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 difference of the roots of the quadratic equation is 5 and the difference of their cubes is 215, then the quadratic equation is $x^2\,\pm\,7x\,+\,m\,=\,0$. The value of $m$ is

  1. 2

  2. 4

  3. 6

  4. 10
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let roots be a and b. Given a-b=5 and a^3-b^3=215. Since a^3-b^3 = (a-b)(a^2+ab+b^2) = (a-b)((a-b)^2 + 3ab), we have 215 = 5(25 + 3ab), so 43 = 25 + 3ab, meaning ab = 6. The equation is x^2 - (a+b)x + ab = 0. Given the form x^2 +/- 7x + m = 0, m must be 6.

Multiple choice

How many real roots does the equation $x^2+3|x|+2=0$ have ?

  1. zero

  2. one

  3. Two

  4. Four
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For x^2 + 3|x| + 2 = 0, note that x^2 = |x|^2. The equation becomes |x|^2 + 3|x| + 2 = 0. Let u = |x|, then u^2 + 3u + 2 = 0, which factors to (u+1)(u+2) = 0. Since |x| must be non-negative, there are no real solutions for u = -1 or u = -2.

Multiple choice

193245 Two candidates attempt to solve a quadratic of the form $\displaystyle x^{2}+px+q= 0.$ One starts with a wrong value of $p$ and finds the roots to be $2$ and $6$. The other starts with a wrong value of $q$ and finds the roots to be $2$ and $-9$. Find the correct roots.
  1. $3, 4$
  2. $-2,-5$
  3. $-3,-4$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

First candidate: wrong p, correct q. Roots 2, 6. Product = 12 = q. Second candidate: wrong q, correct p. Roots 2, -9. Sum = -7 = -p, so p = 7. Equation is x^2 + 7x + 12 = 0. Roots are (x+3)(x+4) = 0, so -3, -4.

Multiple choice

Let $m, n$ be positive integers and the quadratic equation $\displaystyle 4x^2 + mx + n = 0$ has two distinct real roots $p$ and $q$ $(p \leq q)$. Also, the quadratic equations $\displaystyle x^2 - px + 2q = 0$ and $\displaystyle x^2 - qx + 2p = 0$ have a common root say $\displaystyle \alpha$. Number of possible ordered pairs $(m, n)$ is equal to
  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots p, q of 4x^2 + mx + n = 0 satisfy p+q = -m/4 and pq = n/4. Common root alpha satisfies alpha^2 - p*alpha + 2q = 0 and alpha^2 - q*alpha + 2p = 0. Subtracting: (q-p)alpha + 2(p-q) = 0 => (q-p)(alpha-2) = 0. Since roots are distinct, p != q, so alpha = 2. Substituting alpha=2: 4 - 2p + 2q = 0 => q-p = -2 => p = q+2. Also 4 - 2q + 2p = 0 => p-q = -2 => p = q-2. This implies p, q are related. Testing integer values for m, n leads to 3 pairs.

Multiple choice

If the equations $\displaystyle ax^{2}+bx+c=0$ and $cx^{2}+bx+a=0,:a\neq c$ have a negative common root, then the value of $a-b+c$ is
  1. $0$
  2. $2$
  3. $1$

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

Let alpha be the common root. Then a*alpha^2 + b*alpha + c = 0 and c*alpha^2 + b*alpha + a = 0. Subtracting gives (a-c)alpha^2 + (c-a) = 0, so (a-c)(alpha^2 - 1) = 0. Since a != c, alpha^2 = 1, so alpha = 1 or -1. Given the root is negative, alpha = -1. Substituting into the first equation: a(-1)^2 + b(-1) + c = 0, which gives a - b + c = 0.

Multiple choice

Let $m, n$ be positive integers and the quadratic equation $\displaystyle 4x^2 + mx + n = 0$ has two distinct real roots $p$ and $q$ $(p \leq q)$. Also, the quadratic equations $\displaystyle x^2 - px + 2q = 0$ and $\displaystyle x^2 - qx + 2p = 0$ have a common root say $\displaystyle \alpha$. If $p$ and $q$ are rational, then uncommon root of the equation $\displaystyle x^2 - px + 2q = 0$ and $\displaystyle x^2 - qx + 2p = 0$ is equal to
  1. $-q$
  2. $\displaystyle \frac {-q}{4}$
  3. $\displaystyle \frac {-q}{2}$
  4. $\displaystyle \frac {q}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Subtracting the two quadratic equations gives (q - p)(x + 2) = 0. Since p and q are distinct roots, the common root is -2. The other root of the first equation is p + 2, and the common-root condition gives p + q = -2, so this root equals -q.

Multiple choice

If each pair of the following three equations $ { x }^{ 2 }+ax+b=0$, ${ x }^{ 2 }+cx+d=0$, ${ x }^{ 2 }+ex+f=0$ has exactly one root in common, then
  1. $ \left( { a+c+e } \right) ^{ 2 }=3\left( ac+ce+ea-b-d-f \right) $
  2. $ \left( { a+c+e } \right) ^{ 2 }=8\left( ac+ce+ea-b-d-f \right) $
  3. $ \left( { a+c+e } \right) ^{ 3 }=4\left( ac+ce+ea-b-d-f \right) $
  4. $ \left( { a+c+e } \right) ^{ 2 }=4\left( ac+ce+ea-b-d-f \right) $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If each pair has a common root, let the roots be alpha, beta, gamma. The equations are (x-alpha)(x-beta)=0, (x-beta)(x-gamma)=0, (x-gamma)(x-alpha)=0. Expanding these gives coefficients related to the sums and products of roots. The condition leads to the identity (a+c+e)^2 = 4(ac+ce+ea-b-d-f).