(i) $2x^{2} – 3x + 5$ = 0
(ii) $3x^{2} – 4\sqrt{3}x + 4$ = 0
(iii) $2x^{2} – 6x + 3$ = 0
(i) Given,$2x^{2} – 3x + 5$ = 0
Comparing the equation with $ax^{2} + bx + c $= 0, we get
a = 2, b = -3 and c = 5
We know, discriminant = $b^{2} – 4ac$
= $( – 3)^{2}– 4 (2) (5)$ = 9 – 40
= – 31
As you can see, $b^{2} – 4ac$ < 0
Therefore, no real root is possible for the given equation, $2x^{2}$ – 3x + 5 = 0.
(ii) $3x^{2} – 4\sqrt{3}x$ + 4 = 0
Comparing the equation with $ax^{2}$ + bx + c = 0, we get
a = 3, b = $-4\sqrt{3}$ and c = 4
We know, Discriminant = $b^{2}$ – 4ac
= $(-4\sqrt{3})^{2}$ – 4(3)(4)
= 48 – 48 = 0
As $b^{2} – 4ac$ = 0,
Real roots exist for the given equation, and they are equal to each other.
Hence, the roots will be $\dfrac{–b}{2a}$ and $\dfrac{–b}{2a}$.
$\dfrac{–b}{2a}$
=$ \dfrac{-(-4\sqrt{3})}{2×3}$
= $\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) $2x^{2} – 6x + 3$ = 0
Comparing the equation with $ax^{2} + bx + c$ = 0, we get
a = 2, b = -6, c = 3
As we know, discriminant = $b^{2} – 4ac$
= $(-6)^{2}$ – 4 (2) (3)
= 36 – 24 = 12
As$ b^{2} – 4ac$ > 0,
Therefore, there are distinct real roots that exist for this equation, $2x^{2} – 6x + 3$ = 0.
x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
= $\dfrac{(-(-6) ± \sqrt{(-62-4(2)(3)) )}}{ 2(2)}$
=$ \dfrac{(6±2\sqrt{3} )}{4}$
=$\dfrac{(3±\sqrt{3})}{2}$
Therefore, the roots for the given equation are $\dfrac{(3+\sqrt{3})}{2}$and $\dfrac{(3-\sqrt{3})}{2}$
(i) $2x^{2} + kx$ + 3 = 0
(ii) kx (x – 2) + 6 = 0
(i) $2x^{2} + kx + 3$ = 0
Comparing the given equation with $ax^{2} + bx + c$ = 0, we get
a = 2, b = k and c = 3
As we know, discriminant =$ b^{2} – 4ac$
= $(k)^{2} $– 4(2) (3)
=$ k^{2}$ – 24
For equal roots, we know,
Discriminant = 0
$k^{2}$ – 24 = 0
$k^{2}$ = 24
k = ±$\sqrt{24}$ = ±$2\sqrt{6}$
(ii) kx(x – 2) + 6 = 0
or $kx^{2}$ – 2kx + 6 = 0
Comparing the given equation with $ax^{2}+ bx + c$ = 0, we get
a = k, b = – 2k and c = 6
We know, Discriminant = $b^{2} – 4ac$
=$ ( – 2k)^{2} $– 4 (k) (6)
= $4k^{2}$ – 24k
For equal roots, we know,
$b^{2} – 4ac$ = 0
$4k^{2} – 24k$ = 0
4k (k – 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘$x^{2}$‘and ‘x‘.
Therefore, if this equation has two equal roots, k should be 6 only.
Let the breadth of the mango grove be $l$.
The length of the mango grove will be $2l$.
Area of the mango grove = $(2l) (l)$= $ 2l^{2}$
$ 2l^{2}$ = 800
$ l^{2}$ = $ \dfrac{800}{2}$ = 400
$ l^{2} $ – 400 =0
Comparing the given equation with $ ax^{2} + bx + c$ = 0, we get
a = 1, b = 0, c = 400
As we know, discriminant =$ b^{2} – 4ac$
=> (0)2 – 4 × (1) × ( – 400) = 1600
Here, $ b^{2} – 4ac $ > 0
Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.
$l $= ±20
As we know, the value of length cannot be negative.
Therefore, the breadth of the mango grove = 20 m.
Length of the mango grove = 2 × 20 = 40 m.
Let’s say the age of one friend is x years.
Then, the age of the other friend will be (20 – x) years.
Four years ago,
Age of first friend = (x – 4) years
Age of second friend = (20 – x – 4) = (16 – x) years
As per the given question, we can write,
(x – 4) (16 – x) = 48
16x – $ x^{2} – 64 + 4x$ = 48
– $ x^{2}+ 20x – 112$ = 0
$ x^{2} – 20x + 112$ = 0
Comparing the equation with $ ax^{2} + bx + c$ = 0, we get
a = 1, b = -20 and c = 112
Discriminant = $ b^{2} – 4ac$
= $ (-20)^{2} – 4$ × 112
= 400 – 448 = -48
$ b^{2} – 4ac$ < 0
Therefore, there will be no real solution possible for the equations. Hence, the condition doesn’t exist.
Let the length and breadth of the park be $l$ and b.
Perimeter of the rectangular park = $2 (l + b)$ = 80
So, $l + b$ = 40
Or, b = $40 – l$
Area of the rectangular park = $l×b$ =$ l(40 – l)$ = $40l – l^{2} $ = 400
$ l^{2} – 40l + 400 $= 0, which is a quadratic equation.
Comparing the equation with $ ax^{2} + bx + c $ = 0, we get
a = 1, b = -40, c = 400
Since discriminant = $ b^{2} – 4ac$
=$ (-40)^{2} – 4 × 400$
= 1600 – 1600 = 0
Thus, $ b^{2} – 4ac$ = 0
Therefore, this equation has equal real roots. Hence, the situation is possible.
The root of the equation,
$l$ = $ \dfrac{–b}{2a}$
$l$ =$ \dfrac{ -(-40)}{2(1)}$ = $ \dfrac{40}{2}$ = 20
Therefore, the length of the rectangular park, $l $= 20 m
And the breadth of the park, b = 40 –$ l$ = 40 – 20 = 20 m.