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 arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

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

Since $a,b,c$ are in A.P., so,

$2b = a + c$

$4{b^2} = {\left( {a + c} \right)^2}$

The discriminant of the given function$f\left( x \right) = 3a{x^2} - 4bx + c$ is,

$D = 16{b^2} - 12ac$

$ = 4{\left( {a + c} \right)^2} - 12ac$

$ = 4\left[ {\left( {{a^2} + {c^2} + 2ac} \right) - 3ac} \right]$

$ = 4\left( {{a^2} + {c^2} - ac} \right)$

$ = 4\left( {{a^2} + {c^2} - 2ac + ac} \right)$

$ = 4\left( {{{\left( {a - c} \right)}^2} + ac} \right)$

Case 1: If $a$ and$c$ are of opposite signs, then, $D = \left(  +  \right){\rm{ve}}$.

Case 2: If $a$ and$c$ are of same signs, then, $D = \left(  +  \right){\rm{ve}}$.

This shows that $f\left( x \right) = 0$ has two unequal real roots.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

If roots of the equations $(b-c)x^2+(c-a)x+a-b=0$, where $b\neq c$, are equal, then a, b, c are in?

  1. G.P.

  2. H.P.

  3. A.P.

  4. A.G.P.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(b-c)x^{2}+(c-a)x+a-b=0$
Root are equal, so $D=0$
$\Rightarrow (c-a)^{2}-4(b-c)(a-b)=0$
$\Rightarrow c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0$
$\Rightarrow c^2+a^2+4b^2+2ac-4ab-4bc=0$
$\Rightarrow (a+c-2b)^{2}=0$
$a+c=2b$

























Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

If the roots of ${ x }^{ 2 }-k{ x }^{ 2 }+14x-8=0$ are in geometric progression, then $k=$

  1. $-3$
  2. $7$
  3. $4$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow \frac { a }{ r } ,a,ar=8\quad \quad \Rightarrow { a }^{ 3 }=8\quad \quad \Rightarrow a=2$
$a=2$ is a root of the given equation 
$\Rightarrow 8-4k+28-8=0\quad \quad \Rightarrow k=7$

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

If $\alpha, \beta, \gamma$ are non-constant terms in G.P and equations $\alpha { x }^{ 2 }+2\beta x+\gamma =0\quad $ and ${x}^{2}+x-1=0$ has a common root then $\left( \gamma -\alpha  \right) ,\beta $ is

  1. $\alpha \beta $
  2. $\beta \gamma $
  3. $\gamma \alpha $
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the common ratio of G.P is $r$ Therefore $\quad \beta =\alpha t,\alpha { t }^{ 2 }$
Equation $\alpha { x }^{ 2 }+2\alpha rx+\alpha { t }=0\quad 
\Rightarrow { x }^{ 2 }+2rx+{ t }^{ 2 }=0....(i)$
Given equation (i) and ${ x }^{ 2 }+x-1=0....(ii)$ has a common root
$(i)-(ii)\Rightarrow (2e-1)x+({ r }^{ 2 }+1)=0\Rightarrow x=\cfrac { -\left( { r }^{ 2 }+1 \right)  }{ 2r-1 } ....(iii)\quad $
Putting (iii) in equation (ii) $\Rightarrow { \left( { r }^{ 2 }+1 \right)  }^{ 2 }-\left( { r }^{ 2 }+1 \right) (2r-1)-{ \left( { 2r }^{ 2 }-1 \right)  }^{ 2 }=0\Rightarrow { r }^{ 4 }-2{ r }^{ 3 }-{ r }^{ 2 }+2r+1=0....(iv)$
dividing equation (iv) by ${r}^{2}$ $\Rightarrow { \left( r-\cfrac { 1 }{ r }  \right)  }^{ 2 }-2{ \left( r-\cfrac { 1 }{ r }  \right)  }+1=0\Rightarrow { \left( r-\cfrac { 1 }{ r } -1 \right)  }^{ 2 }=0\Rightarrow \cfrac { r-1 }{ r } =1....(v)\quad $
$\left( \gamma -\alpha  \right) \beta =\left( \alpha { r }^{ 2 }-\alpha  \right) \times \alpha r={ \alpha  }^{ 2 }\left( { \alpha  }^{ 2 }-1 \right) r={ \alpha  }^{ 2 }(r-1)={ \alpha  }^{ 2 }{ r }^{ 2 }$
(using $(v)=\alpha \times \alpha { t }^{ 2 }\quad $

Multiple choice

What is the quadratic formula?

  1. A formula for solving quadratic equations

  2. A formula for solving cubic equations

  3. A formula for solving quartic equations
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The quadratic formula is a formula for solving quadratic equations. It is named after al-Khwarizmi, who derived it in the 9th century.

Multiple choice

Bhaskara II's formula for solving quadratic equations is given by: $$ax^2 + bx + c = 0$$. What is the value of x in this formula?

  1. $$x = (-b ± √(b^2 - 4ac)) / 2a$$
  2. $$x = (-b ± √(b^2 + 4ac)) / 2a$$
  3. $$x = (-b ± √(b^2 - 2ac)) / 2a$$
  4. $$x = (-b ± √(b^2 + 2ac)) / 2a$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bhaskara II's formula for solving quadratic equations is given by $$x = (-b ± √(b^2 - 4ac)) / 2a$$. This formula is still used today to solve quadratic equations.

Multiple choice

What was Brahmagupta's formula for solving a quadratic equation?

  1. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  2. $x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$
  3. $x = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}$
  4. $x = \frac{-b \pm \sqrt{b^2 + 2ac}}{2a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Brahmagupta's formula for solving a quadratic equation is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Multiple choice

What is the formula for solving a quadratic equation $ax^2 + bx + c = 0$ according to Bhaskara I?

  1. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  2. $x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$
  3. $x = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}$
  4. $x = \frac{-b \pm \sqrt{b^2 + 2ac}}{2a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bhaskara I's formula for solving a quadratic equation is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Multiple choice

What is the name of the equation that states that the product of the two roots of a quadratic equation is equal to the constant term?

  1. Vieta's formula

  2. Aryabhata's formula

  3. Brahmagupta's formula

  4. Bhaskara's formula
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Vieta's formula is a well-known formula in algebra that states that the product of the two roots of a quadratic equation is equal to the constant term.

Multiple choice

Brahmagupta's formula for solving quadratic equations is:

  1. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  2. $x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$
  3. $x = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}$
  4. $x = \frac{-b \pm \sqrt{b^2 + 2ac}}{2a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Brahmagupta's formula for solving quadratic equations is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Multiple choice

Bhaskara II's work on algebra includes the study of quadratic equations. What is the formula for solving a quadratic equation of the form (ax^2 + bx + c = 0) using the method described by Bhaskara II?

  1. \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  2. \(x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}\)
  3. \(x = \frac{-b \pm \sqrt{b^2 - 2ac}}{2a}\)
  4. \(x = \frac{-b \pm \sqrt{b^2 + 2ac}}{2a}\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bhaskara II's formula for solving a quadratic equation is (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where (a), (b), and (c) are the coefficients of the quadratic equation.

Multiple choice

Brahmagupta's formula for solving quadratic equations is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. What is the discriminant of this quadratic equation?

  1. $b^2 - 4ac$
  2. $b^2 + 4ac$
  3. $4ac - b^2$
  4. $4ac + b^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The discriminant of a quadratic equation is given by the expression $b^2 - 4ac$. It determines the nature of the roots of the equation.

Multiple choice

What was Bhaskara II's formula for solving quadratic equations?

  1. $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
  2. $x = (-b \pm \sqrt{b^2 + 4ac}) / 2a$
  3. $x = (-b \pm \sqrt{b^2 - 4ac}) / a$
  4. $x = (-b \pm \sqrt{b^2 + 4ac}) / a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bhaskara II's formula for solving quadratic equations is $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$.

Multiple choice

What was Bhaskara II's formula for solving cubic equations?

  1. $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
  2. $x = (-b \pm \sqrt{b^2 + 4ac}) / 2a$
  3. $x = (-b \pm \sqrt{b^2 - 4ac}) / a$
  4. $x = (-b \pm \sqrt{b^2 + 4ac}) / a$
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

Bhaskara II did not develop a formula for solving cubic equations.

Multiple choice

What is Brahmagupta's formula for solving quadratic equations?

  1. $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
  2. $x = (-b \pm \sqrt{b^2 + 4ac}) / 2a$
  3. $x = (-b \pm \sqrt{b^2 - 4ac}) / a$
  4. $x = (-b \pm \sqrt{b^2 + 4ac}) / a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Brahmagupta's formula for solving quadratic equations is $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$.