Question

Which of the following is an extraneous solution of StartRoot negative 3 x minus 2 EndRoot = x + 2? x = –6 x = –1 x = 1 x = 6

Answer 1

Answer:

A on ed

Step-by-step explanation:

Answer 2

The extraneous solution is x=-1 and x=-6

Step-by-step explanation:

Given the question as;

$\sqrt{-3x-2} =x+2$

Eliminate the root in the left hand side

$(\sqrt{-3x-2}) =x+2\\\\\\(\sqrt{-3x-2} )^2=(x+2)^2\\\\-3x-2=x^2+4x+4\\\\0=x^2+4x+3x+4+2\\\\\\0=x^2+7x+6$

solve the quadratic equation by factorization

Find multiples of 6 that add up to 7-------- 1*6

Rewrite the equation as;

x²+7x+6=0

x²+6x+x+6=0

x(x+6) +1 (x+6) =0

(x+1)(x+6) =0

x+1=0

x=-1

or

x+6=0

x=-6

The roots are, x=-1, and x=-6

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Quadratic equations : brainly.com/question/1332667

Keywords : extraneous, solution

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