Example 3.8
Find the monic polynomial equation of minimum degree with real coefficients having $2-\sqrt{3}i$ as a root

Ans :
We know that if we have p+qi as a root, then p-qi is also a root.
Since we are going to find a polynomial having $2-\sqrt{3}i$ as a root, By the property $2+\sqrt{3}i$ is also a root.
From, $x^2-(\alpha+\beta)x+(\alpha\beta)=0$ where $\alpha,\beta$ are the roots,
Here $\alpha =2-\sqrt{3}i$ , $\beta=2+\sqrt{3}i$.
$\alpha+\beta = 2-\sqrt{3}i + 2+\sqrt{3}i = 4$
$\alpha\beta = (2-\sqrt{3}i)(2+\sqrt{3}i)$= $2^{2} + 3=4+3=7$
Substitute the values, we get $x^2-4x+1=0$ with the rational coefficients.

Example 3.9
Find a polynomial equation of minimum degree with rational coefficients, having $2-\sqrt{3}$ as a root.

Ans:
we know that If we have $p+\sqrt{q}$ as a root, then $p-\sqrt{q}$ is also a root.
By this property, and given $2-\sqrt{3}$ as a root. Then $2+\sqrt{3}$ is also a root.
From, $x^2-(\alpha+\beta)x+(\alpha\beta)=0$ where $\alpha,\beta$ are the roots,
here $\alpha = 2-\sqrt{3}, \beta=2+\sqrt{3}$.
$\alpha+\beta = 2-\sqrt{3} + 2+\sqrt{3} = 2+2$
$\hspace{1.0cm}=4$
$\alpha\beta = (2-\sqrt{3})(2+\sqrt{3})$
$\hspace{1.0cm}=2^2 - 3$
$\hspace{1.0cm}=1$
Substitute the values, we get $x^2-4x+1=0$ with the rational coefficients.

Example 3.10
Form a polynomial equation with integer coefficients with $\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}$ as a root.

Ans :
Since $\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}$ is a root, ( x- $\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}$ ) is a factor.
And we know that If x-$\sqrt{a}$ is a factor then x+$\sqrt{a}$ is also a factor.
From this property ( x+$\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}$ ) also a factor.
$$(x+\sqrt{\frac{\sqrt{2}}{\sqrt{3}}} ) (x+\sqrt{\frac{\sqrt{2}}{\sqrt{3}}} ) = x^2 - \frac{\sqrt{2}}{\sqrt{3}} -----(1)$$
$\hspace{0.5in}$Since we need integer coefficients, we didn’t achieve our goal. So we include another factor $x^2 + \frac{\sqrt{2}}{\sqrt{3}}$ for eqn(1) by the same property and to get a product,
$$( x^2 + \frac{\sqrt{2}}{\sqrt{3}} )( x^2 - \frac{\sqrt{2}}{\sqrt{3}} ) = x^4 - \frac{2}{3}-----(2)$$
$So, 3x^4 - 4=0$ is a required polynomial equation with the integer coefficients.

Example 3.11
Show that the equation $2x^2-6x+7=0$ cannot be satisfied by any real values of x.

Ans :
To find the nature of roots, we have to find the Discriminent $\Delta$ = $b^2-4ac$.
From the equation, a=2, b=-6, c=7,
$\Delta=(-6)^2-4(2)(7)$
$\Delta=36-56=-20<0$
This implies the nature of the roots in Imaginary,. So values of cannot be real.

Example 3.12
If $x^2+2(k+2)x+9k=0$ has a equal roots,find k.

Ans :
We kow that if a polynomial has real and equal roots, then the value of discriminant is 0.
i.e $\Delta$ = $b^2-4ac$=0
Here a=1, b=2(k+2)=2k+4, c= 9k
$(2k+4)^2-4(1)(9k)$=0
$4k^2+4^2+2(2k)(4)-36k=0$
$4k^2+16+16k-36k=0$
$4k^2-20k+16=0$
$\div 4$ $\Rightarrow k^2-5k+4=0$
By factorizing, (k-1)(k-4)=0 $\Rightarrow$ k=1 or 4

Example 3.13
Show that, if p,q,r are rational the roots of the equation $x^2-2px+p^2-q^2+2qr-r^2$ are rational.

Ans :
To find the nature of roots, we have to find the Discriminent $\Delta$ = $b^2-4ac$.
From the equation, a=1, b=-2p, c=$p^2-q^2+2qr-r^2$,
$\Delta = (-2p)^2$-4(1)($p^2-q^2+2qr-r^2$)
$\hspace{0.5cm}$ = $4p^2-4p^2+4q^2-8qr+4r^2$
$\hspace{0.5cm}$ = $4q^2-8qr+4r^2$ = 4($q^2-2qr+r^2$)
$\hspace{0.5cm}$ = $4(q-r)^2$ which is a perfect square.Hence the roots are rational .

Example 3.14
Prove that a line cannot intersect a circle at more than two points.

Ans :
By choosing the coordinate axes suitably, we take the equation of the circle as$x^2 + y^2 = r^2$ and the equation of the straight line as y = mx + c . We know that the points of intersections of the circle and the straight line are the points which satisfy the simultaneous equations.
So we can substitute y=mx+c in the circle equation.
This implies, $x^2 + (mx + c)^2 − r^2 = 0$
$x^2+(mx)^2+c^2+2(mx)(c)-r^2=0$
$x^2+m^2x^2+c^2+2mxc-r^2=0$
$x^2(1+m^2)+(2mc)x(c^2-r^2)=0$
This quadratic equation cannot have more than two solutions, and hence a line and a circle cannot intersect at more than two points.

Exercise 3.2

1) If k is real, discuss the nature of the roots of the polynomial equation 2x2 + kx + k = 0, in terms of k.

Ans :
To discuss the nature of the roots we have to find the discriminant value,
Δ = b² – 4ac
Here a = 2, b = k, c = k
Δ = k² – 4(2)(k)
Δ = k² – 8k
when Δ=0 (Real and equal roots) $\Rightarrow$ k² – 8k=0 $\Rightarrow$ k(k-8)=0 $\Rightarrow$ k=0 or 8.
when Δ > 0(Real and unequal roots) $\Rightarrow$ k² – 8k>0 $\Rightarrow$ k(k-8)>0 $\Rightarrow$ k>0 or k>8 $\Rightarrow$ the values of k>8
when Δ < 0(Imaginary roots) $\Rightarrow$ k² – 8k<0 $\Rightarrow$ k(k-8)<0 $\Rightarrow$ k<0 or k<8 $\Rightarrow$ the values between $0<k<8$

2) Find a polynomial equation of minimum degree with rational coefficients, having $2-\sqrt{3}i$ as a root.

Ans :
Given root is ($2-\sqrt{3}i$)
We know that complex roots occurs in pairs
i.e If p+qi is a root, then p-qi is also a root.
From this,the other root is ($2+\sqrt{3}i$)
From, $x^2-(\alpha+\beta)x+(\alpha\beta)=0$ where $\alpha,\beta$ are the roots,
here $\alpha = 2+\sqrt{3}i, \beta=2-\sqrt{3}i$
$\alpha+\beta = 2-\sqrt{3}i + 2+\sqrt{3}i = 2+2$
$\hspace{1.0cm}=4$
$\alpha\beta = (2-\sqrt{3}i)(2+\sqrt{3}i)$
$\hspace{1.0cm}=2^2 + 3$
$\hspace{1.0cm}=7$
Substitute the values, we get $x^2-4x+7=0$ with the rational coefficients.

3) Find a polynomial equation of minimum degree with rational coefficients, having 2i + 3 as a root.

Ans :
The given root is 3+2i,
We know that complex roots occurs in pairs
i.e If p+qi is a root, then p-qi is also a root.
From this,the other root is 3 – 2i.
From, $x^2-(\alpha+\beta)x+(\alpha\beta)=0$ where $\alpha,\beta$ are the roots,
here $\alpha = 3+2i, \beta=3-2i$
$\alpha+\beta = 3+2i + 3-2i = 3+3$
$\hspace{1.0cm}=6$
$\alpha\beta = (2-\sqrt{3}i)(2+\sqrt{3}i)$
$\hspace{1.0cm}=2^2 + 9$
$\hspace{1.0cm}=13$
The required equation is x² – 6x + 13 = 0

4) Find a polynomial equation of minimum degree with rational coefficients, having √5 – √3 as a root.

Ans :
The given one roots of the polynomial equation are (√5 – √3).
We know that, if we have $\sqrt{p}$+$\sqrt{q}$ as a root then we will have another three roots that are $\sqrt{p}$-$\sqrt{q}$, -$\sqrt{p}$+$\sqrt{q}$, -$\sqrt{p}$-$\sqrt{q}$.
(Hint : ++,+-, -+, --)
From this the other roots are (√5 + √3), (-√5 + √3) and (-√5 – √3).
The quadratic factor with roots (√5 – √3) and (√5 + √3) is ;
From, $x^2-(\alpha+\beta)x+(\alpha\beta)=0$ where $\alpha,\beta$ are the roots,
here $\alpha = (√5 – √3), \beta=(√5 + √3)$
$\alpha+\beta = (√5 - √3) + (√5 + √3) = 2√5$
$\alpha\beta = (√5 – √3)(√5 + √3)$
$\hspace{1.0cm}=(√5)^2 - (√3)^2$=5-3
$\hspace{1.0cm}=2$
The required equation is x² – 2√5x + 2 = 0
The quadratic factor with roots (-√5 + √3) and (-√5 – √3) is ;
From, $x^2-(\alpha+\beta)x+(\alpha\beta)=0$ where $\alpha,\beta$ are the roots,
here $\alpha = (-√5 +√3), \beta=(-√5 – √3)$
$\alpha+\beta = (-√5 +√3) + (-√5 – √3) = -2√5$
$\alpha\beta = (-√5 + √3)(-√5 - √3)$
$\hspace{1.0cm}=(-√5)^2 - (√3)^2$=5-3
$\hspace{1.0cm}=2$
The required equation is x² + 2√5x + 2 = 0
To rationalize the co-efficients with minimum degree
$(x^2 – 2√5 x + 2) (x^2 + 2√5 x + 2) = 0$
Rearrange, $(\underline{x^2+2}– 2√5 x) (\underline{x^2+2}+ 2√5 x ) = 0$
⇒ $(x^2 + 2)^2 – (2√5 x)^2 = 0$
⇒ $x^4 + 4 + 4x^2 – 20x^2 = 0$
⇒ $x^4 – 16x^2 + 4$ = 0

5) Prove that a straight line and parabola cannot intersect at more than two points.

Ans :
Let be the equation of a straight line y = mx + c.
Let be the equation of a parabola y² = 4ax
We know that the points of intersections of the line and the parabola are the points which satisfy the simultaneous equations.
So we can substitute y=mx+c in the parabola equation.
$\Rightarrow$(mx + c)² = 4ax
$\Rightarrow$m²x² + 2mcx + c² – 4ax = 0
$\Rightarrow$m²x² + (2x)(mc – 2a) + c² = 0 Which is a quadratic equation. Hence It cannot have more than two solutions.