are the roots of the quadratic equation $x^{2} + 6 x - 7 = 0$ .

x=-7 and 1

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x=7 and 1

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x=7 and -1

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Solution

The correct option is A x=-7 and 1
$x^{2} + 6 x - 7 = 0$

$x^{2} + 6 x + 9 - 16 = 0$
$(x + 3)^{2} = 16$

$(x + 3) = \sqrt{16}$

$(x + 3) = + 4 o r (x + 3) = - 4$

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

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