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02 Aug 2026, 19:00
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A
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The sum of the roots of a quadratic equation is 2 and the sum of their cubes is 26. What is the equation?
A. \(x^2 - 2x - 3 = 0\)
B. \(x^2 + 2x - 3 = 0\)
C. \(x^2 - 2x + 3 = 0\)
D. \(x^2 + 2x + 3 = 0\)
E. \(x^2 - 2x - 1 = 0\)
A. \(x^2 - 2x - 3 = 0\)
B. \(x^2 + 2x - 3 = 0\)
C. \(x^2 - 2x + 3 = 0\)
D. \(x^2 + 2x + 3 = 0\)
E. \(x^2 - 2x - 1 = 0\)
02 Aug 2026, 21:41
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Let the roots be α and β.
Given:
α + β = 2
α^3 + β^3 = 26
Use the identity:
α^3 + β^3 = (α + β)^3 − 3αβ(α + β)
Substitute the given values:
26 = 23 − 3αβ(2)
26 = 8 − 6αβ
6αβ = 8 − 26 = −18
αβ = −3
The quadratic equation with roots α and β is:
x^2 − (α + β)x + αβ = 0
x^2 − 2x − 3 = 0
Given:
α + β = 2
α^3 + β^3 = 26
Use the identity:
α^3 + β^3 = (α + β)^3 − 3αβ(α + β)
Substitute the given values:
26 = 23 − 3αβ(2)
26 = 8 − 6αβ
6αβ = 8 − 26 = −18
αβ = −3
The quadratic equation with roots α and β is:
x^2 − (α + β)x + αβ = 0
x^2 − 2x − 3 = 0
02 Aug 2026, 21:46
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This one's satisfying once you see the trick, so let me walk through it.
Let the two roots be r and s. We're told r + s = 2, and r^3 + s^3 = 26.
1. The identity to reach for here is r^3 + s^3 = (r+s)^3 - 3rs(r+s). That's basically the whole problem. If that identity isn't cold in your memory, this question turns into a guessing game fast.
2. Plug in what we know: 26 = (2)^3 - 3rs(2) = 8 - 6rs.
3. Solve for rs: 6rs = 8 - 26 = -18, so rs = -3.
4. Now you have both symmetric functions of the roots, sum = 2 and product = -3, so the quadratic is just x^2 - (sum)x + (product) = 0, meaning x^2 - 2x - 3 = 0.
5. Quick sanity check, since I don't trust myself under time pressure: factor it. (x-3)(x+1) = 0, roots 3 and -1. Sum is 2, checks out. Cubes: 27 + (-1) = 26, checks out too.
Answer is A.
The thing I'd flag if you got stuck here: this isn't really an algebra problem, it's a "do you recognize the identity" problem. Once you see sum and product of roots, think Vieta's and skip solving for the individual roots entirely. I used to grind through the quadratic formula on questions like this early in my prep and it cost me way more time than it should have. Memorize r^3+s^3 and (r+s)^2 = r^2+2rs+s^2 cold, they show up constantly on root and symmetric-function questions.
Let the two roots be r and s. We're told r + s = 2, and r^3 + s^3 = 26.
1. The identity to reach for here is r^3 + s^3 = (r+s)^3 - 3rs(r+s). That's basically the whole problem. If that identity isn't cold in your memory, this question turns into a guessing game fast.
2. Plug in what we know: 26 = (2)^3 - 3rs(2) = 8 - 6rs.
3. Solve for rs: 6rs = 8 - 26 = -18, so rs = -3.
4. Now you have both symmetric functions of the roots, sum = 2 and product = -3, so the quadratic is just x^2 - (sum)x + (product) = 0, meaning x^2 - 2x - 3 = 0.
5. Quick sanity check, since I don't trust myself under time pressure: factor it. (x-3)(x+1) = 0, roots 3 and -1. Sum is 2, checks out. Cubes: 27 + (-1) = 26, checks out too.
Answer is A.
The thing I'd flag if you got stuck here: this isn't really an algebra problem, it's a "do you recognize the identity" problem. Once you see sum and product of roots, think Vieta's and skip solving for the individual roots entirely. I used to grind through the quadratic formula on questions like this early in my prep and it cost me way more time than it should have. Memorize r^3+s^3 and (r+s)^2 = r^2+2rs+s^2 cold, they show up constantly on root and symmetric-function questions.
