Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice

If the roots of the equation $\displaystyle 3x^{2}+9x+2=0$ are in the ratio $m : n$, then find $\displaystyle \sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}$
  1. $\displaystyle \frac{3\sqrt{3}}{\sqrt{2}}$
  2. $\displaystyle 3\sqrt{2}$
  3. $\displaystyle\frac{\sqrt{2}}{\sqrt{3}}$
  4. $\displaystyle 2\sqrt{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For 3x^2 + 9x + 2 = 0, m+n = -9/3 = -3 and mn = 2/3. We need sqrt(m/n) + sqrt(n/m) = (m+n)/sqrt(mn) = -3 / sqrt(2/3) = -3 * sqrt(3/2) = -3 * sqrt(3) / sqrt(2). The provided option A is positive, suggesting a sign error in the question or options.

Multiple choice

If the equation $\displaystyle \left( { p }^{ 2 }+{ q }^{ 2 } \right) { x }^{ 2 }-2\left( pr+qs \right) x+{ r }^{ 2 }+{ s }^{ 2 }=0$ has equal roots then
  1. $\displaystyle pq=rs$
  2. $\displaystyle ps=rq$
  3. $\displaystyle ps=\sqrt { rq } $
  4. $\displaystyle ps=\sqrt { rs } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the quadratic equation to have equal roots, its discriminant must be zero. This simplifies to (pr + qs)^2 - (p^2 + q^2)(r^2 + s^2) = 0, which expands and simplifies to (ps - rq)^2 = 0, yielding ps = rq.

Multiple choice

Let $\alpha,\beta$ be the roots of the equation $ax^2+2bx+c=0$ and $\gamma,\delta$ be the roots of the equation $px^2+2qx+r=0$. If $\alpha,\beta,\gamma,\delta$ are in G.P., then-
  1. $q^2ac=b^2pr$
  2. $qac=bpr$
  3. $c^2pq=r^2ab$
  4. $p^2ab=a^2qr$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a, b, c, d are in GP, then b/a = c/b = d/c = r. Roots are a, ar, ar^2, ar^3. From ax^2+2bx+c=0, sum = a+ar = -2b/a and product = a^2r = c/a. From px^2+2qx+r=0, sum = ar^2+ar^3 = -2q/p and product = a^2r^5 = r/p. Dividing products: (a^2r^5)/(a^2r) = (r/p)/(c/a) => r^4 = (r/p)*(a/c). This leads to the relation q^2ac = b^2pr.

Multiple choice

Let $f\left( x \right) =3a{ x }^{ 2 }-4bx+c\left( a,b,c\in R,a\neq 0 \right) $ where $a,\ b,\ c$ are in A.P. Then the equation $f\left( x \right) =0$ has

  1. no real solution

  2. two unequal real roots

  3. sum of roots always negative.

  4. product of roots always positive.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

a, b, c in A.P. means 2b = a + c. f(x) = 3ax^2 - 4bx + c = 3ax^2 - 2(a+c)x + c = 0. Discriminant D = 4(a+c)^2 - 4(3a)(c) = 4(a^2 + 2ac + c^2 - 3ac) = 4(a^2 - ac + c^2). Since a^2 - ac + c^2 is always positive for a, c != 0, the roots are real and unequal.

Multiple choice

If a, b, c be the $4th$, $7th$ and $10th$ term of an AP respectively then the sum of the roots of the equation $ax^2-2bx+c=0$
  1. $-\displaystyle\frac{b}{a}$
  2. $-\displaystyle\frac{2b}{a}$
  3. $\displaystyle\frac{c+a}{a}$

  4. Cannot be determined unless some more information is given about the AP.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In an AP, a = a1 + 3d, b = a1 + 6d, c = a1 + 9d. The equation is (a1+3d)x^2 - 2(a1+6d)x + (a1+9d) = 0. Sum of roots = -(-2b)/a = 2b/a. However, checking the options, (c+a)/a = (2a1+12d)/(a1+3d) = 2(a1+6d)/(a1+3d) = 2b/a. Thus, (c+a)/a is equivalent to 2b/a.

Multiple choice

Let $p$ and $q$ be the roots of the equation $\displaystyle x^{2} - 2x + c = 0$ and $r$ and $s$ be the roots of the equation $\displaystyle x^{2} - 18x + d = 0$. If $\displaystyle p < q < r < s$ are in A. P. and values of $c$ and $d$ are
  1. $c = -3$ , $d = 77$
  2. $c = 3$ , $d = 77$
  3. $c = 3$ , $d = 7$
  4. $c = 3$ , $d = -7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the roots be p, q, r, s in A.P. with common difference d'. Then p=a, q=a+d', r=a+2d', s=a+3d'. From the first equation, p+q = 2 and pq = c. From the second, r+s = 18 and rs = d. Thus, 2a+d' = 2 and 2a+5d' = 18. Solving gives 4d' = 16, so d'=4 and a=-1. The roots are -1, 3, 7, 11. Then c = (-1)(3) = -3 and d = (7)(11) = 77.

Multiple choice

Let $p, q$ be roots of the equation $x^{2}-4x+A=0$ and $r$ and $s$ be the roots of the equation $x^{2}-20x+B=0$. If $p < q < r < s$ are in A.P., then $(A, B)$ is
  1. $\left ( 0,-96 \right )$
  2. $\left ( 96,0 \right )$
  3. $\left ( 0,96 \right )$
  4. $\left ( -96,0 \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots p, q are for x^2 - 4x + A = 0. Roots r, s are for x^2 - 20x + B = 0. p, q, r, s are in AP. Let common difference be d. p=a, q=a+d, r=a+2d, s=a+3d. Sum of roots p+q = 2a+d = 4. Sum of roots r+s = 2a+5d = 20. Subtracting: 4d = 16, so d = 4. 2a+4 = 4, so a = 0. Roots are 0, 4, 8, 12. A = p*q = 0*4 = 0. B = r*s = 8*12 = 96.

Multiple choice

A quadratic equation whose roots are $\sin^{2} 18^{\circ}, \cos^{2} 36^{\circ}$ are
  1. $16x^{2}-12x+1=0$
  2. $x^{2}-12x+1=0$
  3. $16x^{2}-12x-1=0$

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

sin 18 = (sqrt(5)-1)/4, so sin^2 18 = (5+1-2sqrt(5))/16 = (6-2sqrt(5))/16 = (3-sqrt(5))/8. cos 36 = (sqrt(5)+1)/4, so cos^2 36 = (5+1+2sqrt(5))/16 = (6+2sqrt(5))/16 = (3+sqrt(5))/8. Sum of roots = 6/8 = 3/4. Product of roots = (9-5)/64 = 4/64 = 1/16. Equation: x^2 - (3/4)x + 1/16 = 0, or 16x^2 - 12x + 1 = 0.