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Given that the sum of two positive real variables is x+y=9x + y = 9 and the sum of their squares is x2+y2=53x^2 + y^2 = 53, use special product formulas to evaluate the exact numerical value of the difference of fourth powers x4−y4x^4 - y^4, given that x>yx > y.

Answer and steps

Final answer

4775477\sqrt{5}

Steps

01

Factor the difference of fourth powers

Apply the difference of squares formula, a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b), repeatedly to expand x4−y4x^4 - y^4:

x4−y4=(x2)2−(y2)2=(x2+y2)(x2−y2)x^4 - y^4 = (x^2)^2 - (y^2)^2 = (x^2 + y^2)(x^2 - y^2)

Factoring the second term x2−y2x^2 - y^2 further gives:

x4−y4=(x2+y2)(x+y)(x−y)x^4 - y^4 = (x^2 + y^2)(x + y)(x - y)

02

Substitute the known values

We are given:

  • x2+y2=53x^2 + y^2 = 53
  • x+y=9x + y = 9

Multiplying these two known factors:

(x2+y2)(x+y)=53⋅9=477(x^2 + y^2)(x + y) = 53 \cdot 9 = 477

Thus, the target expression simplifies to:

x4−y4=477(x−y)x^4 - y^4 = 477(x - y)

03

Determine the difference factor (x−y)(x - y)

To find x−yx - y, relate it to the given sums using the algebraic identity connecting the square of a sum and the square of a difference:

(x−y)2=2(x2+y2)−(x+y)2(x - y)^2 = 2(x^2 + y^2) - (x + y)^2

Substituting the given values:

(x−y)2=5(x - y)^2 = 5

Since it is given that x>yx > y, the difference must be strictly positive:

x−y=5x - y = \sqrt{5}

Common Pitfall: When taking the square root to find x−yx - y, always verify the sign condition. Since x>yx > y, we select the positive root +5+ \sqrt{5} rather than −5- \sqrt{5}.

04

Calculate the final value

Multiply the combined components together:

x4−y4=477⋅(x−y)=4775x^4 - y^4 = 477 \cdot (x - y) = 477\sqrt{5}

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to subtract 2xy when computing (x - y)², mistakenly calculating 53 - 14 instead of 53 - 28.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    (x + y)^2 = x^2 + 2xy + y^2   ⟹  \implies 81 = 53 + 2xy   ⟹  \implies 2xy = 28   ⟹  \implies xy = 14

  2. Step 2

    (x - y)^2 = x^2 + y^2 - xy = 53 - 14 = 39   ⟹  \implies x - y = 39\sqrt{39}

    Correct

    (x - y)^2 = x^2 + y^2 - 2xy = 53 - 28 = 25   ⟹  \implies x - y = 5

    The perfect square of a binomial is (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2. You omitted the factor of 22 in the middle term, subtracting xy=14xy = 14 rather than 2xy=282xy = 28.

  3. Step 3

    x^4 - y^4 = (x^2 + y^2)(x + y)(x - y)   ⟹  \implies x^4 - y^4 = 53 ⋅\cdot 9 ⋅\cdot 39\sqrt{39} = 47739\sqrt{39}

    Correct

    x^4 - y^4 = (x^2 + y^2)(x + y)(x - y) = 53 ⋅\cdot 9 ⋅\cdot 5 = 2385

    Your factorization and substitution plan were completely correct, but substituting the incorrect value of x−y=39x - y = \sqrt{39} from Step 2 led to an affected final result.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly factored the difference of fourth powers into (x2+y2)(x+y)(x−y)(x^2 + y^2)(x + y)(x - y).

You correctly applied the expansion of (x+y)2(x + y)^2 to find the product 2xy=282xy = 28.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Missing coefficient in binomial square expansion

You wrote (x−y)2=x2+y2−xy(x - y)^2 = x^2 + y^2 - xy instead of (x−y)2=x2+y2−2xy(x - y)^2 = x^2 + y^2 - 2xy, subtracting 1414 instead of 2828.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the middle term of the square of a binomial with the unfactored quadratic factor from the difference of cubes identity, a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2), where the cross term has a coefficient of 11 instead of 22.

REASONING PATH
Your ACEpath teacher’s suggested approach
Binomial Square Identity

  • (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2

  • Always write out the explicit factor of 22 in the cross term whenever expanding or relating squares of binomials.

Direct Square-to-Square Bridge

  • (x−y)2=2(x2+y2)−(x+y)2(x - y)^2 = 2(x^2 + y^2) - (x + y)^2

  • When both x2+y2x^2 + y^2 and x+yx + y are given, use this identity directly to bypass calculating xyxy separately.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Apply the full expansion: (x−y)2=(x2+y2)−2xy=53−28=25(x - y)^2 = (x^2 + y^2) - 2xy = 53 - 28 = 25.

  2. Since x>yx > y, take the principal square root: x−y=25=5x - y = \sqrt{25} = 5.

  3. Multiply the factors: x4−y4=(x2+y2)(x+y)(x−y)=53⋅9⋅5=2385x^4 - y^4 = (x^2 + y^2)(x + y)(x - y) = 53 \cdot 9 \cdot 5 = 2385.

Final answer

23852385

NEXT STEPS
Final Takeaways from Your Teacher

  • Double-check that every binomial square (a±b)2(a \pm b)^2 includes the doubled product ±2ab\pm 2ab.

  • Practice using the dual identity (x−y)2=2(x2+y2)−(x+y)2(x - y)^2 = 2(x^2 + y^2) - (x + y)^2 to streamline symmetric calculations.

OverviewACEpath Teacher's Diagnosis

02 The student misremembers the difference of squares identity for fourth powers and mistakenly writes x⁴ - y⁴ as (x² - y²)².

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    (x+y)2=x2+2xy+y2  ⟹  81=53+2xy  ⟹  2xy=28  ⟹  xy=14(x+y)^2 = x^2 + 2xy + y^2 \implies 81 = 53 + 2xy \implies 2xy = 28 \implies xy = 14

  2. Step 2

    (x−y)2=53−28=25  ⟹  x−y=5(x-y)^2 = 53 - 28 = 25 \implies x-y = 5

  3. Step 3

    x^2 - y^2 = (x+y)(x-y) = 9 ⋅\cdot 5 = 45

  4. Step 4

    x^4 - y^4 = (x^2 - y^2)^2 = 45^2 = 2025

    Correct

    x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = 53 ⋅\cdot 45 = 2385

    The expression x4−y4x^4 - y^4 is a difference of squares, (x2)2−(y2)2(x^2)^2 - (y^2)^2, which factors into (x2+y2)(x2−y2)(x^2 + y^2)(x^2 - y^2). Squaring (x2−y2)(x^2 - y^2) yields x4−2x2y2+y4x^4 - 2x^2y^2 + y^4, which is not equivalent to x4−y4x^4 - y^4.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly related (x+y)2(x+y)^2 to x2+y2x^2 + y^2 and determined the product xy=14xy = 14.

You used (x−y)2=x2+y2−2xy(x-y)^2 = x^2+y^2-2xy to find the positive difference x−y=5x - y = 5.

You correctly factored and evaluated x2−y2=(x+y)(x−y)=45x^2 - y^2 = (x+y)(x-y) = 45.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Misapplied Special Product Identity

You treated the difference of fourth powers x4−y4x^4 - y^4 as (x2−y2)2(x^2 - y^2)^2 instead of factoring it as (x2+y2)(x2−y2)(x^2 + y^2)(x^2 - y^2).

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the difference of squares pattern A2−B2=(A+B)(A−B)A^2 - B^2 = (A + B)(A - B) with the square of a binomial (A−B)2=A2−2AB+B2(A - B)^2 = A^2 - 2AB + B^2 when dealing with higher powers.

REASONING PATH
Your ACEpath teacher’s suggested approach
Identify Difference of Squares vs. Square of a Difference

  • For any terms AA and BB, A2−B2=(A+B)(A−B)A^2 - B^2 = (A+B)(A-B), whereas (A−B)2=A2−2AB+B2(A-B)^2 = A^2 - 2AB + B^2.

  • Express x4−y4x^4 - y^4 as (x2)2−(y2)2(x^2)^2 - (y^2)^2 to clearly see that A=x2A = x^2 and B=y2B = y^2, giving the sum-and-difference product (x2+y2)(x2−y2)(x^2 + y^2)(x^2 - y^2).

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Rewrite x4−y4x^4 - y^4 as a difference of squares: x4−y4=(x2)2−(y2)2=(x2+y2)(x2−y2)x^4 - y^4 = (x^2)^2 - (y^2)^2 = (x^2 + y^2)(x^2 - y^2).

  2. Substitute the given sum of squares x2+y2=53x^2 + y^2 = 53 and your evaluated difference x2−y2=45x^2 - y^2 = 45.

  3. Compute the product: 53⋅45=238553 \cdot 45 = 2385.

Final answer

2385

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice rewriting higher-even-power binomials like a4−b4a^4 - b^4 or a8−b8a^8 - b^8 directly into nested difference-of-squares factors.

  • Whenever you write (A−B)2(A - B)^2, mentally check for the presence of the middle cross-term −2AB-2AB to ensure it is not being confused with A2−B2A^2 - B^2.

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly adds 2xy instead of subtracting it when expressing (x - y)², leading to an erroneous value for x - y.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    (x+y)^2 = x^2 + 2xy + y^2   ⟹  \implies 81 = 53 + 2xy   ⟹  \implies 2xy = 28   ⟹  \implies xy = 14

  2. Step 2

    (x-y)^2 = x^2 + y^2 + 2xy = 53 + 28 = 81   ⟹  \implies x - y = 9

    Correct

    (x-y)^2 = x^2 + y^2 - 2xy = 53 - 28 = 25   ⟹  \implies x - y = 5

    In the expansion of the square of a binomial difference, the middle term must be subtracted: (x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2. You added 2xy2xy instead of subtracting it.

  3. Step 3

    x^4 - y^4 = (x^2+y^2)(x+y)(x-y)   ⟹  \implies x^4 - y^4 = 53 ⋅\cdot 9 ⋅\cdot 9 = 4293

    Correct

    x^4 - y^4 = (x^2+y^2)(x+y)(x-y) = 53 ⋅\cdot 9 ⋅\cdot 5 = 2385

    Your factorization of x4−y4x^4 - y^4 is completely correct, but the final calculation is affected by substituting x−y=9x - y = 9 from Step 2.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly applied the expansion (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2 to find the product term 2xy=282xy = 28.

You correctly factored the difference of fourth powers into (x2+y2)(x+y)(x−y)(x^2 + y^2)(x + y)(x - y).

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign error in special product expansion

You wrote +2xy+2xy instead of −2xy-2xy when expanding (x−y)2(x - y)^2.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the sign of the cross-term in the square of a difference identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 with the square of a sum identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check Binomial Signs

  • (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2, whereas (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.

  • Always verify that the middle cross-term matches the sign inside the parentheses before substituting numerical values.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Apply the correct difference formula: (x−y)2=(x2+y2)−2xy=53−28=25(x - y)^2 = (x^2 + y^2) - 2xy = 53 - 28 = 25.

  2. Take the principal square root since x>yx > y: x−y=25=5x - y = \sqrt{25} = 5.

  3. Substitute into the factored expression: x4−y4=(x2+y2)(x+y)(x−y)=53⋅9⋅5=2385x^4 - y^4 = (x^2 + y^2)(x + y)(x - y) = 53 \cdot 9 \cdot 5 = 2385.

Final answer

2385

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the basic expansion formulas for (a+b)2(a + b)^2 and (a−b)2(a - b)^2, paying close attention to the sign of the middle term.

  • Practice computing (a−b)2(a - b)^2 directly from (a+b)2−4ab(a + b)^2 - 4ab and a2+b2−2aba^2 + b^2 - 2ab to solidify both representations.

Lock it in

Question

What is the complete algebraic expansion of the binomial square (a−b)2(a - b)^2?

Answer

a2−2ab+b2a^2 - 2ab + b^2. Remember that squaring a two-term difference always produces a middle term with a coefficient of −2-2, not −1-1.

Similar question

01

Given two positive numbers aa and bb such that a+b=7a + b = 7 and a2+b2=29a^2 + b^2 = 29, find the value of a−ba - b given that a>ba > b.

Answer

33. Key steps: Compute 2ab=(a+b)2−(a2+b2)=49−29=202ab = (a + b)^2 - (a^2 + b^2) = 49 - 29 = 20. Then (a−b)2=(a2+b2)−2ab=29−20=9(a - b)^2 = (a^2 + b^2) - 2ab = 29 - 20 = 9. Since a>ba > b, a−b=9=3a - b = \sqrt{9} = 3.

02

Given that u+v=10u + v = 10 and uv=21uv = 21, find the exact value of u3−v3u^3 - v^3, given that u>vu > v.

Answer

316316. Key steps: Find (u−v)2=(u+v)2−4uv=100−84=16(u - v)^2 = (u + v)^2 - 4uv = 100 - 84 = 16, so u−v=4u - v = 4. Next, compute u2+uv+v2=(u+v)2−uv=100−21=79u^2 + uv + v^2 = (u + v)^2 - uv = 100 - 21 = 79. Thus u3−v3=(u−v)(u2+uv+v2)=4⋅79=316u^3 - v^3 = (u - v)(u^2 + uv + v^2) = 4 \cdot 79 = 316.

03

Given two positive real numbers xx and yy such that x+y=8x + y = 8 and x2+y2=34x^2 + y^2 = 34 with x>yx > y, evaluate the difference of fourth powers x4−y4x^4 - y^4.

Answer

544544. Key steps: Factor x4−y4=(x2+y2)(x+y)(x−y)x^4 - y^4 = (x^2 + y^2)(x + y)(x - y). Compute (x−y)2=2(x2+y2)−(x+y)2=2(34)−64=68−64=4(x - y)^2 = 2(x^2 + y^2) - (x + y)^2 = 2(34) - 64 = 68 - 64 = 4, which gives x−y=2x - y = 2. Multiply the components: 34⋅8⋅2=54434 \cdot 8 \cdot 2 = 544.

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