sada asked in Science & MathematicsMathematics · 1 year ago

# 100 − 121k² = 0 What are the solutions to the equation above?

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• mizoo
Lv 7
1 year ago

100 - 121k^2 = 0

k^2 = 100/121

k = ± √(100/121)

k = ± 10/11

• 1 year ago

I don't know, but the one below looks interesting!

• 1 year ago

100-121k^2=0

100/121=(121k^2)/121

k^2=100/121

sqrt(k^2)=sqrt(100/121)

k=10/11

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• C
Lv 5
1 year agoReport

The questioner called it an equation, NOT a function. Had the equation been stated to be a function, then, yes, you would be correct that there is only one answer.

• Como
Lv 7
1 year ago

:-

k² = 100 / 121

k = ± 10/11

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• 1 year ago

100 - 121k^2 = 0 can be written as:

100 = 121k^2. So 100/121 = k^2

If we square root both sides we get:

10/11 = k or 0.909 recurring.

• Jeff Aaron
Lv 7
1 year agoReport

You forgot that k can also be -10/11

• 1 year ago

Factor the left side as a difference of squares: 100 - 121k^2 = (10 + 11k)(10 - 11k). Then set each factor to 0 and continue.

• 1 year ago

100 − 121k² = 0

Add 112 k^2 to both sides of this equation.

100 = 112 * k^2

112 * k^2 = 100

Divide both sides by 112.

k^2 = 100 ÷ 112

Let’s take the square root of both sides of this equation.

k = ± (10 ÷ 11)

This is approximately +0.909 or -0.909. I hope this is helpful for you.

• ted s
Lv 7
1 year agoReport

good thinking is ruined by poor math/typing....add 122 k²