Solution

We know that for a quadratic equation $a x^{2}+b x+c=0$, discriminant is $b^{2}-4 a c$.

(A) If $b^{2}-4 a c > 0 $ two distinct real roots

(B) If $b^{2}-4 a c=0 $ two equal real roots

(C) If $b^{2}-4 a c < 0 $ no real roots

(I) $2 x^{2}-3 x+5=0$

Comparing this equation with $a x^{2}+b x+c=0$, we obtain

$a=2, b=-3, c=5$

Discriminant $=b^{2}-4 a c=(-3)^{2}-4(2)(5)=9-40$

$=-31$

As $b^{2}-4 a c<0$,

Therefore, no real root is possible for the given equation.

(II) $3 x^{2}-4 \sqrt{3} x+4=0$

Comparing this equation with $a x^{2}+b x+c=0$, we obtain

$a=3, b=-4 \sqrt{3}, c=4$

Discriminant $=b^{2}-4 a c=(-4 \sqrt{3})^{2}-4(3)(4)$

$=48-48=0$

As $b^{2}-4 a c=0$,

Therefore, real roots exist for the given equation and they are equal to each other.

And the roots will be $\dfrac{-b}{2 a}$ and $\dfrac{-b}{2 a}$.

$\dfrac{-b}{2 a}=\dfrac{-(-4 \sqrt{3})}{2 \times 3}=\dfrac{4 \sqrt{3}}{6}=\dfrac{2 \sqrt{3}}{3}=\dfrac{2}{\sqrt{3}}$

Therefore, the roots are $\dfrac{2}{\sqrt{3}}$ and $\dfrac{2}{\sqrt{3}}$.

(III) $2 x^{2}-6 x+3=0$

Comparing this equation with $a x^{2}+b x+c=0$, we obtain

$a=2, b=-6, c=3$

Discriminant $=b^{2}-4 a c=(-6)^{2}-4(2)(3)$

$=36-24=12$

As $b^{2}-4 a c>0$,

Therefore, distinct real roots exist for this equation as follows.

$ \begin{aligned} x & =\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ & =\dfrac{-(-6) \pm \sqrt{(-6)^{2}-4(2)(3)}}{2(2)} \\ & =\dfrac{6 \pm \sqrt{12}}{4}=\dfrac{6 \pm 2 \sqrt{3}}{4} \\ & =\dfrac{3 \pm \sqrt{3}}{2} \end{aligned} $

Therefore, the roots are $\dfrac{3+\sqrt{3}}{2}$ or $\dfrac{3-\sqrt{3}}{2}$.

Solution

We know that if an equation $a x^{2}+b x+c=0$ has two equal roots, its discriminant $(b^{2}-4 a c)$ will be 0 .

(I) $2 x^{2}+k x+3=0$

Comparing equation with $a x^{2}+b x+c=0$, we obtain

$a=2, b=k, c=3$

Discriminant $=b^{2}-4 a c=(k)^{2}-4(2)(3)$

$=k^{2}-24$

For equal roots,

Discriminant $=0$

$k^{2}-24=0$

$k^{2}=24$

$k= \pm \sqrt{24}= \pm 2 \sqrt{6}$

(II) $k x(x-2)+6=0$

or $k x^{2}-2 k x+6=0$

Comparing this equation with $a x^{2}+b x+c=0$, we obtain

$a=k, b=-2 k, c=6$

Discriminant $=b^{2}-4 a c=(-2 k)^{2}-4(k)(6)$

$=4 k^{2}-24 k$

For equal roots,

$b^{2}-4 a c=0$

$4 k^{2}-24 k=0$

$4 k(k-6)=0$

Either $4 k=0$ or $k=6=0$

$k=0$ or $k=6$

However, if $k=0$, then the equation will not have the terms ’ $x{ }^{21}$ and ’ $x$ ‘.

Therefore, if this equation has two equal roots, $k$ should be 6 only.

Solution

Let the breadth of mango grove be $I$.

Length of mango grove will be 21 .

Area of mango grove $=(2 I)(I)$

$=2 l^{2}$

$2 l^{2}=800$

$l^{2}=\dfrac{800}{2}=400$

$l^{2}-400=0$

Comparing this equation with $a l^{2}+b l+c=0$, we obtain

$a=1 b=0, c=400$

Discriminant $=b^{2}-4 a c=(0)^{2}-4 \times(1) \times(-400)=1600$

Here, $b^{2}-4 a c>0$

Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

$l= \pm 20$

However, length cannot be negative.

Therefore, breadth of mango grove $=20 m$

Length of mango grove $=2 \times 20=40 m$

Solution

Let the age of one friend be $x$ years.

Age of the other friend will be (20 - $x$ ) years.

4 years ago, age of $1^{\text{st }}$ friend $=(x-4)$ years

And, age of $2^{\text{nd }}$ friend $=(20-x-4)$

$=(16-x)$ years

Given that,

$(x-4)(16-x)=48$

$16 x-64-x^{2}+4 x=48$

$-x^{2}+20 x-112=0$

$x^{2}-20 x+112=0$

Comparing this equation with $a x^{2}+b x+c=0$, we obtain

$a=1, b=-20, c=112$

Discriminant $=b^{2}-4 a c=(-20)^{2}-4(1)(112)$

$=400-448=-48$

As $b^{2}-4 a c<0$,

Therefore, no real root is possible for this equation and hence, this situation is not possible.

Solution

Let the length and breadth of the park be $/$ and $b$.

Perimeter $=2(I+b)=80$

$l+b=40$

Or, $b=40-1$

Area $=I \times b=I(40-I)=40 I-I^{2}$ $40 l-r^{2}=400$

$l^{2}-40 l+400=0$

Comparing this equation with

$a l^{2}+b l+c=0$, we obtain

$a=1, b=-40, c=400$

Discriminate $=b^{2}-4 a c=(-40)^{2}-4(1)(400)$

$=1600-1600=0$

As $b^{2}-4 a c=0$,

Therefore, this equation has equal real roots. And hence, this situation is possible.

Root of this equation,

$l=-\dfrac{b}{2 a}$

$l=-\dfrac{(-40)}{2(1)}=\dfrac{40}{2}=20$

Therefore, length of park, $I=20 m$

And breadth of park, $b=40-I=40-20=20 m$