How do you solve ${(x - 5)}^{2} = 4$ $(x-5)^2=4$ ?

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How do you solve ${(x - 5)}^{2} = 4$ $(x-5)^2=4$ ?

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Answer 1 · 2017-03-07T08:06:51

$the answer : x=3 or x=7$ $"the answer : x=3 or x=7"$

Explanation:

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

$\sqrt{{(x - 5)}^{2}} = \sqrt{4}$ $sqrt ((x-5)^2)=sqrt 4$

$\pm (x - 5) = \pm 2$ $+- (x-5)=+-2$

$if + (x - 5) = + 2 then x - 5 = 2, x = 7$ $"if "+(x-5)=+2" then "x-5=2" , "x=7$

$if + (x - 5) = - 2 then x - 5 = - 2, x = 3$ $"if "+(x-5)=-2" then "x-5=-2" , "x=3$

$if - (x - 5) = + 2 then - x + 5 = 2, x = 3$ $"if "-(x-5)=+2" then "-x+5=2" , "x=3$

$if - (x - 5) = - 2 then - x + 5 = - 2, x = 7$ $"if "-(x-5)=-2" then "-x+5=-2" , "x=7$

$thus ;$ $"thus ;"$

$x=3 or x=7$ $"x=3 or x=7"$

Answer 2 · 2017-03-07T08:09:47

$x = 7$ $x= 7$ $&$ $&$ $x = 3$ $x= 3$

Explanation:

Method 1-

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

$\Rightarrow x - 5 = \pm 2$ $rArr x-5= +-2$ ----(taking square root both sides)

$\Rightarrow x - 5 = 2$ $rArr x-5= 2$ $&$ $&$ $\Rightarrow x - 5 = - 2$ $rArr x-5= -2$

$\Rightarrow x = 7$ $rArr x= 7$ $&$ $&$ $\Rightarrow x = 3$ $rArr x= 3$

Method 2-

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

$\Rightarrow x^{2} - 10 x + 25 = 4$ $rArrx^2-10x+25= 4$

$\Rightarrow x^{2} - 10 x + 21 = 0$ $rArrx^2-10x+21= 0$

$\Rightarrow x^{2} - 7 x - 3 x + 21 = 0$ $rArrx^2-7x-3x+21= 0$

$\Rightarrow x (x - 7) - 3 (x - 7) = 0$ $rArrx(x-7)-3(x-7)=0$

$\Rightarrow (x - 3) (x - 7) = 0$ $rArr(x-3)(x-7)=0$

$\Rightarrow x = 7$ $rArr x= 7$ $&$ $&$ $\Rightarrow x = 3$ $rArr x= 3$

Method 3-

Graphical Approach

Plot $y = {(x - 5)}^{2}$ $y=(x-5)^2$ and $y = 4$ $y=4$ on graph paper. Their intersection points are its solutions.

$y = {(x - 5)}^{2}$ $y=(x-5)^2$ $\to$ $rarr$ Parabola
$y = 4$ $y=4$ $\to$ $rarr$ Straight line parallel to x-axis.

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Answer 3 · 2017-03-07T08:27:39

$x = 3 or x = 7$ $x=3" or "x=7$

Explanation:

To 'undo' the square on the left side.

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

$\sqrt{{(x - 5)}^{2}} = \pm \sqrt{4}$ $sqrt((x-5)^2)=+-sqrt4$

$\Rightarrow x - 5 = \pm 2 \leftarrow there are 2 solutions$ $rArrx-5=+-2larrcolor(red)" there are 2 solutions"$

$∙ x - 5 = + 2$ $• x-5=+2$

$\Rightarrow x = 2 + 5$ $rArrx=2+5$

$\Rightarrow x = 7 \to (1)$ $rArrx=7to(color(red)(1))$

$∙ x - 5 = - 2$ $• x-5=-2$

$\Rightarrow x = - 2 + 5$ $rArrx=-2+5$

$\Rightarrow x = 3 \to (2)$ $rArrx=3to(color(red)(2))$

$As a check$ $color(blue)"As a check"$

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

$x = 7 \to {(7 - 5)}^{2} = 4 = right side$ $x=7to(7-5)^2=4=" right side"$

$x = 3 \to {(3 - 5)}^{2} = 4 = right side$ $x=3to(3-5)^2=4=" right side"$

$\Rightarrow x = 3 or x = 7 are the solutions$ $rArrx=3" or "x=7" are the solutions"$

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