How do you solve sqrt(8-x)=x+6√8−x=x+6 and identify any restrictions?

1 Answer
Douglas K.
May 27, 2017

The right side cannot be negative and neither can the argument of the radical.
Square both sides.
Solve the quadratic.
Discard any roots that violate the restrictions.

Explanation:

The restrictions are that the right side cannot be negative and the same for the argument of the radical:

x+6 >= 0x+6≥0 and 8-x >=08−x≥0

x >= -6x≥−6 and 8 >=x8≥x

Combining them:

-6 <= x <= 8−6≤x≤8

Add this restriction to the given equation:

sqrt(8-x)=x+6; -6 <= x <= 8√8−x=x+6;−6≤x≤8

Square both sides:

8-x = x^2+ 12x+ 36; -6 <= x <= 88−x=x2+12x+36;−6≤x≤8

Combine like terms:

0 = x^2+ 13x+ 28; -6 <= x <= 80=x2+13x+28;−6≤x≤8

Use the quadratic formula:

x = (-13+-sqrt(13^2-4(1)(28)))/(2(1)); -6 <= x <= 8x=−13±√132−4(1)(28)2(1);−6≤x≤8

x = (-13+-sqrt(57))/2; -6 <= x <= 8x=−13±√572;−6≤x≤8

The negative root us outside the restriction, therefore, we discard it:

x = (-13+sqrt(57))/2x=−13+√572

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