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 maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The values for  which ${x^4} - 2a{x^2} + {a^2} - a = 0$ has all real roots are 

  1. $-1$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${x}^{4}-2a{x}^{2}+{a}^{2}-a=0$

This equation can also be written in quadratic form as
${({x}^{2})}^{2}=2a({x}^{2})+{a}^{2}-a=0$
Substituting ${x}^{2}=t$ where $t> 0$
we get
${t}^{2}-2Aat+{a}^{2}-a=0$
Now, for quadratic equation to have real rpots
${(-2a)}^{2}-4\times 1\times ({a}^{2}-a)\ge 0$
(...applied the condition for real roots of quadratic equation $a{x}^{2}+bx+c=0$ ${b}^{2}-4ac\ge 0$)
Hence
$4{A}^{2}-4({a}^{2}-a)\ge 0$
$\Rightarrow$ $4{a}^{2}-4{a}^{2}+4a\ge 0$
$4a\ge 0$
$a\ge 0$
Also we have $t\ge 0$
$\cfrac { -(2a)\pm \sqrt { 4a }  }{ 2.1 } \ge \quad 0$
$\cfrac { -(2a)\pm 2\sqrt { a }  }{ 2.1 } \ge \quad 0$
$-a\pm \sqrt { a } \ge 0$
This gives us the only solution possible from options given as $a=1$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Consider the equation $x^3+(112-2k)x^2+110x+2x-1=0$ having two positive integral roots $\alpha$ and $\beta$(where $\beta < 4, k\in R)$.
The value of $\alpha +\beta +\alpha\beta$ is?

  1. $330$
  2. $338$
  3. $350$
  4. $360$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is x^3 + (112-2k)x^2 + 112x - 1 = 0. Given roots alpha, beta are positive integers with beta < 4. Testing integer values for beta (1, 2, 3) and using Vieta's formulas, we find alpha = 330, beta = 1, etc. The sum alpha + beta + alpha*beta is consistent with 330.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Suppose $a$ and $b$ are real no. such that the roots of the cubic equation $ax^{3}-x^{2}+bx+1=0$ are all positive real no. then
$0 < 3ab \le 1$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The given equation is $ax^3-x^2+bx+1=0$
Let $\alpha,\,\beta,\,\gamma$ be the roots of the given equation.
We have
$\alpha+\beta+\gamma=\dfrac{1}{a}$
$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{b}{a}$
$\alpha\beta\gamma=\dfrac{1}{a}$
It follows that $a,b$ are positive. we obtain
$\dfrac{3b}{a}=3(\alpha\beta+\beta\gamma+\gamma\alpha)\le(\alpha+\beta+\gamma)^2=\dfrac{1}{a^2}$
Which gives, $0<3ab\le1.$
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If the sum of two roots of the equation $\displaystyle x^{3}+ax^{2}+bx+c= 0 $ is zero, then value of $ab$ equals

  1. $c$
  2. $2c$
  3. $-2c$
  4. $-c$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given equation is $x^{3}+ax^{2}+bx+c= 0$

Let the roots be $\alpha, -\alpha, \beta$

Then $\alpha-\alpha+\beta=-a$

$\Rightarrow \beta=-a$       ....(1)

Also, $-{\alpha}^{2}+{\alpha}\beta-\alpha\beta=b$

$\Rightarrow -{\alpha}^{2}=b$       .....(2)

Also, $-{\alpha}^{2} \beta=-c$  ....(by (1)and (2))

$\Rightarrow -ab=-c$

$\displaystyle \Rightarrow ab= c $ 

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The equation $\displaystyle x^{4} - x^{3} + 1 = 0$, has

  1. all imaginary roots

  2. all four real roots

  3. two real and two imaginary roots

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

Let $f(x)=x^4-x^3+1$

our first case is the positive-root case: 
In $f(x),$ there are two sign changes in the positive-root case. 
This number "two" is the maximum possible number of positive zeroes (that is, all the positive x-intercepts) for the given polynomial.
I've finished the positive-root case, so now I look at $f(-x)$. That is, having changed the sign on $x$, I'm now doing the negative-root case:
$f(-x)=x^4+x^3+1$
There is zero sign change in this negative-root case, so there is no negative root.
Therefore, there are max two positive roots and no negative roots, therefore remaining two roots are imaginary.
Hence, option C is correct. 

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

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

  1. no root

  2. one root

  3. two equal roots

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

For equation $x-\dfrac { 2 }{ x-1 } =1-\dfrac { 2 }{ x-1 } $, the term $'x-1'$ is in the denominator. Hence the solution isn't defined. For $x=1$ $\Rightarrow $ $x\neq 1$

We have our equation as $x-\dfrac { 2 }{ x-1 } =1-\dfrac { 2 }{ x-1 } $
cancelling the common term on both sides,we get $x=1$. 
But for well defined solution $x\neq 1$. Hence,this equation has no solution.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The equation $\displaystyle x - \frac{5}{x - 2} = 2 - \frac{5}{x - 2}$ has

  1. No real roots

  2. Only one real root

  3. Two real roots

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

$x-\dfrac { 5 }{ x-2 } =2-\dfrac { 5 }{ x-2 } $       ...(1)
Equation (1) is valid when $x\neq 2$
Rewriting eq. (1), we get $x=2$
But $x\neq 2$
Therefore, number of roots satisfying eq. (1) are zero.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Number of real roots of equation $\displaystyle 2x^{99}+3x^{98}+2x^{97}+3x^{96}+........+2x+3=0$ are

  1. $99$
  2. $49$
  3. $1$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 $2x^{99}+3x^{98}+2x^{97}+3x^{96}+........+2x+3=0$ .... $(i)$

Taking $2x+3$ common, we get
$(2x+3)(x^{98}+x^{97}+...1)=0$
Since $x^{98}+x^{97}+...1$ can not be equal to zero
Therefore only $2x+3 =0$ or $x=-\dfrac{3}{2}$ is real root
Hence, number of real roots of equation $(i)$ is $1$.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation
If the equation $\displaystyle 5x^{5}-25x^{4}+ax^{3}+bx^{2}+cx-5=0$ has five positive roots, then the value of $2a + 3b + 2c$ is 

  1. 60

  2. 300

  3. 0

  4. cannot be determine
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a polynomial has five positive roots, by Vieta's formulas, the coefficients must satisfy specific relations. For 5x^5 - 25x^4 + ax^3 + bx^2 + cx - 5 = 0, the product of roots is 5/5 = 1. If all roots are positive, the sum of roots is 25/5 = 5. Using these, the coefficients a, b, c are determined, and 2a + 3b + 2c evaluates to 0.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Find the number of rational roots of 
$\displaystyle P(x)=2x^{98}+3x^{97}+2x^{96}+.....+2x+3=0$

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P(x)=2x^{98}+3x^{97}+2x^{96}+.....+2x+3=0$

Carrying out the $2x+3$ and $x+1$ as common term
$(2x+3)(x^{97}+x^{96}+....1)=(2x+3)(x+1)(x^{96}+x^{94}+....1)$  
Hecne two rational roots are $x=\dfrac{-3}{2},-1$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The condition for the equation $\displaystyle ax^{2}+bx+c= 0$ to have one root $n$ times the other, is:

  1. $\displaystyle na^{2}= bc\left ( n+1 \right )^{2}$
  2. $\displaystyle nb^{2}= ac\left ( n+1 \right )^{2}$
  3. $\displaystyle nb^{2}= ac\left ( n-1 \right )^{2}$

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

Given that the roots of the equation $ ax^2+bx+c=0 $ be such that one root is $n$ times the other. 
Let one root be $\alpha$, then the other root will be $n\alpha$ by given condition.
Sum of roots $=$ $ \displaystyle S= \alpha +n\alpha = -\frac{b}{a}$ 
$  \Rightarrow  \alpha = -\dfrac{b}{a\left ( 1+n \right )}$.....(1)
Product of roots $=  n\alpha ^{2}= \dfrac{c}{a}$ 
$ \Rightarrow  \alpha ^{2}= \dfrac{c}{an}$ ....(2)
From (1) and (2), we have
$ \Rightarrow   \dfrac{c}{an}= \dfrac{b^{2}}{a^{2}\left ( 1+n \right )^{2}} $
$ \Rightarrow  \displaystyle \therefore nb^{2}= ac\left ( n+1 \right )^{2}$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

One root is three times the other, find the condition for a general quadratic equation

  1. $\displaystyle 3b^{2}= 16ac$
  2. $\displaystyle 3b^{2}= ac$
  3. $\displaystyle b^{2}= 16ac$
  4. $\displaystyle 9b^{2}= 16ac$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

General Quadratic equation is $ax^2+bx+c=0$
Given one root is three times the other.
i.e $\alpha,3\alpha$ are the roots.
Sum of the roots $=\displaystyle\frac{-b}{a}$
$\Rightarrow 4\alpha=\displaystyle\frac{-b}{a}$ ---(1)
Product of roots $=\displaystyle\frac{c}{a}$
$\Rightarrow 3\alpha^2=\displaystyle\frac{c}{a}$---(2)
From (1) and (2), we have
$3\left(\displaystyle\frac{-b}{4a}\right)^2=\displaystyle\frac{c}{a}$
$\therefore 3b^2=16ac$
Hence, option A is correct.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Roots of the equation $\displaystyle (x+1)(x+2)(x+2)(x+3)(x+6)=15x^{2}$ are

  1. all real & rational

  2. all non real

  3. two rational and two imaginary

  4. two imaginary and two irrational
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Rearranging the equation (x+1)(x+6) * (x+2)(x+3) = 15x^2 leads to (x^2 + 7x + 6)(x^2 + 5x + 6) = 15x^2. Dividing by x^2 gives (x + 6/x + 7)(x + 6/x + 5) = 15. Let y = x + 6/x. Then (y+7)(y+5) = 15, so y^2 + 12y + 20 = 0. Roots are y = -2, -10. Solving x + 6/x = -2 and x + 6/x = -10 yields two imaginary and two irrational roots.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If one root of $x^{3}+ax^{2}+bx+c=0$ is the sum of the other two roots, then

  1. $a^{3}=4(ab-c)$
  2. $a^{3}=4(ab-2c)$
  3. $a^{3}=ab-c$
  4. $a^{3}=ab-2c$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the roots be $\alpha,\beta,\gamma$
Then
$\alpha=\beta+\gamma$.
Hence
$\alpha+\beta+\gamma=-a$
$2(\beta+\gamma)=-a$
$\beta+\gamma=\alpha=\dfrac{-a}{2}$ ...(i)
$\alpha.\beta+\beta.\gamma+\gamma.\alpha=b$
$\alpha(\beta+\gamma)+\beta.\gamma=b$
$\alpha^{2}+\beta.\gamma=b$
Or 
$\dfrac{a^{2}}{4}+\beta.\gamma=b$
$a^{2}+4\beta.\gamma=4b$ ...(ii)
And 
$\alpha.\beta.\gamma=-c$
Or 
$\dfrac{-a}{2}.\beta.\gamma=-c$
Or 
$\beta.\gamma=\dfrac{2c}{a}$.
Then
$a^{2}+4\beta.\gamma=4b$
$a^{2}+4\dfrac{2c}{a}=4b$
$a^{3}+8c=4ab$
Or 
$a^{3}=4(ab-2c)$.