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How do you solve the inequality #(x+6)^2 <=8#?

Answer 1

Avery Alexander

#-6-2sqrt2<=x<= -6+2sqrt2#

#"subtract 8 from both sides"#

#(x+6)^2-8<=0#

#"solve "(x+6)^2-8=0#

#"using "color(blue)"difference of squares"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#"with "a=x+6" and "b=sqrt8=2sqrt2#

#=(x+6-2sqrt2)(x+6+2sqrt2)=0#

#x+6-2sqrt2=0rArrx=-6+2sqrt2#

#x+6+2sqrt2=0rArrx=-6-2sqrt2#

#"since coefficient of "x^2>0" then minimum " uuu#

#-6-2sqrt2<=x<=-6+2sqrt2# graph{(x+6)^2-8 [-16.02, 16.01, -8.01, 8.01]}

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Answer 2

Genesis Allen

To solve the inequality ((x+6)^2 \leq 8), you would first take the square root of both sides and then solve for (x). However, since the square root of 8 is not a nice integer, you would first simplify the expression inside the inequality. Then, solve for (x) by considering both the positive and negative roots.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve the inequality #(x+6)^2 &lt;=8#?

How do you solve the inequality #(x+6)^2 <=8#?