Tag: algebraic functions, equations and inequalities

Quizzes (2) Questions (62)

Questions Related to algebraic functions, equations and inequalities

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 
$a^{3}=4(ab-2c)$.

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 of $x^{3}+x^{2}-2x-1=0$ lies between 

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

$x^{ 3 }+x^{ 2 }-2x-1=0$
$f\left( 0 \right) =-1\ f\left( -1 \right) =1\ f\left( -2 \right) =-1\ f\left( -3 \right) =-13\ f\left( -4 \right) =-71$
As $f\left( -1 \right) >0$ and $f\left( -2 \right) <0$
Therefore one roots lies between $-2$ 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 two roots $\alpha,\beta$ of the equation $x^{4}-5x^{3}+11x^{2}-13x+6=0$ are connected by the relation $2\alpha+3\beta=7$, then the roots of the equation are

  1. $-1,3,1\pm i\sqrt{2}$
  2. $-1,3,1\pm i\sqrt{3}$
  3. $2, 1,1\pm i\sqrt{2}$
  4. $2, 1,1\pm i\sqrt{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $\alpha ,\beta ,\gamma ,\delta $ are roots of $x^{ 4 }-5x^{ 3 }+11x^{ 2 }-13x+6=0$
${ s } _{ 1 }=\alpha +\beta +\gamma +\delta =5\ { s } _{ 4 }=\alpha \beta \gamma \delta =6$


For $\gamma ,\delta =1\pm i\sqrt { 2 } $ or $1\pm i\sqrt { 3 } \quad $
${ s } _{ 1 }\Rightarrow \alpha +\beta +2=5\Rightarrow \alpha +\beta =3$
Solving this with $2\alpha +3\beta =7$ we get
$\alpha =2$ and $\beta =1$

Now for $\gamma ,\delta =1\pm i\sqrt { 2 } $
$\alpha \beta \gamma \delta =2\left( 1+2 \right) =6$

And for $\gamma ,\delta =1\pm i\sqrt { 3 } $
$\alpha \beta \gamma \delta =2\left( 1+3 \right) =8$, not possible

Therefore, roots are $2, 1, 1\pm\sqrt{2}$
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

lf the difference of the squares of the roots of equation ${x}^{2} -6x+q=0$ is $24$, then the value of ${q}$ is:

  1. $ -7$
  2. $8$
  3. $5$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $\alpha,\beta$ are roots of ${x}^{2}-6x+q=0,$ then

${ S } _{ 1 }=\alpha +\beta =6$
And ${ S } _{ 2 }=\alpha \beta =q$

Given ${ \alpha  }^{ 2 }-{ \beta  }^{ 2 }=24$
Now from ${ \left( \alpha -\beta  \right)  }^{ 2 }={ \left( \alpha +\beta  \right)  }^{ 2 }-4\alpha \beta $
$\Rightarrow { \left( \alpha -\beta  \right)  }^{ 2 }=36-4q\Rightarrow \left( \alpha -\beta  \right) =\sqrt { 36-4q } $

As ${ \alpha  }^{ 2 }-{ \beta  }^{ 2 }=24\Rightarrow \left( \alpha -\beta  \right) \left( \alpha +\beta  \right) =24$
$\Rightarrow \sqrt { 36-4q } \left( 6 \right) =24\Rightarrow \sqrt { 36-4q } =4$
$\Rightarrow 36-4q=16\Rightarrow 4q=20\Rightarrow q=5$

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 $\mathrm{a} _{\mathrm{n}}\mathrm{x}^{\mathrm{n}}+\mathrm{a} _{\mathrm{n}-1}\mathrm{x}^{\mathrm{n}-1}+\ldots\ldots+\mathrm{a} _{1}\mathrm{x}=0,\ \mathrm{a} _{1}\neq 0,\ \mathrm{n}\geq 2$, has a positive root $\mathrm{x}=\alpha$, then the equation $\mathrm{n}\mathrm{a} _{\mathrm{n}}\mathrm{x}^{\mathrm{n}-1}+(\mathrm{n}-1)\mathrm{a} _{\mathrm{n}-1}\mathrm{x}^{\mathrm{n}-2}+\ldots..+\mathrm{a} _{1}=0$ has a positive root, which is 

  1. greater than $\alpha$
  2. smaller than $\alpha$
  3. greater than or equal to $\alpha$
  4. equal to $\alpha$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

=$ \because { a } _{ n }{ x }^{ 2\  }+{ a } _{ n }+{ x }^{ n-1 }+............+{ a } _{ 1 }x=\quad 0\quad \quad \quad { a } _{ 1 }\neq 0\quad n\ge 2 $

= has the root $x=\infty$ 
= ${ f }^{ 1 }(x)=\quad x{ a } _{ n }{ x }^{ n-1 }+\quad (x-1)\quad { a } _{ n-1 }{ x }^{ n-2 }+.......{ a } _{ n }$
= $\because f(x)=0$
Let us take an example to see 
Let a quadratic equation ${ x }^{ 2 }+2x-3=0$
${ x }^{ 2 }+3x-x-3=0$
$x(x+3)-1(x+3)=0 ........(i)$
$x=1\quad x=-3$
Now ${ f }^{ 1 }(x)=\quad 2x+1$
${ f }^{ 1 }(x)=\quad 0\quad =>\quad x=\quad -\cfrac { 1 }{ 2 } ..........(ii) $
From (i) and (ii) we can see that
The root of ${ f }^{ 1 }(x)$ is always less than the root of $f(x)$
Hence we can conclude
for $n{ a } _{ n }{ x }^{ n-1 }+(n-1){ a } _{ n-1 }{ x }^{ n-2 }+......{ a } _{ 1 }$
has roots always less than $\alpha $ for the value of $\alpha$.


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 $x^{3}+ax+b=0$ is zero, then the value of $b$, is:

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

$let\quad \alpha ,-\alpha ,\beta \quad be\quad the\quad roots\ Given\quad sum\quad of\quad the\quad roots\quad is\quad zero\ \alpha -\alpha +\beta =0\ \beta =0\ Therefore\quad product\quad of\quad the\quad roots\quad is\quad zero,\quad i.e.,\quad b=0$

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

lf one root of $\mathrm{x}^{2}-\mathrm{x}-\mathrm{k}=0(\mathrm{k}>0)$ is the square of the other root, then $\mathrm{k}=$ 

  1. $ 2\pm\sqrt{5}$
  2. $ 2+\sqrt{5}$
  3. $ 2-\sqrt{5}$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ Let\quad \alpha \quad and\quad \alpha ^{ 2 }\quad be\quad the\quad roots\quad as\quad per\quad the\quad given\quad condition.\ \therefore \quad \alpha +\alpha ^{ 2 }=1\quad and\quad { \alpha  }^{ 3 }=-k\ Now\quad (\alpha +\alpha ^{ 2 })^{ 3 }=1\ \Rightarrow { \alpha  }^{ 3 }+({ \alpha  }^{ 2 })^{ 3 }+3{ \alpha  }^{ 3 }(\alpha +\alpha ^{ 2 })=1\ Replacing\quad { \alpha  }^{ 3 }\quad by\quad -k\quad we\quad get\quad \ -k+k^{ 2 }-3k-1=0\ \Rightarrow { k }^{ 2 }-4k-1=0\ \Rightarrow k=\frac { 4+\sqrt { 20 }  }{ 2 } =2+\sqrt { 5 } \ or\quad k=\frac { 4-\sqrt { 20 }  }{ 2 } =2-\sqrt { 5 } <0\ But\quad k>0\quad therefore\quad we\quad reject\quad this\quad value.\ \therefore \quad k=2+\sqrt { 5 } \ Answer-\quad Option\quad B. $