Mathematics · Quantitative Aptitude

Polynomial and Quadratic Equations

239 Questions

Polynomial and quadratic equations involve finding the roots of expressions with varying degrees. The focus is on the relationship between coefficients and roots, symmetric functions, and solving higher degree polynomials. This topic is a core component of algebra in competitive testing.

Roots of quadratic equationsSymmetric root functionsCubic polynomialsRoot approximation methodsSum and product of roots

Polynomial and Quadratic Equations Questions

Multiple choice general knowledge science & technology

the formulae in quadratic equation to find roots are: (-b+sqrt(b^2-4ac))/2a and (-b+sqrt(b^2-4ac))/2a what is it known as??

  1. pythagoras' quadratic second function

  2. pulini's hypothesis

  3. shreedharacharya's formula

  4. ramakant root formula
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The quadratic formula x = (-b ± √(b²-4ac))/2a is correctly known as Shreedharacharya's formula, named after the ancient Indian mathematician who discovered it. Pythagoras is associated with the Pythagorean theorem, not quadratic equations. The other options are fictional names.

Multiple choice general knowledge

Soultion for Quadratic Equations were Given by?

  1. Gellilio

  2. Gauss

  3. Shridhracharya

  4. Ramanjuan
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Shridharacharya (c. 870-930 CE), an ancient Indian mathematician, provided the solution for quadratic equations in his work 'Bijaganita'. The quadratic formula is known as 'Shridharacharya's formula' in Indian mathematics.

Multiple choice general knowledge math & puzzles

The sum of the reciprocals of the roots of the equation ax^2+bx+c is 22. The product of the roots of the equation cx^2+bx+a is 11. What is the difference of the roots of the equation ax^2+bx+c

  1. (20/11)^0.5

  2. (40/11)^0.5

  3. (30/11)^0.5

  4. (50/11)^0.5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For ax²+bx+c: sum of roots = -b/a, product = c/a. Sum of reciprocals = (sum)/(product) = (-b/a)/(c/a) = -b/c = 22. For cx²+bx+a: product = a/c = 11, so a = 11c. From -b/c = 22, b = -22c. In ax²+bx+c: difference of roots = √[(sum)² - 4(product)] = √[(-b/a)² - 4(c/a)] = √[(22c/11c)² - 4(c/11c)] = √[(2)² - 4/11] = √[4 - 4/11] = √(40/11).

Multiple choice general knowledge math & puzzles

If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5, then what is the value of m?

  1. -22.5

  2. -17.5

  3. -10.5

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If 1.5 is a root, substitute x = 1.5 into x² + mx + 24 = 0: (1.5)² + m(1.5) + 24 = 0, giving 2.25 + 1.5m + 24 = 0, so 1.5m = -26.25, therefore m = -26.25/1.5 = -17.5. Verification: x² - 17.5x + 24 = 0 has roots 1.5 and 16 (product = 24, sum = 17.5).

Multiple choice

The following is comment written for a C function

/* This function computes the roots of a quadratic equation
a.x<sup>2</sup>+b.x+c = 0. The function stores two real roots
in *root1 and *root2 and returns the status of validity of
roots. It handles four different kinds of cases.
(i) When coefficient a is zero irrespective of discriminant
(ii) When discriminant is positive
(iii) When discrimanant is zero
(iv) When discrimanant is negative
Only in cases (ii) and (iii), the stored roots are valid.
Otherwise 0 is stored in the roots, the function returns 0 when the roots are valid and -1 otherwise.
The function also ensures root1 greater than or equal to root2.
int get_QuadRoots float a, float b, float c, float *root1, float *root2;
*/

A software test engineer is assigned the job of doing black box testing. He comes up with the following test cases, many of which are redundant.

Test Case Input set Expected Output set
--- --- ---
a b
T1 0.0 0.0
T2 0.0 1.0
T3 1.0 2.0
T4 4.0 -12.0
T5 1.0 -2.0
T6 1.0 1.0

Which one of the following options provide the set of non-redundant tests using equivalence class, partitioning approach from input perspective for black box testing?

  1. T1, T2, T3, T6

  2. T1, T3, T4, T5

  3. T2, T4, T5, T6

  4. T2, T3, T4, T5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

T1 and T2 checking same condition a = 0 Hence, any one of T1 and T2 is redundant. T3, T4: in both case discriminant (D) = b2 – 4ac = 0. Hence, any one of it is redundant. T5 : D>0 T6 : D<0

Multiple choice

Let p and q be the real roots of the equation x2 - 4x + A = 0 and let r and s be the real roots of the equation x2 - 12x + B = 0, where A and B are positive integers. If p, q, r, s are in arithmetic progression and p < q < r < s, then find the value of A*B.

  1. 105

  2. 115

  3. 85

  4. 95

  5. 185
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First equation, x2 - 4x + A = 0 Discriminant D = 16 - 4A Roots are (4 - (16 - 4A)1/2)/2 and (4 + (16 - 4A)1/2)/2, i.e. the roots are (2-(4-A)1/2)and (2+(4-A)1/2). Obviously, q = (2+(4-A)1/2) and p = (2-(4-A)1/2) because it is written that q>p. So, q-p = 2(4-A)1/2 Second equation, x2 - 12x + B =0 Discriminant D = 144 - 4B Roots are (12 - (144 - 4B)1/2)/2 and (12 + (144 - 4B)1/2)/2, i.e. the roots are (6-(36-B)1/2)and (6+(36-B)1/2). Obviously s = (6+(36-B)1/2) and r = (6-(36-B)1/2) because it is written that s>r. So, s-r = 2(36-B)1/2 Since p, q, r and s are in AP, therefore q - p = s - r (In AP, common difference is the same) 2(4-A)1/2 = 2(36-B)1/2 Solving, we get 4-A = 36-B, i.e. B-A = 32 Now, possible combinations of (B,A) are (36,4), (35,3), (34,2), (33,1) because A<=4 as per the roots of first equation {(4-A)1/2}, otherwise roots will be complex. Also, B<=36 as per the roots of the second equation. Hence, this option is correct.  

Multiple choice

If α, β, γ are roots of the equation x3 - 3x2 + 3x + 7 = 0 (and ω is imaginary cube root of unity), then find the value of (α - 1)/(β - 1) + (β - 1)/(γ - 1) + (γ - 1)/(α - 1).

  1. ω2

  2. ∞ (infinity)

  3. 3

  4. 3ω2

  5. 0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

x3-3x2+3x+7=0 (x-1)3+8=0 (x-1)3=-8 (x-1)3=(-2)3 {(x-1)/(-2)}3=(1) Taking the cube root, we get (x-1)/(-2)=1,ω,ω2 Solving, we get three different values of x or α, β, γ = -1, 1-2ω, 1-2ω2 Putting the values of α, β, γ in the asked equation, (1/ω)+(1/ω)+ω2=3ω2 (As 1 can be written as ω3)(Correct Answer)

Multiple choice

The roots of x2 - 3x + 16 = 0 are

  1. equal

  2. rational

  3. irrational

  4. imaginary

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Discriminant = 9 - 64 < 0 Hence, there are  imaginary roots only.