX")x^2-6xcolor(red)(+9)+5color(red)(-9)=0$

Now we can write the first 3 terms as a squared binomial:
$XXX {(x - 3)}^{2} + 5 - 9 = 0$ $color(white)("XXX")(x-3)^2+5-9=0$

Simplifying $(5 - 9) = - 4$ $(5-9)=-4$ and adding $4$ $4$ to both sides gives
$XXX {(x - 3)}^{2} = 4$ $color(white)("XXX")(x-3)^2=4$

Taking the square root of both sides
$XXX x - 3 = \pm 2$ $color(white)("XXX")x-3=+-2$

Then add $3$ $3$ to both sides
$XXX x = 3 \pm 2$ $color(white)("XXX")x=3+-2$

$\Rightarrow x = 5 or x = 1$ $rArr x=5 or x=1$

Jul 15, 2016

$x = 5, or x = 1$ $x = 5, " or " x =1$

Explanation:

It seems strange to want to use completing the square, because this quadratic trinomial factorises to give $(x - 5) (x - 1) = 0$ $(x-5)(x-1) =0$

But let's proceed... Completing the square is based on the fact that
the square of a binomial gives a standard answer

${(x - 8)}^{2} = x^{2} - 16 x + 64$ $(x -8)^2 = x^2 -color(blue)(16)x +color(magenta)64$

There is ALWAYS a relationship between the $16 and 64$ $color(blue)16 and color(magenta)64$

This is: $16 \div 2 and then squared gives 64 8^{2} = 64$ $color(blue)(16 div 2)" and then squared gives "color(magenta)64 " "8^2 =64$

In $x^{2} - 6 x + 5 = 0 the relationship is not correct$ $x^2 -6x + 5 = 0 " the relationship is not correct"$

This means that 5 is not the correct constant, move it to the RHS

$x^{2} - 6 x + 9 = - 5 + 9 add the correct constant to both sides$ $x^2 -color(blue)(6)x color(magenta)(+9)= -5color(magenta)(+9)" add the correct constant to both sides"$

${(x - 3)}^{2} = 4 now the LHS is a perfect square$ $(x-3)^2 = 4 " now the LHS is a perfect square"$

$x - 3 = \pm \sqrt{4} square root both sides$ $x-3 = +-sqrt4" square root both sides"$

$x = 2 + 3, or x = - 2 + 3$ $x = 2+3, " or " x = -2+3$

$x = 5, or x = 1$ $x = 5, " or " x =1$

Jul 15, 2016

Another example of method. Have a look at:

https://socratic.org/s/awchNPwZ

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