# Video: SAT Practice Test 1 β’ Section 4 β’ Question 10

If π = 2/π and π + 2π = 0, what is the value of π?

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### Video Transcript

If π is equal to two divided by π and π plus two π is equal to zero, what is the value of π?

In order to solve these two simultaneous equations, to calculate the value of π, we need to eliminate the variable π. Letβs consider the first equation: π is equal to two divided by π. If we multiply both sides of this equation by π, weβre left with ππ is equal to two. If we then divide both sides of this new equation by π, we are left with π is equal to two divided by π, or two over π. We can now call this equation one.

Our second equation was π plus two π is equal to zero. We can now substitute π is equal to two over π into equation two to eliminate π. Substituting this in gives us π plus two multiplied by two over π is equal to zero. Two multiplied by two over π is equal to four over π. In order to simplify this equation, we need to multiply each of the three terms, π, four over π, and zero, by π. π multiplied by π is equal to π squared. Four over π multiplied by π is equal to four. And zero multiplied by π is equal to zero. We are left with π squared plus four equals zero.

If we subtract four from both sides of this equation, weβre left with π squared is equal to negative four. Our final step to solve for π would be to square-root both sides of the equation. The square root of π squared is equal to π. So we are left with π is equal to the square root of negative four.

Square-rooting any negative number gives an imaginary number. We know that the square root of negative one is equal to π. This means that the square root of negative four is equal to two π. This is because the square root of negative four can be written as the square root of four multiplied by negative one. And the square root of four equals two.

We can therefore conclude that, in the simultaneous equations π is equal to two over π and π plus two π equals zero, the value of π is not a real number. In fact, it is the imaginary number two π.