Question Video: Using a Variation Function to Find an Unknown | Nagwa Question Video: Using a Variation Function to Find an Unknown | Nagwa

# Question Video: Using a Variation Function to Find an Unknown Mathematics • Second Year of Secondary School

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If the variation function π(π₯) = ππ₯Β² + ππ₯ at π₯ = π is π(β) = πβΒ² + πβ, what is the value of π?

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### Video Transcript

If the variation function π of π₯ is equal to ππ₯ squared plus ππ₯ at π₯ is equal to π is given by π of β is equal to πβ squared plus πβ, what is the value of π?

In this question, weβre given a quadratic function π of π₯ which is equal to ππ₯ squared plus ππ₯. And weβre told when π₯ is equal to π, the variation function of this quadratic function π of π₯ is given by π of β is equal to πβ squared plus πβ for some unknown constants π and π. We need to use this information to determine the value of π.

To do this, letβs start by recalling what we mean by the variation function of a function π of π₯ at π₯ is equal to π. We can recall itβs the function π of β which is equal to π evaluated at π plus β minus π evaluated at π. In other words, this function measures how much the function π changes when its input values change from π to π plus β. And in this question, weβre given an expression for the variation function π of β. Itβs πβ squared plus πβ.

We can substitute this expression into the left-hand side of the equation. And we can also find an expression for the right-hand side of this equation by noting weβre given the function π of π₯. Itβs ππ₯ squared plus ππ₯. So we can substitute π₯ is equal to π plus β and π₯ is equal to π into this function to find an expression for the right-hand side of this equation. First, if we substitute π₯ is equal to π plus β into the function π of π₯, we get π times π plus β all squared plus π multiplied by π plus β. Then, we need to subtract π evaluated at π, which we can find by substituting π₯ is equal to π into π of π₯. We subtract ππ squared plus ππ. This then gives us an equation in terms of π, β, π, and π.

Letβs now rearrange this equation. Letβs start by distributing the exponent over the parentheses. And we can do this by using the binomial formula or by expanding the brackets π plus β multiplied by π plus β. We get π squared plus two πβ plus β squared. And we need to multiply this expression by π. We can also distribute the negative over the two terms at the end of this expression. This gives us that πβ squared plus πβ is equal to π times π squared plus two πβ plus β squared plus π times π plus β minus ππ squared minus ππ.

We can simplify the right-hand side further by distributing the factor of π and the factor of π over the parentheses. Doing this gives us the following equation, which we can simplify. We have ππ squared minus ππ squared, which is equal to zero. And we also have π times π minus π times π, which is equal to zero. We can also note that we have πβ squared on both the left- and right-hand side of the equation and π times β on both the left- and right-hand side of the equation. So we can subtract πβ squared from both sides of the equation and π plus β from both sides of the equation. This gives us that zero is equal to two times π times π multiplied by β.

So, for this equation to hold true, one of our unknowns must be equal to zero. Itβs worth noting this canβt be β because β is a variable. It can take any value for our variation function. And π is a parameter of our original equation; itβs not equal to zero. Otherwise, our original function would not be quadratic. So the only way for this equation to hold true for all values of β would be for π to be equal to zero. Therefore, we were able to show if the variation function of π of π₯ is equal to ππ₯ squared plus ππ₯ at π₯ is equal to π is π of β is equal to πβ squared plus πβ, then the value of π must be zero.

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