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Solve:

1. Square both sides of the equation.
2. Expand the left side. (3 − k)(3 − k) = 3k + 1
3. Multiply 9 − 6k + k2 = 3k + 1
4. Write the quadratic equation in standard form. k2 − 9k + 8 = 0
5. Factor the quadratic equation. (k − 8)(k − 1) = 0
6. Use the zero product property.
The solutions to the quadratic equation are
.

The true solution(s) to the radical equation


Sagot :

The solutions to the equation [tex]3-k=\sqrt{3k+1}[/tex] are k = 1 and 8

The given equation is:

[tex]3-k=\sqrt{3k+1}[/tex]

Square both sides of the equation

[tex](3-k)^2=3k+1[/tex]

Expand the left side of the equation above

[tex](3-k)^2=3k+1\\\\(3-k)(3-k)=3k+1\\\\3^2-3k-3k+k^2=3k+1\\\\9-6k+k^2=3k+1[/tex]

Write the quadratic equation in standard form

[tex]k^2-6k-3k+9-1=0\\\\k^2-9k+8=0\\\\[/tex]

Factor the quadratic equation

[tex](k-8)(k-1)=0[/tex]

Use the zero product property

k - 8 = 0

k = 8

k - 1 = 0

k = 1

The solutions to the equation [tex]3-k=\sqrt{3k+1}[/tex] are k = 1 and 8

Learn more here: https://brainly.com/question/25840704

answer

first one: 8 and 1

second one: is 1

Step-by-step explanation:

slay

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