# Answer to Question #22411 in Trigonometry for Kristen Woods

Question #22411

Solve the given equation.

11a. x^(2/3)+6x^(1/3)=7

11b. (y+3)^2-8(y+3)+12=0

11a. x^(2/3)+6x^(1/3)=7

11b. (y+3)^2-8(y+3)+12=0

Expert's answer

11a. x^(2/3)+6x^(1/3)=7

As we have here rational degrees, x must be non-negative:x>=0 Make a substitution

y = x^(1/3)> 0

Then

y^2 + 6y - 7 =0

D = 36 + 4*7 =64 = 8^2

y1 = (-6+8)/2 =1

y2 = (-6-8)/2 =-7 < 0

Since y>0, we obtain only one solution

y1 = 1 =x^(1/3)

whence

x=1

Answer: 1

---------------------------------

11b. (y+3)^2-8(y+3)+12=0

Make a substitution

t = y+3

Then

t^2 - 8t + 12 =0

D = 64 - 48 =16 = 4^2

t1 = (8+4)/2 =6

t2 = (8-4)/2 =2

Whence

y1 = t1-3 =6-3=3

y2 = t2-3 =2-3=-1

Answer: two roots: {-1, 3}

As we have here rational degrees, x must be non-negative:x>=0 Make a substitution

y = x^(1/3)> 0

Then

y^2 + 6y - 7 =0

D = 36 + 4*7 =64 = 8^2

y1 = (-6+8)/2 =1

y2 = (-6-8)/2 =-7 < 0

Since y>0, we obtain only one solution

y1 = 1 =x^(1/3)

whence

x=1

Answer: 1

---------------------------------

11b. (y+3)^2-8(y+3)+12=0

Make a substitution

t = y+3

Then

t^2 - 8t + 12 =0

D = 64 - 48 =16 = 4^2

t1 = (8+4)/2 =6

t2 = (8-4)/2 =2

Whence

y1 = t1-3 =6-3=3

y2 = t2-3 =2-3=-1

Answer: two roots: {-1, 3}

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