# 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

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

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

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

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

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

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

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

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

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