How do you solve using the completing the square method $x^{2} + 6 x = 7$ $x^2 + 6x = 7$ ?

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Answer 1 · 2017-07-20T16:02:28

$x = 1 or - 7$ $x=1 or -7$

$x^{2} + 6 x = 7$ $x^2+6x=7$

half the coefficient of $x$ $x$ and square it

$6 \to 3^{2}$ $6rarr3^2$

add and subtract this value

$x^{2} + 6 x + 3^{2} - 3^{2} = 7$ $color(red)(x^2+6x+3^2)-3^2=7$

$= = ↑$ $color(white)(==)uarr$ the terms in red form a perfect square

$(x^{2} + 6 x + 3^{2}) - 9 = 7$ $color(red)((x^2+6x+3^2))-9=7$

$= = ↓$ $color(white)(==)darr$

$\Rightarrow {(x + 3)}^{2} = 7 + 9$ $=>color(red)((x+3)^2)=7+9$

${(x + 3)}^{2} = 16$ $(x+3)^2=16$

now solve for $x$ $x$

$\Rightarrow x + 3 = \pm \sqrt{16}$ $=>x+3=+-sqrt16$

$x + 3 = \pm 4$ $x+3=+-4$

$x = - 3 \pm 4$ $x=-3+-4$

$x = 1 or - 7$ $x=1 or -7$

How do you solve using the completing the square method x2+6x=7x^2 + 6x = 7?