Solved on Jan 21, 2024

Solve for $y$ in simplest radical form: $x^2 + y^2 + 2x - 3y + 4 = 0$

STEP 1

Assumptions
1. The given equation is $x^{2}+y^{2}+2x-3y+4=0$ .
2. We need to solve for $y$ in terms of $x$ .
3. The solution should be in the simplest radical form.

STEP 2

The given equation is a quadratic in $y$ . We will complete the square for both $x$ and $y$ to transform the equation into a form that allows us to solve for $y$ .

STEP 3

Group the $x$ terms and the $y$ terms together.
$(x^{2}+2x) + (y^{2}-3y) = -4$

STEP 4

Complete the square for the $x$ terms. We take half of the coefficient of $x$ (which is 2), square it, and add it inside the parenthesis. We must also subtract it outside the parenthesis to keep the equation balanced.
$\left(x^{2}+2x+\left(\frac{2}{2}\right)^2\right) - \left(\frac{2}{2}\right)^2 + (y^{2}-3y) = -4$

STEP 5

Simplify the equation.
$\left(x^{2}+2x+1\right) - 1 + (y^{2}-3y) = -4$

STEP 6

Complete the square for the $y$ terms. We take half of the coefficient of $y$ (which is -3), square it, and add it inside the parenthesis. We must also subtract it outside the parenthesis to keep the equation balanced.
$\left(x^{2}+2x+1\right) - 1 + \left(y^{2}-3y+\left(\frac{-3}{2}\right)^2\right) - \left(\frac{-3}{2}\right)^2 = -4$

STEP 7

Simplify the equation.
$\left(x^{2}+2x+1\right) - 1 + \left(y^{2}-3y+\frac{9}{4}\right) - \frac{9}{4} = -4$

STEP 8

Recognize that the expressions inside the parentheses are perfect squares.
$\left(x+1\right)^2 - 1 + \left(y-\frac{3}{2}\right)^2 - \frac{9}{4} = -4$

STEP 9

Combine like terms on the left side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -4 + 1 + \frac{9}{4}$

STEP 10

Simplify the right side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -\frac{16}{4} + \frac{4}{4} + \frac{9}{4}$

STEP 11

Combine the fractions on the right side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -\frac{3}{4}$

STEP 12

Notice that the equation has a negative value on the right side, which implies that there are no real solutions for $y$ because the sum of two squares cannot be negative. However, if we are looking for complex solutions, we can proceed.

STEP 13

Isolate the term with $y$ on one side of the equation.
$\left(y-\frac{3}{2}\right)^2 = -\frac{3}{4} - \left(x+1\right)^2$

STEP 14

Take the square root of both sides of the equation to solve for $y$ .
$y-\frac{3}{2} = \pm\sqrt{-\frac{3}{4} - \left(x+1\right)^2}$

STEP 15

Add $\frac{3}{2}$ to both sides of the equation to solve for $y$ .
$y = \frac{3}{2} \pm\sqrt{-\frac{3}{4} - \left(x+1\right)^2}$

STEP 16

Recognize that the expression inside the square root is negative, which indicates the presence of an imaginary number. We can factor out $-1$ to show the imaginary unit $i$ .
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + \left(x+1\right)^2}$

STEP 17

Simplify the expression inside the square root.
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + x^2 + 2x + 1}$

STEP 18

Combine the terms inside the square root.
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + x^2 + 2x + 1}$
Since we cannot further simplify the expression inside the square root without specific values for $x$ , this is the simplest radical form for complex solutions.

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Solve for $y$ in simplest radical form: $x^2 + y^2 + 2x - 3y + 4 = 0$

STEP 1

Assumptions
1. The given equation is $x^{2}+y^{2}+2x-3y+4=0$ .
2. We need to solve for $y$ in terms of $x$ .
3. The solution should be in the simplest radical form.

STEP 2

The given equation is a quadratic in $y$ . We will complete the square for both $x$ and $y$ to transform the equation into a form that allows us to solve for $y$ .

STEP 3

Group the $x$ terms and the $y$ terms together.
$(x^{2}+2x) + (y^{2}-3y) = -4$

STEP 4

STEP 5

Simplify the equation.
$\left(x^{2}+2x+1\right) - 1 + (y^{2}-3y) = -4$

STEP 6

STEP 7

Simplify the equation.
$\left(x^{2}+2x+1\right) - 1 + \left(y^{2}-3y+\frac{9}{4}\right) - \frac{9}{4} = -4$

STEP 8

Recognize that the expressions inside the parentheses are perfect squares.
$\left(x+1\right)^2 - 1 + \left(y-\frac{3}{2}\right)^2 - \frac{9}{4} = -4$

STEP 9

Combine like terms on the left side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -4 + 1 + \frac{9}{4}$

STEP 10

Simplify the right side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -\frac{16}{4} + \frac{4}{4} + \frac{9}{4}$

STEP 11

Combine the fractions on the right side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -\frac{3}{4}$

STEP 12

Notice that the equation has a negative value on the right side, which implies that there are no real solutions for $y$ because the sum of two squares cannot be negative. However, if we are looking for complex solutions, we can proceed.

STEP 13

Isolate the term with $y$ on one side of the equation.
$\left(y-\frac{3}{2}\right)^2 = -\frac{3}{4} - \left(x+1\right)^2$

STEP 14

Take the square root of both sides of the equation to solve for $y$ .
$y-\frac{3}{2} = \pm\sqrt{-\frac{3}{4} - \left(x+1\right)^2}$

STEP 15

Add $\frac{3}{2}$ to both sides of the equation to solve for $y$ .
$y = \frac{3}{2} \pm\sqrt{-\frac{3}{4} - \left(x+1\right)^2}$

STEP 16

Recognize that the expression inside the square root is negative, which indicates the presence of an imaginary number. We can factor out $-1$ to show the imaginary unit $i$ .
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + \left(x+1\right)^2}$

STEP 17

Simplify the expression inside the square root.
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + x^2 + 2x + 1}$

STEP 18

Combine the terms inside the square root.
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + x^2 + 2x + 1}$
Since we cannot further simplify the expression inside the square root without specific values for $x$ , this is the simplest radical form for complex solutions.

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Solve for yyy in simplest radical form: x2+y2+2x−3y+4=0x^2 + y^2 + 2x - 3y + 4 = 0x2+y2+2x−3y+4=0

STEP 1

Assumptions1. The given equation is x2+y2+2x−3y+4=0x^{2}+y^{2}+2x-3y+4=0x2+y2+2x−3y+4=0.2. We need to solve for yyy in terms of xxx.3. The solution should be in the simplest radical form.

STEP 2

The given equation is a quadratic in yyy. We will complete the square for both xxx and yyy to transform the equation into a form that allows us to solve for yyy.

STEP 3

Group the xxx terms and the yyy terms together.(x2+2x)+(y2−3y)=−4(x^{2}+2x) + (y^{2}-3y) = -4(x2+2x)+(y2−3y)=−4

STEP 4

STEP 5

Simplify the equation.(x2+2x+1)−1+(y2−3y)=−4\left(x^{2}+2x+1\right) - 1 + (y^{2}-3y) = -4(x2+2x+1)−1+(y2−3y)=−4

STEP 6

STEP 7

Simplify the equation.(x2+2x+1)−1+(y2−3y+94)−94=−4\left(x^{2}+2x+1\right) - 1 + \left(y^{2}-3y+\frac{9}{4}\right) - \frac{9}{4} = -4(x2+2x+1)−1+(y2−3y+49​)−49​=−4

STEP 8

Recognize that the expressions inside the parentheses are perfect squares.(x+1)2−1+(y−32)2−94=−4\left(x+1\right)^2 - 1 + \left(y-\frac{3}{2}\right)^2 - \frac{9}{4} = -4(x+1)2−1+(y−23​)2−49​=−4

STEP 9

Combine like terms on the left side of the equation.(x+1)2+(y−32)2=−4+1+94\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -4 + 1 + \frac{9}{4}(x+1)2+(y−23​)2=−4+1+49​

STEP 10

Simplify the right side of the equation.(x+1)2+(y−32)2=−164+44+94\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -\frac{16}{4} + \frac{4}{4} + \frac{9}{4}(x+1)2+(y−23​)2=−416​+44​+49​

STEP 11

Combine the fractions on the right side of the equation.(x+1)2+(y−32)2=−34\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -\frac{3}{4}(x+1)2+(y−23​)2=−43​

STEP 12

Notice that the equation has a negative value on the right side, which implies that there are no real solutions for yyy because the sum of two squares cannot be negative. However, if we are looking for complex solutions, we can proceed.

STEP 13

Isolate the term with yyy on one side of the equation.(y−32)2=−34−(x+1)2\left(y-\frac{3}{2}\right)^2 = -\frac{3}{4} - \left(x+1\right)^2(y−23​)2=−43​−(x+1)2

STEP 14

Take the square root of both sides of the equation to solve for yyy.y−32=±−34−(x+1)2y-\frac{3}{2} = \pm\sqrt{-\frac{3}{4} - \left(x+1\right)^2}y−23​=±−43​−(x+1)2​

STEP 15

Add 32\frac{3}{2}23​ to both sides of the equation to solve for yyy.y=32±−34−(x+1)2y = \frac{3}{2} \pm\sqrt{-\frac{3}{4} - \left(x+1\right)^2}y=23​±−43​−(x+1)2​

STEP 16

Recognize that the expression inside the square root is negative, which indicates the presence of an imaginary number. We can factor out −1-1−1 to show the imaginary unit iii.y=32±i34+(x+1)2y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + \left(x+1\right)^2}y=23​±i43​+(x+1)2​

STEP 17

Simplify the expression inside the square root.y=32±i34+x2+2x+1y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + x^2 + 2x + 1}y=23​±i43​+x2+2x+1​

STEP 18

Combine the terms inside the square root.y=32±i34+x2+2x+1y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + x^2 + 2x + 1}y=23​±i43​+x2+2x+1​Since we cannot further simplify the expression inside the square root without specific values for xxx, this is the simplest radical form for complex solutions.

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Solve for $y$ in simplest radical form: $x^2 + y^2 + 2x - 3y + 4 = 0$

Assumptions
1. The given equation is $x^{2}+y^{2}+2x-3y+4=0$ .
2. We need to solve for $y$ in terms of $x$ .
3. The solution should be in the simplest radical form.

The given equation is a quadratic in $y$ . We will complete the square for both $x$ and $y$ to transform the equation into a form that allows us to solve for $y$ .

Group the $x$ terms and the $y$ terms together.
$(x^{2}+2x) + (y^{2}-3y) = -4$

Simplify the equation.
$\left(x^{2}+2x+1\right) - 1 + (y^{2}-3y) = -4$

Simplify the equation.
$\left(x^{2}+2x+1\right) - 1 + \left(y^{2}-3y+\frac{9}{4}\right) - \frac{9}{4} = -4$

Recognize that the expressions inside the parentheses are perfect squares.
$\left(x+1\right)^2 - 1 + \left(y-\frac{3}{2}\right)^2 - \frac{9}{4} = -4$

Combine like terms on the left side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -4 + 1 + \frac{9}{4}$

Simplify the right side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -\frac{16}{4} + \frac{4}{4} + \frac{9}{4}$

Combine the fractions on the right side of the equation.
$\left(x+1\right)^2 + \left(y-\frac{3}{2}\right)^2 = -\frac{3}{4}$

Notice that the equation has a negative value on the right side, which implies that there are no real solutions for $y$ because the sum of two squares cannot be negative. However, if we are looking for complex solutions, we can proceed.

Isolate the term with $y$ on one side of the equation.
$\left(y-\frac{3}{2}\right)^2 = -\frac{3}{4} - \left(x+1\right)^2$

Take the square root of both sides of the equation to solve for $y$ .
$y-\frac{3}{2} = \pm\sqrt{-\frac{3}{4} - \left(x+1\right)^2}$

Add $\frac{3}{2}$ to both sides of the equation to solve for $y$ .
$y = \frac{3}{2} \pm\sqrt{-\frac{3}{4} - \left(x+1\right)^2}$

Recognize that the expression inside the square root is negative, which indicates the presence of an imaginary number. We can factor out $-1$ to show the imaginary unit $i$ .
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + \left(x+1\right)^2}$

Simplify the expression inside the square root.
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + x^2 + 2x + 1}$

Combine the terms inside the square root.
$y = \frac{3}{2} \pm i\sqrt{\frac{3}{4} + x^2 + 2x + 1}$
Since we cannot further simplify the expression inside the square root without specific values for $x$ , this is the simplest radical form for complex solutions.