Question Video: Evaluating an Exponential Expression by Substituting Surds | Nagwa Question Video: Evaluating an Exponential Expression by Substituting Surds | Nagwa

# Question Video: Evaluating an Exponential Expression by Substituting Surds Mathematics • Second Year of Preparatory School

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If π₯ = (β3)/2, π¦ = 1/(β2), and π§ = (β2)/3, then find the value of π₯Β² + (π¦ + π§)Β² β π§β΄ in its simplest form.

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### Video Transcript

If π₯ equals root three over two, π¦ equals one over root two, and π§ equals root two over three, then find the value of π₯ squared plus π¦ plus π§ squared minus π§ to the fourth power in its simplest form.

Rather than substituting the values of π₯, π¦, and π§ into the full expression, we may find it helpful to break it down and evaluate parts of the expression first. Letβs begin with π₯ squared, which is equal to root three over two squared. When squaring a fraction, we can square the numerator and denominator separately. So this is equal to root three squared over two squared.

We can then recall that for nonnegative real values of π, the square root of π squared is equal to π, and so root three squared is equal to three. Two squared is equal to four, and so the value of π₯ squared is three-quarters.

Next, letβs evaluate π¦ plus π§ all squared, which is equal to one over root two plus root two over three all squared. Distributing the parentheses gives one over root two squared plus root two over three multiplied by one over root two plus one over root two multiplied by root two over three plus root two over three squared.

For the two central terms, we can cross cancel factors of root two, so the two terms each simplify to one-third. For the first and last terms, squaring the numerators and denominators gives one-half and two-ninths. So the expression simplifies to one-half plus one-third plus one-third plus two-ninths. Expressing each of these fractions as equivalent fractions with denominators of 18 gives nine over 18 plus six over 18 plus six over 18 plus four over 18, which is 25 over 18.

Next, we evaluate the last part of the expression, π§ to the fourth power, which is root two over three to the fourth power. We can raise the numerator and denominator to the fourth power separately. We should recall from memory that three to the fourth power is 81. And writing the numerator out longhand and then grouping pairs of root two gives two multiplied by two, which is four. Hence, the value of π§ to the fourth power is four over 81. Substituting the value of each part of the expression gives three-quarters plus 25 over 18 minus four over 81.

The first two terms can be combined using a common denominator of 36 to give 27 over 36 plus 50 over 36, which is 77 over 36. The lowest common multiple of 36 and 81 is 324. So the sum can be written as 693 over 324 minus 16 over 324, which is 677 over 324. So, by evaluating each part of the expression separately and then combining, weβve found that for the given values of π₯, π¦, and π§, the value of π₯ squared plus π¦ plus π§ all squared minus π§ to the fourth power is 677 over 324.

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