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# how to solve this radical equation?

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the radical equation: $$\sqrt{{q^2\over 2}+11} = q - 1$$

i then have to square both sides; but how should i square q - 1?

and after i squared both sides, what should i do next?

Oct 11, 2017

#1
+1

Solve for q:
sqrt(q^2/2 + 11) = q - 1

Raise both sides to the power of two:
q^2/2 + 11 = (q - 1)^2

Expand out terms of the right hand side:
q^2/2 + 11 = q^2 - 2 q + 1

Subtract q^2 - 2 q + 1 from both sides:
-q^2/2 + 2 q + 10 = 0

Multiply both sides by -2:
q^2 - 4 q - 20 = 0

q^2 - 4 q = 20

q^2 - 4 q + 4 = 24

Write the left hand side as a square:
(q - 2)^2 = 24

Take the square root of both sides:
q - 2 = 2 sqrt(6) or q - 2 = -2 sqrt(6)

q = 2 + 2 sqrt(6) or q - 2 = -2 sqrt(6)

q = 2 + 2 sqrt(6) or q = 2 - 2 sqrt(6)

sqrt(q^2/2 + 11) ⇒ sqrt(1/2 (2 - 2 sqrt(6))^2 + 11) = sqrt(25 - 4 sqrt(6)) ≈ 3.89898
q - 1 ⇒ (2 - 2 sqrt(6)) - 1 = 1 - 2 sqrt(6) ≈ -3.89898:
So this solution is incorrect

sqrt(q^2/2 + 11) ⇒ sqrt(1/2 (2 + 2 sqrt(6))^2 + 11) = sqrt(25 + 4 sqrt(6)) ≈ 5.89898
q - 1 ⇒ (2 + 2 sqrt(6)) - 1 = 1 + 2 sqrt(6) ≈ 5.89898:
So this solution is correct

The solution is: q = 2 + 2 sqrt(6)

Oct 11, 2017

#1
+1

Solve for q:
sqrt(q^2/2 + 11) = q - 1

Raise both sides to the power of two:
q^2/2 + 11 = (q - 1)^2

Expand out terms of the right hand side:
q^2/2 + 11 = q^2 - 2 q + 1

Subtract q^2 - 2 q + 1 from both sides:
-q^2/2 + 2 q + 10 = 0

Multiply both sides by -2:
q^2 - 4 q - 20 = 0

q^2 - 4 q = 20

q^2 - 4 q + 4 = 24

Write the left hand side as a square:
(q - 2)^2 = 24

Take the square root of both sides:
q - 2 = 2 sqrt(6) or q - 2 = -2 sqrt(6)

q = 2 + 2 sqrt(6) or q - 2 = -2 sqrt(6)

q = 2 + 2 sqrt(6) or q = 2 - 2 sqrt(6)

sqrt(q^2/2 + 11) ⇒ sqrt(1/2 (2 - 2 sqrt(6))^2 + 11) = sqrt(25 - 4 sqrt(6)) ≈ 3.89898
q - 1 ⇒ (2 - 2 sqrt(6)) - 1 = 1 - 2 sqrt(6) ≈ -3.89898:
So this solution is incorrect

sqrt(q^2/2 + 11) ⇒ sqrt(1/2 (2 + 2 sqrt(6))^2 + 11) = sqrt(25 + 4 sqrt(6)) ≈ 5.89898
q - 1 ⇒ (2 + 2 sqrt(6)) - 1 = 1 + 2 sqrt(6) ≈ 5.89898:
So this solution is correct

The solution is: q = 2 + 2 sqrt(6)

Guest Oct 11, 2017
#2
0

once you got -q^2/2 + 2q - 10 = 0, why would you wanna multiply both sides by -2? is a denominator not allowed?

Guest Oct 11, 2017
#3
+98196
+1

Multiplying through by -2  sets it up for completing the square........

Oct 11, 2017