Unproctored Placement Assessment
Question 3

Solve for [tex]$y$[/tex], where [tex]$y$[/tex] is a real number.

[tex]\sqrt{-4y + 21} = y[/tex]

(If there is more than one solution, separate them with a comma.)



Answer :

To solve for [tex]\( y \)[/tex] in the equation [tex]\(\sqrt{-4y + 21} = y\)[/tex], let's follow these steps:

1. Square both sides of the equation:
[tex]\[ \left(\sqrt{-4y + 21}\right)^2 = y^2 \][/tex]
This simplifies to:
[tex]\[ -4y + 21 = y^2 \][/tex]

2. Rearrange the equation to set it to zero:
[tex]\[ y^2 + 4y - 21 = 0 \][/tex]

3. Apply the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to solve the quadratic equation [tex]\( y^2 + 4y - 21 = 0 \)[/tex]. Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -21 \)[/tex]:
[tex]\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{16 + 84}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm 10}{2} \][/tex]

4. Find the solutions by solving the two possible values for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-4 + 10}{2} \][/tex]
[tex]\[ y = \frac{6}{2} \][/tex]
[tex]\[ y = 3 \][/tex]

and
[tex]\[ y = \frac{-4 - 10}{2} \][/tex]
[tex]\[ y = \frac{-14}{2} \][/tex]
[tex]\[ y = -7 \][/tex]

5. Check the solutions in the original equation to ensure they are valid (since we squared the equation, additional solutions might have been introduced):

Check [tex]\( y = 3 \)[/tex]:
[tex]\[ \sqrt{-4(3) + 21} = 3 \][/tex]
[tex]\[ \sqrt{-12 + 21} = 3 \][/tex]
[tex]\[ \sqrt{9} = 3 \][/tex]
[tex]\[ 3 = 3 \quad \text{(This is a valid solution)} \][/tex]

Check [tex]\( y = -7 \)[/tex]:
[tex]\[ \sqrt{-4(-7) + 21} = -7 \][/tex]
[tex]\[ \sqrt{28 + 21} = -7 \][/tex]
[tex]\[ \sqrt{49} = -7 \][/tex]
[tex]\[ 7 = -7 \quad \text{(This is not valid since 7 is not equal to -7)} \][/tex]

Only [tex]\( y = 3 \)[/tex] is a valid solution to the equation.

Therefore, the solution is:
[tex]\[ y = 3 \][/tex]