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

(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 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.
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 $ax^2+bx+c=0$, then the quadratic equation whose roots are $\cfrac{1}{\alpha}$  and $\cfrac{1}{\beta}$ is

  1. $ax^2+bx+c=0$
  2. $bx^2+ax+c=0$
  3. $cx^2+bx+a=0$
  4. $cx^2+ax+c=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The quadratic equation whose roots are  $\dfrac{1}{\alpha }$ & $\dfrac{1}{\beta }$      is  $x^{2}-\left(\dfrac{1}{\alpha }+\dfrac{1}{\beta }\right)x+\dfrac{1}{\alpha \beta }$
 
=> $ x^{2}-\left(\dfrac{\alpha + \beta }{\alpha \beta}\right)x+\dfrac{1}{\alpha \beta } = 0$
 
also we know that $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ as $\alpha,\beta$ are roots of the equation $ax^2+bx+c$
 
=> $ x^{2}-\left(\dfrac{-b }{c}\right)x+\dfrac{a}{c } = 0$
 
=> $ x^{2}+\left(\dfrac{b }{c}\right)x+\dfrac{a}{c } = 0$
 
=>  $  cx^{2}+b x+a = 0 $


Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

The equation $x - \displaystyle{\frac{2}{x - 1}} = 1 - \displaystyle{\frac{2}{x - 1}}$ has

  1. no root

  2. one root

  3. two equal roots

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

consider, $x-\cfrac{2}{x-1}=1-\cfrac{2}{x-1}$

$\Rightarrow x-1=\cfrac{2}{x-1}-\cfrac{2}{x-1}$

$\Rightarrow x-1=0$

$\Rightarrow x=1$.

However, as $x\rightarrow 1$,     

$(\cfrac{2}{x-1})\rightarrow \infty$

Hence, the above equation has no root.

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $\alpha, \beta$ are the roots of the equation $x^2 - 3x + 1=0$, then the equation with roots $\displaystyle \frac{1}{\alpha - 2}, \frac{1}{\beta - 2}$ will be-

  1. $x^2 - x - 1 = 0$
  2. $x^2 + x - 1 = 0$
  3. $x^2 + x + 2 = 0$

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

Let y = 1/(x-2). Then x-2 = 1/y, so x = 2 + 1/y = (2y+1)/y. Substitute into x^2 - 3x + 1 = 0: ((2y+1)/y)^2 - 3((2y+1)/y) + 1 = 0. Simplifying leads to y^2 - y - 1 = 0.