10th Class Mathematics Polynomials Graphical Representation

Graphical Representation

Category : 10th Class

*      Graphical Representation

 

*            Graphical Representation of Different Forms of Quadratic Equation  

 

Characteristic of the function   \[{{b}^{2}}-4\,ac\,<\,0\]   \[{{b}^{2}}-4\,ac\,\,0\]   \[{{b}^{2}}-4\,ac>0\]
When 'a' is positive
When 'a' negative i.e. a < 0

 

 

*            Relationship between the Zeroes of the Polynomials and Coefficient of Polynomials

 

If \[a{{x}^{2}}+bx+c=0\] is a quadratic equation whose roots are a and p, then the relation between the roots of the equation is given by

Sum of the roots = \[\alpha +\beta -\frac{b}{a},\]

Product of the roots = \[\alpha \beta -\frac{c}{a}.\]

Fora cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\], the relation between the roots whose roots are \[\alpha ,\beta \,and\,\gamma ,\] is given by

Sum of roots = \[\alpha +\beta +\gamma =-\frac{b}{a},\]

Sum of the product of roots = \[\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}\]

Product of the roots = \[\alpha \beta \gamma =-\frac{d}{a}.\]    

 

 

 

 

  The graphical representation of the equation \[f\left( x \right)={{x}^{2}}+2x+10\] is:

(a) Straight line                                

(b) Circle

(c) Parabola                                       

(d) Ellipse

(e) None of these  

 

Answer: (c)  

Explanation

The above given equation is a quadratic equation which traces a parabola on the graph.  

 

 

  The zeroes of the polynomials  \[f\left( x \right)=ab{{x}^{2}}+\left( {{b}^{2}}+ac \right)x+bc\,is\_\_\_\_.\]

(a) \[\left( \frac{b}{a}\And -\frac{c}{a} \right)\]                                

(b) \[\left( -\frac{b}{a}\And -\frac{c}{a} \right)\]

(c) \[\left( -\frac{b}{a}\And \frac{c}{a} \right)\]                                                 

(d) \[\left( \frac{b}{a}\And \frac{c}{a} \right)\]

(e) None of these  

 

Answer: (b)  

Explanation

We have,

\[ab{{x}^{2}}+\left( {{b}^{2}}+ac \right)x+bc\]

\[=ab{{x}^{2}}+{{b}^{2}}x+acx+bc\]

\[=bx\left( ax+b \right)+c\left( ax+b \right)\]

\[=\left( ax+b \right)\left( bx+c \right)\]

Therefore, \[x=\left( -\frac{b}{a}\And -\frac{c}{a} \right)\]  

 

 

  If a and (3 are the roots of the given equation \[2\sqrt{3}{{x}^{2}}+4x-3\sqrt{3}\], then the value of \[\frac{1}{{{\alpha }^{3}}}+\frac{1}{{{\beta }^{3}}}is\_\_\_\_\_.\]

(a) 0                                                      

(b) \[-\frac{280\sqrt{3}}{243}\]

(c) \[+\frac{280\sqrt{3}}{243}\]                                 

(d) \[\frac{280}{243}\]

(e) None of these  

 

Answer: (c)

Explanation

The sum of the roots is \[\alpha +\beta =-\frac{b}{a}=-\frac{2}{\sqrt{3}}\]

Product of the roots \[\alpha \beta =\frac{c}{a}=-\frac{3}{2}\]

Now, \[=\frac{1}{{{\alpha }^{3}}}\frac{1}{{{\beta }^{3}}}=\frac{\left( \alpha +{{\beta }^{3}} \right)-3\alpha \beta \left( \alpha +\beta  \right)}{{{\left( \alpha \beta  \right)}^{3}}}\]  

\[=\frac{\frac{-8}{3\sqrt{3}}-\frac{9}{\sqrt{3}}}{-\frac{27}{8}}\]

 

 

  If \[\alpha \,and\,\beta \] are the roots of the polynomials \[k{{y}^{2}}+6y-18\,such\,that\,{{\alpha }^{2}}+{{\beta }^{2}}=36\] then find the value of k.

(a) \[\frac{1\pm \sqrt{2}}{2}\]                                   

(b) \[\frac{1\pm \sqrt{3}}{2}\]

(c) \[\frac{1+\sqrt{3}}{2}\]                                          

(d) \[\frac{1\pm \sqrt{5}}{2}\]

(e) None of these  

 

Answer: (d)

Explanation

We have, \[{{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta  \right)}^{2}}-2\alpha \beta \]

\[\frac{36}{{{k}^{2}}}+\frac{36}{k}=36\]

\[{{k}^{2}}-k-1=0\]

\[k=\frac{1\pm \sqrt{5}}{2}\]  

 

 

  If a and b are the roots of the equation \[{{m}^{2}}+5m-8\], find a polynomial whose roots are \[2a+1\,and\,2b+1\].

(a) \[h(m)=3{{m}^{2}}+4m+1\]

(b) \[k({{m}^{2}}+9m+41)\]

(c) \[({{m}^{2}}+8m-)41\]                            

(d) \[k({{m}^{2}}-9m-41)\]

(e) None of these  

 

Answer: (a)  

Explanation

We have sum of the roots = a + b = - 5

Product of the roots = ab = - 8

Now for the required equation,

Sum of the roots = 2 (a + b) + 2 = - 10 + 2 = - 8

Product of the roots = 4ab + 2(a + b) + 1 = - 41

Therefore, required equation is k \[({{m}^{2}}+8m-)41\].      

 

 

 

*       Based on Cubic Polynomials

 

  The zeroes of the polynomials\[f(y)={{y}^{3}}-8{{y}^{2}}+24+64\], if two zeroes are equal in magnitude but opposite in sign.

(a) (5, 8, & 9)                                     

(b) \[(2\sqrt{2},-2\sqrt{2}\And 8)\]

(c) \[(-2\sqrt{2},5,\And 8)\]                        

(d) \[(2\sqrt{2},5\sqrt{2}\And 8)\]

(e) None of these

 

Answer: (b)

Explanation

If \[\alpha ,\beta \,and\,\gamma \] are the roots of the equation, then sum of the roots \[=\alpha +\beta +\gamma =8\]s

Product of the roots \[=\alpha \beta \,\,\gamma =\,-64\]

Putting the value we have, \[\alpha \beta =-8\]

Therefore, \[\alpha ,2\sqrt{2}=a\,and\,\beta =-2\sqrt{2}\,and\,\gamma =8\]

 

 

Find the values of k for which the zeroes of the polynomial \[f(n)={{n}^{3}}+12{{n}^{2}}+39n+k\] are in A.P.

(a) 60                                                    

(b) 25

(c) - 25                                                 

(d) 28

(e) None of these

 

Answer: (a)

 

 

The cubic polynomial whose three roots are 3, -1 and - 3 is:

(a) \[{{n}^{3}}+{{n}^{2}}-9n-9\]                                

(b) \[{{n}^{3}}-{{n}^{2}}-9n-9\]

(c) \[{{n}^{3}}+{{n}^{2}}+9n+9\]                               

(d) \[{{n}^{3}}-{{n}^{2}}+9n+9\]s

(e) None of these

 

Answer: (a)

 

 

If the zeroes of the polynomial\[f(y){{y}^{3}}-3{{y}^{2}}+9y+8\,are\,(p-q),\,p\,and\,(p+q)\]. The value of p and q is:

(a) (1, 3)                                              

(b) \[(1,\pm 3)\]

(c) \[(-1,\pm 3)\]                                             

(d) \[(-1,-3)\]

(e) None of these

 

Answer: (b)

 

 

If \[2\pm \sqrt{3}\] are the two zeroes of the polynomial\[f(x)={{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35\], then the other two zeroes of the polynomials f(x) are:

(a) (- 5, 3)                                           

(b) (1, 3)

(c) (- 1, 3)                                            

(d) (- 5, 7)

(e) None of these

 

Answer: (d)



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