How do you solve using the completing the square method $3 x^{2} - 67 = 6 x + 55$ $3x^2-67=6x+55$ ?

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How do you solve using the completing the square method $3 x^{2} - 67 = 6 x + 55$ $3x^2-67=6x+55$ ?

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Answer 1 · 2018-02-03T11:16:00

$x = 1 \pm \frac{5 \sqrt{15}}{3}$ $x=1+-(5sqrt15)/3$

Explanation:

$rearrange into standard form$ $"rearrange into standard form"$

$∙ x a x^{2} + b x + c = 0$ $•color(white)(x)ax^2+bx+c=0$

$\Rightarrow 3 x^{2} - 6 x - 122 = 0 \leftarrow in standard form$ $rArr3x^2-6x-122=0larrcolor(blue)"in standard form"$

$to complete the square$ $"to "color(blue)"complete the square"$

$∙ coefficient of x^{2} term must be 1$ $• " coefficient of "x^2" term must be 1"$

$\Rightarrow 3 (x^{2} - 2 x - \frac{122}{3}) = 0$ $rArr3(x^2-2x-122/3)=0$

$∙ add/subtract {(\frac{1}{2} coefficient of x-term)}^{2} to$ $• " add/subtract "(1/2"coefficient of x-term")^2" to"$
$x^{2} - 2 x$ $x^2-2x$

$3 (x^{2} + 2 (- 1) x + 1 - 1 - \frac{122}{3}) = 0$ $3(x^2+2(-1)xcolor(red)(+1)color(red)(-1)-122/3)=0$

$\Rightarrow 3 {(x - 1)}^{2} + 3 (- 1 - \frac{122}{3}) = 0$ $rArr3(x-1)^2+3(-1-122/3)=0$

$\Rightarrow 3 {(x - 1)}^{2} - 125 = 0$ $rArr3(x-1)^2-125=0$

$\Rightarrow 3 {(x - 1)}^{2} = 125$ $rArr3(x-1)^2=125$

$\Rightarrow {(x - 1)}^{2} = \frac{125}{3}$ $rArr(x-1)^2=125/3$

$take square root of both sides$ $color(blue)"take square root of both sides"$

$\Rightarrow x - 1 = \pm \sqrt{\frac{125}{3}} \leftarrow note plus or minus$ $rArrx-1=+-sqrt(125/3)larrcolor(blue)"note plus or minus"$

$\Rightarrow x = 1 \pm \frac{5 \sqrt{5}}{\sqrt{3}}$ $rArrx=1+-(5sqrt5)/sqrt3$

$\Rightarrow x = 1 \pm \frac{5 \sqrt{15}}{3} \leftarrow rationalise denominator$ $rArrx=1+-(5sqrt15)/3larrcolor(blue)"rationalise denominator"$

$\Rightarrow x \approx - 5.45 or x \approx 7.45 to 2 dec. places$ $rArrx~~-5.45" or "x~~7.45" to 2 dec. places"$

Answer 2 · 2018-02-03T11:57:50

$x = 1 \pm \frac{5 \sqrt{15}}{3}$ $x=1+-(5sqrt(15))/3$

Explanation:

Given: $3 x^{2} - 67 = 6 x + 55$ $3x^2-67=6x+55$

By manipulating this we obtain

$3 x^{2} - 6 x - 122 = 0$ $3x^2-6x-122=0$

Set this as $y = 0 = 3 x^{2} - 6 x - 122$ $y=0=3x^2-6x-122$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Completing the square$ $color(blue)("Completing the square")$

Consider the perfect square format of

$y = {(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $y=(a+b)^2 = a^2+2ab+b^2$

Our objective is to manipulate $y = 3 x^{2} - 6 x - 122$ $y=3x^2-6x-122$ to get it as close as we can to the perfect square format. This process introduces an error. This error is compensated for by including an 'adjustment' that takes the final result back to the value it should be. This correction I will show as $+ k$ $+k$

Write as $y = 3 (x^{2} - \frac{6}{3} x) - 122$ $y=3(x^2-6/3x)-122$ do not need $k$ $k$ yet

Now we will

$y=3(x^(cancel(color(white)(.)2))-6/(2xx3) cancel(x) )^2-122+k$

We have halved the $-6/3$ and moved the squared from $x^2$ to outside the brackets. Also we have removed the $x$ from $-6/3x$

$y=3(xcolor(cyan)(-1))^2-122color(red)(+k)$

We need to set $color(white)("d")3color(cyan)((-1)^2)color(red)(+k)=0$
$=> color(red)(k)=-3$ giving

$y=3(x-1)^2-122+kcolor(white)("dddd") ->color(white)("dddd")y=3(x-1)^2-125$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Check")$

$y=3(x-1)^2-123 color(white)("ddd")->color(white)("ddd")y=3(x^2-2x+1)-125$

$color(white)("ddddddddddddddddddd")->color(white)("ddd")y=3x^2-6x-122color(red)(larr" True")$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determining the values that satisfy the equation")$

$y=0=3(x-1)^2-125$

Add $125$ to both sides

$3(x-1)^2=125$

Divide both sides by 3

$(x-1)^2=125/3$

Square root both sides

$x-1=+-sqrt(125/3)$

$x=+1+-sqrt(125/3)$

$125=25xx5 -> 5^2xx5$ giving:

$x=+1+-(5sqrt(5))/sqrt(3)$

Mathematically it is considered that to have a root in the denominator is not good practice. So lets get rid of it.

Multiply the roots by 1 but in the form of $1=sqrt3/sqrt3$

$x=1+-[(5sqrt(5))/sqrt(3)xxsqrt3/sqrt3]$

$x=1+-(5sqrt(15))/3$

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How do you solve using the completing the square method $3 x^{2} - 67 = 6 x + 55$ $3x^2-67=6x+55$ ?

2 Answers

Explanation:

Explanation:

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How do you solve using the completing the square method 3x^2-67=6x+553x2−67=6x+553x^2-67=6x+55?

2 Answers

Explanation:

Explanation:

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How do you solve using the completing the square method $3 x^{2} - 67 = 6 x + 55$ $3x^2-67=6x+55$ ?