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Determine all real values of the constant cc such that the quadratic expression 36x2+(c−5)x+4936x^2 + (c - 5)x + 49 can be written as the square of a single linear binomial (6x+d)2(6x + d)^2 for some real number dd.

Answer and steps

Final answer

c = 89  or \text{ or } c = -79

Steps

01

Expand the binomial square

To find the values of cc, expand the binomial square and equate the coefficients of corresponding terms.

Using the perfect square trinomial formula (A+B)2=A2+2AB+B2(A + B)^2 = A^2 + 2AB + B^2: (6x+d)2=(6x)2+2(6x)(d)+d2=36x2+12dx+d2(6x + d)^2 = (6x)^2 + 2(6x)(d) + d^2 = 36x^2 + 12dx + d^2

02

Equate corresponding coefficients

We are given that: 36x2+(c−5)x+49=36x2+12dx+d236x^2 + (c - 5)x + 49 = 36x^2 + 12dx + d^2

Matching the constant terms: d2=49d^2 = 49 d=7ord=−7d = 7 \quad \text{or} \quad d = -7

Matching the linear (xx) coefficients: c−5=12dc - 5 = 12d c=12d+5c = 12d + 5

03

Solve for cc in each case

  • Case 1: If d=7d = 7: c=12(7)+5=84+5=89c = 12(7) + 5 = 84 + 5 = 89
  • Case 2: If d=−7d = -7: c=12(−7)+5=−84+5=−79c = 12(-7) + 5 = -84 + 5 = -79

Common Pitfall: Forgetting the negative root when solving d2=49d^2 = 49. Since squaring a negative number yields a positive result, both d=7d = 7 and d=−7d = -7 produce valid real binomial squares: (6x+7)2(6x + 7)^2 and (6x−7)2(6x - 7)^2.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student only considers the positive square root of 49, missing the negative root d = -7 and therefore omitting c = -79.

Sample Student Work

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

    (6x+d)2=36x2+12dx+d2(6x + d)^2 = 36x^2 + 12dx + d^2

  2. Step 2

    d^2 = 49   ⟹  \implies d = 7

    Correct

    d^2 = 49   ⟹  \implies d = ±\pm 7

    When solving d2=49d^2 = 49 for a real number dd, remember to include both the positive and negative square roots: d=7d = 7 and d=−7d = -7.

  3. Step 3

    c - 5 = 12d   ⟹  \implies c - 5 = 12(7) = 84

    Correct

    c - 5 = 12d   ⟹  \implies c - 5 = 12(7) = 84   or\text{or}   c - 5 = 12(-7) = -84

    You set up the coefficient equation correctly, but you only evaluated it for d=7d = 7 because the negative root was omitted earlier.

  4. Step 4

    c = 89

    Correct

    c = 89   or\text{or}   c = -79

    Your final answer includes only one of the two real values for cc.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly used the algebraic pattern (A+B)2=A2+2AB+B2(A + B)^2 = A^2 + 2AB + B^2 to expand the binomial.

You correctly equated the corresponding coefficients of identical polynomial expressions.

You accurately solved the linear equation c−5=84c - 5 = 84 to find c=89c = 89.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Omission of Negative Square Root

When solving the quadratic equation d2=49d^2 = 49, you wrote d=7d = 7 instead of d=±7d = \pm 7.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Treating the square root operation in an algebraic equation x2=kx^2 = k as the principal square root k\sqrt{k}, forgetting that squaring a negative real number also yields a positive result.

REASONING PATH
Your ACEpath teacher’s suggested approach
Square Root Property

  • For any real constant k>0k > 0, the equation x2=kx^2 = k has two real solutions: x=±kx = \pm \sqrt{k}.

  • Whenever you take the square root of both sides to solve an equation, immediately write the ±\pm symbol before simplifying.

Binomial Symmetry

  • Both (ax+b)2(ax + b)^2 and (ax−b)2(ax - b)^2 share the same leading and constant terms, producing two possible linear coefficients: ±2ab\pm 2ab.

  • Check whether the problem allows both addition and subtraction forms inside the squared binomial.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Solve d2=49d^2 = 49 completely: d=7d = 7 or d=−7d = -7.

  2. For d=7d = 7: c−5=12(7)=84  ⟹  c=89c - 5 = 12(7) = 84 \implies c = 89.

  3. For d=−7d = -7: c−5=12(−7)=−84  ⟹  c=−84+5=−79c - 5 = 12(-7) = -84 \implies c = -84 + 5 = -79.

Final answer

c = 89  or \text{ or } c = -79

NEXT STEPS
Final Takeaways from Your Teacher

  • Write out both branches whenever you solve d2=kd^2 = k with k>0k > 0.

  • Verify each resulting value of cc by substituting it back into the original quadratic expression.

OverviewACEpath Teacher's Diagnosis

02 The student forgets the factor of 2 in the middle term of the expanded square, equating c - 5 to 6d instead of 12d.

Sample Student Work

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

    (6x + d)^2 = 36x^2 + 6dx + d^2

    Correct

    (6x + d)^2 = 36x^2 + 12dx + d^2

    When expanding (A+B)2=A2+2AB+B2(A + B)^2 = A^2 + 2AB + B^2, the middle cross-term requires multiplying by 22: 2(6x)(d)=12dx2(6x)(d) = 12dx, not 6dx6dx.

  2. Step 2

    d^2 = 49, so \text{ so } d = ±\pm 7

  3. Step 3

    c - 5 = 6d   ⟹  \implies If \text{If } d = 7,\ c - 5 = 42, so \text{ so } c = 47   ⟹  \implies If \text{If } d = -7,\ c - 5 = -42, so \text{ so } c = -37

    Correct

    c - 5 = 12d   ⟹  \implies If \text{If } d = 7,\ c - 5 = 84   ⟹  \implies c = 89;  If \text{If } d = -7,\ c - 5 = -84   ⟹  \implies c = -79

    Your evaluation method of testing both values of dd was carried out consistently, but the values were affected by setting the linear coefficient equal to 6d6d instead of 12d12d.

  4. Step 4

    c = 47, -37

    Correct

    c = 89  or \text{ or } c = -79

    Because the middle term was missing a factor of 22, the resulting values of cc are incorrect.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified both the positive and negative roots when solving d2=49d^2 = 49.

You set up an equation matching the linear coefficients to solve for cc in both cases.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Omitted factor of 2 in binomial expansion

You wrote the middle term of (6x+d)2(6x + d)^2 as (6x)(d)=6dx(6x)(d) = 6dx instead of doubling the product to get 2(6x)(d)=12dx2(6x)(d) = 12dx.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Incomplete recall of the perfect square trinomial formula (A+B)2=A2+2AB+B2(A + B)^2 = A^2 + 2AB + B^2, confusing (A)(B)(A)(B) with 2AB2AB.

REASONING PATH
Your ACEpath teacher’s suggested approach
Middle Term Expansion Check

  • The middle term of (A+B)2(A + B)^2 is always twice the product of the two terms: 2⋅A⋅B2 \cdot A \cdot B.

  • Write out the explicit multiplication 2(Ax)(B)2(Ax)(B) before combining constants.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Match the xx-coefficients: c−5=12dc - 5 = 12d.

  2. Substitute d=7d = 7: c−5=12(7)=84  ⟹  c=84+5=89c - 5 = 12(7) = 84 \implies c = 84 + 5 = 89.

  3. Substitute d=−7d = -7: c−5=12(−7)=−84  ⟹  c=−84+5=−79c - 5 = 12(-7) = -84 \implies c = -84 + 5 = -79.

Final answer

c = 89  or \text{ or } c = -79

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice reciting and writing the identity (A+B)2=A2+2AB+B2(A + B)^2 = A^2 + 2AB + B^2 whenever you encounter a binomial square.

  • Always write out the intermediate step 2⋅(first term)⋅(second term)2 \cdot (\text{first term}) \cdot (\text{second term}) to avoid dropping the coefficient 22.

OverviewACEpath Teacher's Diagnosis

03 The student subtracts 5 instead of adding 5 when solving the linear equation c - 5 = 12d for c.

Sample Student Work

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

    (6x + d)^2 = 36x^2 + 12dx + d^2   ⟹  \implies d^2 = 49,  so \text{ so } d = ±\pm 7

  2. Step 2

    c - 5 = 12d   ⟹  \implies 12(±\pm 7) = ±\pm 84

  3. Step 3

    c = 84 - 5 = 79  or \text{ or } c = -84 - 5 = -89

    Correct

    c = 12d + 5   ⟹  \implies c = 84 + 5 = 89  or \text{ or } c = -84 + 5 = -79

    To isolate cc in the equation c−5=12dc - 5 = 12d, you must add 55 to both sides using the addition property of equality. Subtracting 55 instead led to the inverted offsets 7979 and −89-89.

  4. Step 4

    c = 79, -89

    Correct

    c = 89  or \text{ or } c = -79

    This conclusion follows directly from the sign error in isolating cc in the previous step.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly used the expansion identity (A+B)2=A2+2AB+B2(A + B)^2 = A^2 + 2AB + B^2.

You remembered to consider both the positive and negative roots of d2=49d^2 = 49.

You accurately matched the corresponding algebraic coefficients of the two trinomials.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Inverse Operation Sign Error

You subtracted 55 from the right-hand side instead of adding 55 when solving c−5=12dc - 5 = 12d.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Applying an incorrect inverse operation when isolating a variable: a subtracted term −5-5 must be canceled by adding +5+5 to both sides.

REASONING PATH
Your ACEpath teacher’s suggested approach
Use the Addition Property of Equality

  • If x−a=bx - a = b, then x=b+ax = b + a.

  • Always apply the inverse operation: undo subtraction with addition.

Perform a Quick Mental Check

  • Substitute your solved value back into the original linear equation.

  • Check: 89−5=8489 - 5 = 84 and −79−5=−84-79 - 5 = -84, whereas 79−5=74≠8479 - 5 = 74 \neq 84.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Start from the equated linear coefficients: c−5=12dc - 5 = 12d.

  2. Add 55 to both sides to isolate cc: c=12d+5c = 12d + 5.

  3. Substitute d=7d = 7: c=12(7)+5=84+5=89c = 12(7) + 5 = 84 + 5 = 89.

  4. Substitute d=−7d = -7: c=12(−7)+5=−84+5=−79c = 12(-7) + 5 = -84 + 5 = -79.

Final answer

c = 89  or \text{ or } c = -79

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice solving two-step linear equations with signed constants to solidify inverse operations.

  • Always plug the candidate values of cc back into the original quadratic expression to verify it factors into a perfect square.

Lock it in

Question

What are all real solutions to the equation u2=64u^2 = 64?

Answer

u=8u = 8 or u=−8u = -8. By the square root property, an equation u2=ku^2 = k (k>0k > 0) always has two real roots: u=±ku = \pm \sqrt{k}.

Similar question

01

Find all real values of kk such that 4x2+kx+814x^2 + kx + 81 can be expressed in the form (2x+m)2(2x + m)^2 for some real number mm.

Answer

k=36k = 36 or k=−36k = -36. Expanding gives (2x+m)2=4x2+4mx+m2(2x + m)^2 = 4x^2 + 4mx + m^2. Equating constant terms gives m2=81m^2 = 81, so m=±9m = \pm 9. Thus k=4m=4(±9)=±36k = 4m = 4(\pm 9) = \pm 36.

02

Find all real values of bb such that the trinomial 25x2+(2b+1)x+925x^2 + (2b + 1)x + 9 is the square of a linear expression (5x+n)2(5x + n)^2.

Answer

b=14.5b = 14.5 or b=−15.5b = -15.5 (or b=292, −312b = \frac{29}{2},\, -\frac{31}{2}). Equating coefficients with 25x2+10nx+n225x^2 + 10nx + n^2: n2=9  ⟹  n=±3n^2 = 9 \implies n = \pm 3. For n=3n = 3, 2b+1=30  ⟹  2b=29  ⟹  b=2922b + 1 = 30 \implies 2b = 29 \implies b = \frac{29}{2}. For n=−3n = -3, 2b+1=−30  ⟹  2b=−31  ⟹  b=−3122b + 1 = -30 \implies 2b = -31 \implies b = -\frac{31}{2}.

03

The polynomial 49x2−42x+(3k−2)49x^2 - 42x + (3k - 2) is a perfect square trinomial equal to (7x+d)2(7x + d)^2. Determine the values of dd and kk.

Answer

d=−3d = -3 and k=113k = \frac{11}{3}. Expanding yields 49x2+14dx+d249x^2 + 14dx + d^2. Match linear terms: 14d=−42  ⟹  d=−314d = -42 \implies d = -3. Then match constant terms: d2=(−3)2=9d^2 = (-3)^2 = 9. Setting 3k−2=93k - 2 = 9 gives 3k=113k = 11, so k=113k = \frac{11}{3}.

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