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 theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Two students Ragini and Gourav were asked to solve a quadratic equation $\displaystyle ax^{2}+bx+c=0,a\neq 0$ Ragini made some mistake in writing b and found the roots as 3 and $\displaystyle -\frac{1}{2}$ Gourav too made mistake in writing c and found the roots -1 and $\displaystyle -\frac{1}{4}$ The correct roots of the given equation should be

  1. $-2,$ $\displaystyle \frac{3}{4}$
  2. $3, -1$
  3. $\displaystyle -\frac{1}{2}$, -1
  4. $3,$ $\displaystyle -\frac{1}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: Ragini found roots as $3, -\dfrac 12$ when copied the wrong coefficient of $x$ and Gourav found roots as $-1, -\dfrac 14$ when copied wrong constant term.

To find the correct roots of the given equation
Sol: Viete's formula for the roots $x _1$ and $x _2$ of equation $ax^2+bx+c=0$:  $x _1+x _2=−\dfrac ba$ and $ x _1\times x _2=\dfrac ca$
According to Ragini, she copied the constant term and the coefficient of $ x^2$ correctly. Hence $3\times -\dfrac 12=-\dfrac 32=\dfrac ca$
And according to Gourav, he copied coefficient of $x$ and $x^2$ correctly. Hence $-1+\left(-\dfrac 14\right)=-\dfrac 54=-\dfrac ba$
Hence the equation becomes,
$x^2+\dfrac 54x-\dfrac 32=0\implies 4x^2+5x-6=0$
$\implies 4x^2+8x-3x-6=0\\implies 4x(x+2)-3(x+2)=0\\implies (4x-3)(x+2)=0\\implies x _1=-2, x _2=\dfrac 34$
are the correct roots of the given equation.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Rohan and Sohan were attempting to solve the quadratic equation  $\displaystyle x^{2}-ax+b=0$. Rohan copied the coefficient of x wrongly and obtained the roots as 4 and 12 . Sohan copied the constant term wrongly and obtained the roots as -19 and 3. Find the correct roots

  1. -8, -10

  2. -8, -6

  3. -4, -12

  4. 4, 12
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

With the roots $ 4, 12 $, the equation was $ (x-4))(x-12) ={x}^{2} -4x  -12x + 48 = {x}^{2} -16x + 48 $

As Rohan made a mistake in noting the coffecient of $ x $ , in the original equation, coefficient of $ {x}^{2} = 1 $ and constant $ = 48 $

Now, with the roots $ -19, 3 $, the equation was $ (x-(-19))(x-3) = (x+19)(x-3) = {x}^{2} + 19x -3x -57 = {x}^{2} + 16x -57 $

As Sohan made a mistake in noting just the constant term in the original  equation, coefficient of $ {x}^{2} = 1 $ and of $ x = 16 $

So, we get the original equation as $ {x}^{2} + 16x + 48 = 0 $
Solving it, we get $ {x}^{2} + 4x +12x + 48 = 0 $
$ => x(x+4) + 12(x+4) = 0 $
$ => (x+4)(x+12) = 0 $
$ => x = -4, -12 $

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the equation formed by decreasing each root of $ax^{2}+bx+c=0$ by $1$ is $2x^{2}+8x+2=0$, then

  1. $\mathrm{a}=-\mathrm{b}$
  2. $\mathrm{b}=-\mathrm{c}$
  3. $\mathrm{c}=-\mathrm{a}$
  4. $\mathrm{b}=\mathrm{a}+\mathrm{c}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the equation $2{ x }^{ 2 }+8x+2=0$ has roots which are 1 less than those of the equation $a{ x }^{ 2 }+bx+c=0$ then if we replace $ x $ by $ x+1 $ in latter we'll get the former.
$\Rightarrow a(x+1)^{ 2 }+b(x+1)+c=0$
$\Rightarrow  a{ x }^{ 2 }+(2a+b)x+a+b+c=0$
comparing this equation with that of $2{ x }^{ 2 }+8x+2=0$
we get option (b) as the correct answer

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Umesh and Varun are solving an equation of the form $\displaystyle x^{2}+bx+c=0$. In doing so Umesh commits a mistake in noting down the constant term and finds the roots as $-3$ and $-12$. And Varun commits a mistake in noting down the coefficient of $x$ and find the roots as $-27$ and $-2$. If so find the original equation

  1. $\displaystyle x^{2}-15x+36=0$
  2. $\displaystyle x^{2}+15x+36=0$
  3. $\displaystyle x^{2}-15x+54=0$
  4. $\displaystyle x^{2}+15x+54=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

With the roots $ -3, -12 $, the equation was $ (x-(-3))(x-(-12) = (x+3)(x+6) = {x}^{2} + 3x + 12x + 36 = {x}^{2} + 15x + 36 $

As Umesh made a mistake in noting just the constant term, in the original equation, coefficient of $ {x}^{2} = 1 $ and of $ x = 15 $

Now, with the roots $ -27, -2 $, the equation was $ (x-(-27))(x-(-2)) = (x+27)(x+2) = {x}^{2} + 27x + 2x + 54 = {x}^{2} + 29x + 54 $

As Varun made a mistake in noting the coffecient of $ x $ in the original  equation, coefficient of $ {x}^{2} = 1 $ and constant $ = 54 $

So, we get the original equation as $ {x}^{2} + 15x + 54 = 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

Let $f(x)=1+2x+3x^2+.....+(n+1)x^n,$ where n is even. Then the number of real roots of the equation $f(x)=0$ is 

  1. $0$
  2. $1$
  3. $n$
  4. $None$ $of$ $these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

f(x) = 1 + 2x + 3x^2 + ... + (n+1)x^n. This is the derivative of 1 + x + x^2 + ... + x^(n+1). Since n is even, the sum is (x^(n+2) - 1) / (x - 1). The derivative of this for x not equal to 1 is positive for all x > 0. For x < 0, the terms alternate, and analysis shows no real 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 $1+\surd {3}i/2$ is a root of equation $x^{4}-x^{3}+x1=0$ then its real roots are 

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

Given that 1 + i√3/2 is a root, its complex conjugate 1 - i√3/2 is also a root (coefficients are real). The sum of all roots is 1 (from x³ coefficient with opposite sign). If the remaining two roots are real and equal to r, then: 2(1) + 2r = 1, giving r = ±1. Testing shows x = 1 and x = -1 satisfy the equation. Therefore, the real roots are 1 and -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 a,b,c and d are the real roots of the equation : $x^{4}+p _{1}x^{1}+p _{2}x^{2}+p _{3}x+p _{4}=0$ and $(1+a^{2})(1+b^{2})(1+c^{2})(1+d^{2})=k(1-p _{2}+p _{4})^{2}+(p _{3}-p _{1})^{2}$ then the value f k is:

  1. -1

  2. 1

  3. 2

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

Using the relationship between roots and coefficients, and the identity that (1+a²)(1+b²)(1+c²)(1+d²) can be expressed in terms of the polynomial's coefficients, we can derive that k = 1. This involves substituting the roots into the given expression and using Vieta's formulas to relate it to the coefficients p₁, p₂, p₃, p₄.

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 $\alpha $ and $\beta $ are the roots of ${ x }^{ 2 }+px+q=0$ and ${ \alpha  }^{ 4 } , { \beta  }^{ 4 }$ are the roots of ${ x }^{ 2 }-rx+s=0$, then the equation ${ x }^{ 2 }-4qx+2{ q }^{ 2 }-r=0$ has always two real roots.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If alpha, beta are roots of x^2 + px + q = 0, then alpha^4, beta^4 are roots of x^2 - rx + s = 0. The discriminant of the final equation x^2 - 4qx + 2q^2 - r = 0 is D = (4q)^2 - 4(2q^2 - r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r. Since r = alpha^4 + beta^4, which is always positive, D > 0, so the roots are real.

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

How is the Descartes rule used to find the number of roots in an equation?

  1. By counting the number of times the equation changes signs

  2. By counting positive signs in the equation

  3. By counting negative signs in the equation

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

Descartes rule counts the number of times the sign changes from either $+$ to $-$ or from $-$ to $+$

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

Equation $12x^4-56x^3+89x^2-56x+12=0$ has 

  1. four real and roots

  2. two irrational roots

  3. one integer roots

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

The equation is a reciprocal equation of the first kind. Dividing by x^2 and substituting y = x + 1/x, the equation becomes 12(y^2 - 2) - 56y + 89 = 0, or 12y^2 - 56y + 65 = 0. Solving for y gives y = 5/2 and y = 13/6. Solving x + 1/x = y for these values yields four real 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 a cubic equation is real and second root is imaginary, then what can be said about the third root?

  1. Can be imaginary or real

  2. Must be real

  3. Must be Imaginary

  4. must be zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Cubic Equations can have at max $3$ roots. Also, imaginary roots always occur in a pair of conjugates.
Since here one root is real and the other is imaginary, the third one must be imaginary and it will be the conjugate of the second root.

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 roots of the cubic $x^{3} - (\pi - 1)x^{2} - \pi = 0$, are

  1. All three real and distinct

  2. One real and two coincident
  3. One real and two imaginary with product of the imaginary roots being $\pi$
  4. One real, two imaginary with sum of the imaginary roots being $(-\pi)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let f(x) = x^3 - (pi-1)x^2 - pi. Testing values: f(1) = 1 - pi + 1 - pi = 2 - 2pi < 0. f(pi) = pi^3 - (pi-1)pi^2 - pi = pi^3 - pi^3 + pi^2 - pi = pi^2 - pi > 0. There is a real root between 1 and pi. Further analysis shows the other two roots are imaginary.

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 real value of $\lambda $ for which the equation, $3{x^3} + {x^2} - 7x + \lambda  = 0$, has two distinct real roots in $[0,\,1]$ lie in the interval $(s)$.

  1. $(-2,\,0)$
  2. $[0,\,1]$
  3. $[1,\,2]$
  4. $\left( { - \infty ,\,\infty } \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a cubic to have two distinct real roots in [0, 1], the function must have a local maximum and minimum within or near the interval, and the values at the endpoints must satisfy specific conditions. Analysis of the derivative 9x^2 + 2x - 7 = 0 gives roots at x = -1 and x = 7/9. Testing these in the cubic leads to the interval [0, 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

lf the equation $4 x ^ { 2 } + 2 x ^ { 3 }-4 x - 2 = 0$ has two real roots $\alpha \text { and } \beta$ then between $\alpha \text { and } \beta$ the equation $8 x ^ { 3 } + 3 x ^ { 2 } - 2 = 0$ has 

  1. At least one root

  2. No root

  3. Exactly one root

  4. At most two roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let f(x) = 2x^3 + 4x^2 - 4x - 2. By Rolle's Theorem, if f(alpha) = f(beta) = 0, then f'(x) = 6x^2 + 8x - 4 has a root between alpha and beta. The second equation is g(x) = 8x^3 + 3x^2 - 2. Analysis of the derivatives and signs shows exactly one root exists between the roots of f(x).