Mathematics · Quantitative Aptitude

Polynomial and Quadratic Equations

223 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

(k + 4)x2 + (k + 1)x + 1 = 0 has equal roots if

  1. k = 5

  2. k = 3

  3. k = -5

  4. k > 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the equation has equal roots,  D = 0. b2 - 4ac = 0 (k + 1)2 - 4 (k + 4) (1) = 0 k2 + 2k + 1 - 4k - 16 = 0 $\Rightarrow$ k2 - 2k - 15 = 0 k2 - 5k + 3k - 15 = 0 k(k - 5) +3(k - 5) = 0 $\Rightarrow$ (k - 5) (k + 3) = 0 k = 5, -3, so option (1) is the answer.

Multiple choice business maths limits and continuity of a function graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

If all the roots of $z^3 +az^2 +bz+c=0$ are of unit modulus, then

  1. $|a| \le 3$
  2. $|b| > 3$
  3. $|c| < 3$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots z1, z2, z3 have unit modulus, then |z1|=|z2|=|z3|=1. By Vieta's formulas, a = -(z1+z2+z3). By triangle inequality, |a| = |z1+z2+z3| <= |z1|+|z2|+|z3| = 1+1+1 = 3.

Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

The harmonic mean of the roots of equation $(5+\sqrt {2})x^{2}-(4+\sqrt {5})x+8+2\sqrt {5}=0$ is

  1. $2$
  2. $4$
  3. $6$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} \left( { 5+\sqrt { 2 }  } \right) { x^{ 2 } }-\left( { 4+\sqrt { 5 }  } \right) x+8+2\sqrt { 5 } =0 \ a=5+\sqrt { 2 }  \ b=-\left( { 4+\sqrt { 5 }  } \right)  \ c=8+2\sqrt { 5 }  \ Harmonic\, \, mean\, \, of\, \, \lambda ,\beta  \ =\frac { { 2\lambda \beta  } }{ { \lambda +\beta  } }  \ \lambda \beta =\frac { c }{ a } =\frac { { 8+2\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  \ \lambda +\beta =\frac { { -b } }{ a } =\frac { { 4+\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  \ Harmonic\, \, mean=\,  \ \frac { { 2\frac { { \left( { 8+2\sqrt { 5 }  } \right)  } }{ { 5\sqrt { 2 }  } }  } }{ { \frac { { 4+\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  } }  \ =\frac { { 2\left( { 8+2\sqrt { 5 }  } \right)  } }{ { 4+\sqrt { 5 }  } }  \ =4\, \, \,  \end{array}$

Multiple choice range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

If $n> 1, x> -1, x\neq 0$, then the statement $\left ( 1+x \right )^{n}> 1+nx$ is true for

  1. $ \;n\;\epsilon \;N$
  2. $\forall \;n\;> 1$
  3. $x> -1 \;and\; x\neq 0$

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

$P(1)$ is not true 


For $n=2,P\left( 2 \right) :{ \left( 1+x \right)  }^{ 2 }>1+2x$ is true if $x\neq 0$

Let $P(k):{ \left( 1+x \right)  }^{ k }>1+kx$ be two 

$\therefore{ \left( 1+x \right)  }^{ k+1 }=\left( 1+x \right) { \left( 1+x \right)  }^{ k}>\left( 1+x \right) \left( 1+kx \right)> 1+\left( k+1 \right) x+k{ x }^{ 2}>1+\left(k+1\right) x$

$\left( \because k{ x }^{ 2 }>0 \right) $
$\therefore$ By PMI
Given statement is true for every $n\in N$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx+c=0, a \neq 0$ then the wrong statement is

  1. $\alpha ^2+\beta ^2=\dfrac{b^2-2ac}{a^2}$
  2. $\alpha \beta =\frac{c}{a}$
  3. $\alpha +\beta =\frac{b}{a}$
  4. $\frac{1}{\alpha }+\frac{1}{\beta }=-\frac{b}{c}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We know that sum of roots $= - \dfrac{b}{a}$ and product of roots is  $\dfrac{c}{a}$
Hence option (C) is wrong statement.