IPMPrepFind the minimum value of x2 − 4x − 4 for real x.
Solution
The minimum of ax2 + bx + c is at x = −b/2a = 2.
Value at x = 2: (2)2 − 4(2) − 4 = −8.
Answer: −8
Quantitative Ability · Algebra
Quadratic Equations: IPMAT practice questions
24 questions, 11 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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Find the minimum value of x2 − 6x − 11 for real x.
Solution
The minimum of ax2 + bx + c is at x = −b/2a = 3.
Value at x = 3: (3)2 − 6(3) − 11 = −20.
Answer: −20
Find the maximum value of −x2 − 8x − 34 for real x.
Solution
The maximum of ax2 + bx + c is at x = −b/2a = −4.
Value at x = −4: −(−4)2 − 8(−4) − 34 = −18.
Answer: −18
For what positive value of k does x2 + kx + 25 = 0 have equal roots?
Solution
Equal roots need the discriminant to be zero: k2 − 4ac = 0.
k2 = 4 × 1 × 25 = 100, so k = 10.
Answer: (D) 10
For what positive value of k does x2 + kx + 81 = 0 have equal roots?
Solution
Equal roots need the discriminant to be zero: k2 − 4ac = 0.
k2 = 4 × 1 × 81 = 324, so k = 18.
Answer: (A) 18
For how many integer values of k does x2 + kx + 4 = 0 have no real roots?
Solution
No real roots need k2 − 4 × 1 × 4 < 0, so k2 < 16.
−4 < k < 4: the integers from −3 to 3, which is 7 values.
Answer: 7
Find the product of all real roots of x2 − 7|x| + 6 = 0.
Solution
Put y = |x|: y2 − 7y + 6 = 0, so y = 1 or 6.
Then x = ±6 or ±1: four real roots.
Product = (6)(−6)(1)(−1) = 36.
Answer: 36
If α and β are the roots of x2 + 8x + 12 = 0, find (α − β)2.
Solution
α + β = −8 and αβ = 12.
(α − β)2 = (α + β)2 − 4αβ = 16.
Answer: 16
For how many integer values of k does x2 + kx + 16 = 0 have no real roots?
Solution
No real roots need k2 − 4 × 1 × 16 < 0, so k2 < 64.
−8 < k < 8: the integers from −7 to 7, which is 15 values.
Answer: (C) 15
If α and β are the roots of x2 − 15x + 56 = 0, which equation has roots 2α and 2β?
Solution
The roots are 8 and 7, so the new roots are 16 and 14.
Sum = 30, product = 224: the equation is x2 − 30x + 224 = 0.
Answer: (A) x2 − 30x + 224 = 0
If α and β are the roots of x2 − 12x + 35 = 0, which equation has roots α − 1 and β − 1?
Solution
The roots are 7 and 5, so the new roots are 6 and 4.
Sum = 10, product = 24: the equation is x2 − 10x + 24 = 0.
Answer: (C) x2 − 10x + 24 = 0
For what positive value of k does 9x2 + kx + 49 = 0 have equal roots?
Solution
Equal roots need the discriminant to be zero: k2 − 4ac = 0.
k2 = 4 × 9 × 49 = 1764, so k = 42.
Answer: 42
Find the sum of the squares of all real roots of x2 − 7|x| + 6 = 0.
Solution
Put y = |x|: y2 − 7y + 6 = 0, so y = 1 or 6.
Then x = ±6 or ±1: four real roots.
Sum of squares = 2(36 + 1) = 74.
Answer: (C) 74
For how many integer values of k does x2 + kx + 60 = 0 have both roots integers?
Solution
If the roots are integers m and n, then mn = 60 and k = −(m + n).
Factor pairs of 60: (1, 60), (2, 30), (3, 20), (4, 15), (5, 12), (6, 10). Each pair can be both positive or both negative.
The sums are ±61, ±32, ±23, ±19, ±17, ±16: 12 distinct values of k.
Answer: (D) 12
For what positive value of k does x2 + kx + 64 = 0 have equal roots?
Solution
Equal roots need the discriminant to be zero: k2 − 4ac = 0.
k2 = 4 × 1 × 64 = 256, so k = 16.
Answer: 16
If α and β are the roots of x2 + 12x + 32 = 0, find α^2 + β^2.
Solution
α + β = −12 and αβ = 32.
α^2 + β^2 = (α + β)2 − 2αβ = 80.
Answer: 80
If α and β are the roots of x2 + 3x − 28 = 0, find α^3 + β^3.
Solution
α + β = −3 and αβ = −28.
α^3 + β^3 = (α + β)3 − 3αβ(α + β) = −279.
Answer: (A) −279
Find the minimum value of 2x2 + 8x + 14 for real x.
Solution
The minimum of ax2 + bx + c is at x = −b/2a = −2.
Value at x = −2: 2(−2)2 + 8(−2) + 14 = 6.
Answer: (A) 6
For how many integer values of k does x2 + kx + 30 = 0 have both roots integers?
Solution
If the roots are integers m and n, then mn = 30 and k = −(m + n).
Factor pairs of 30: (1, 30), (2, 15), (3, 10), (5, 6). Each pair can be both positive or both negative.
The sums are ±31, ±17, ±13, ±11: 8 distinct values of k.
Answer: (C) 8
For how many integer values of k does x2 + kx + 48 = 0 have both roots integers?
Solution
If the roots are integers m and n, then mn = 48 and k = −(m + n).
Factor pairs of 48: (1, 48), (2, 24), (3, 16), (4, 12), (6, 8). Each pair can be both positive or both negative.
The sums are ±49, ±26, ±19, ±16, ±14: 10 distinct values of k.
Answer: (D) 10
If α and β are the roots of x2 − 9 = 0, find α^3 + β^3.
Solution
α + β = 0 and αβ = −9.
α^3 + β^3 = (α + β)3 − 3αβ(α + β) = 0.
Answer: (B) 0
For how many integer values of k does x2 + kx + 25 = 0 have no real roots?
Solution
No real roots need k2 − 4 × 1 × 25 < 0, so k2 < 100.
−10 < k < 10: the integers from −9 to 9, which is 19 values.
Answer: 19
If α and β are the roots of x2 + 5x − 24 = 0, which equation has roots 3α and 3β?
Solution
The roots are −8 and 3, so the new roots are −24 and 9.
Sum = −15, product = −216: the equation is x2 + 15x − 216 = 0.
Answer: (A) x2 + 15x − 216 = 0
Find the sum of the squares of all real roots of x2 − 10|x| + 24 = 0.
Solution
Put y = |x|: y2 − 10y + 24 = 0, so y = 4 or 6.
Then x = ±6 or ±4: four real roots.
Sum of squares = 2(36 + 16) = 104.
Answer: 104
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.