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#### Find whether the following equations have real roots. If real roots exist, find them.$(i)8x^{2}+2x-3=0$$(ii)-2x^{2}+3x+2=0$$(iii)5x^{2}-2x-10=0$$(iv)\frac{1}{2x-3}+\frac{1}{x-5}=1,x\neq \frac{3}{2},5$$(v)x^{2}+5\sqrt{5}x-70=0$

(i) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $8x^{2}+2x-3=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

a=8,b=2,c=-3

$\\b^{2} - 4ac =(2)^{2} - 4(8)(-3)\\ =4+96\\ =100\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\8x^{2}+2x-3=0\\ 8x^{2}+6x-4x-3=0\\ 2x(4x+3)-1(4x+3)=0\\ (4x+3)(2x-1)=0\\ 4x+3=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x-1=0\\ 4x=-3 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x=1\\ x=\frac{-3}{4} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x= \frac{1}{2}\\$

Roots of the equation $8x^{2}+2x-3=0$  are $\frac{-3}{4}, \frac{1}{2}\\$

(ii) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $-2x^{2}+3x+2=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

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

$\\b^{2} - 4ac =(3)^{2} - 4(-2)(2)\\ =9+16\\ =25\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\-2x^{2}+3x+2=0\\ -2x^{2}+4x-x+2=0\\-2x(x-2)-1(x-2)=0\\(x-2)(-2x-1)=0\\ -2x-1=0 \;\;\;\;\;\;\;\;\;\;\;\;\; x-2=0\\ -2x=1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=2\\ x=\frac{-1}{2}$

Roots of the equation $-2x^{2}+3x+2=0$  are $2, \frac{-1}{2}\\$

(iii) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $5x^{2}-2x-10=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

a=5,b=-2,c=-10

$\\b^{2} - 4ac =(-2)^{2} - 4(5)(-10)\\ =4+200\\ =204\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\5x^{2}-2x-10=0\\ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{2 \pm \sqrt{4-4(5)(-10)}}{2(5)}\\ x=\frac{2 \pm \sqrt{4+200}}{10}\\ x=\frac{2 \pm \sqrt{204}}{10}\\ x=\frac{2(1 \pm \sqrt{51})}{10}\\ x=\frac{1 \pm \sqrt{51}}{5}\\ x=\frac{1 + \sqrt{51}}{5}, x=\frac{1 - \sqrt{51}}{5}\\$

Roots of the equation $5x^{2}-2x-10=0$  are $\frac{1 + \sqrt{51}}{5}, \frac{1 - \sqrt{51}}{5}$

(iv) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $\frac{1}{2x-3}+\frac{1}{x-5}=1$

$\\\frac{(x-5)+(2x-3)}{(2x-3)(x-5)}=1\\ x-5+2x-3=2x(x-5)3(x-5)\\ 3x-8=2x^{2}-10x-3x+15\\ 2x^{2}-13x+15-3x+8=0\\ 2x^{2}-16x+23=0\\$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

a=2,b=-16,c=23

$\\b^{2}-4ac=(-16)^{2}-4(2)(23)\\ =256-184\\ =72\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{16 \pm \sqrt{72}}{2(2)}\;\;\;\;\;\;\;\;\;\;(\because b^{2}-4ac=72)\\ x=\frac{16 \pm 6\sqrt{2}}{4}=\frac{8 \pm 3\sqrt{2}}{2}\\ x=\frac{8 + 3\sqrt{2}}{2},\frac{8 - 3\sqrt{2}}{2}$

Roots of the equation $\frac{1}{2x-3}+\frac{1}{x-5}=1$ are $\frac{8 + 3\sqrt{2}}{2},\frac{8 - 3\sqrt{2}}{2}$

(v) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $x^{2}+5\sqrt{5}x-70=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

$\\a=1, b=5\sqrt{5},c=-70\\ b^{2} - 4ac =(5\sqrt{5})^{2} - 4(1)(-70)\\ =125+280\\ =405\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\x^{2}+5\sqrt{5}x-70=0\\x^{2}+7\sqrt{5}x-2\sqrt{5}x-70=0\\x(x+7\sqrt{5})-2\sqrt{5}(x+7\sqrt{5})=0\\(x+7\sqrt{5})(x-2\sqrt{5}) \\ x+7\sqrt{5}=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x+7\sqrt{5}=0\\ x=-7\sqrt{5} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=2\sqrt{5}\\$

Roots of the equation $x^{2}+5\sqrt{5}x-70=0$  are $-7\sqrt{5}, 2\sqrt{5}\\$

#### Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number

12

Solution

Let the natural number = x

It is given that when the square of x is diminished by 84 then it is equal to thrice of 8 more than x
$x^{2}-84=3(x+8)\\ x^{2}-84=3x+24\\ x^{2}-3x-84-24=0\\ x^{2}-3x-108=0\\ x^{2}-12x+9x-108=0\\ x(x-12)+9(x-12)=0\\ (x-12)(x+9)=0$

x = 12, – 9

We can’t take – 9 because it is given that x is a natural number.

Hence x = 12 is the required natural number.

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#### A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

8

Solution

Let the natural number = x

It is given that x is increased by 12, equal to 160 times its reciprocal
$x+12=\frac{160}{x}\\ x(x+12)=160\\ x^{2}+12x-160=0\\ x^{2}+20x-8x-160=0\\ x(x+20)-8(x+20)=0\\ (x+20)(x-8)=0\\ x+20=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x-8=0\\ x=-20\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=8$

We can’t take x = – 20 because it is given that x is a natural number.

Hence x = 8 is the required natural number.

#### A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

45 km per hour.

Solution

Let the original speed of the train = x km per hour.

Total distance = 360 km

$\text{Time taken by train to reach at final point =}\left ( \frac{360}{x} \right )\left ( \because speed=\frac{distance}{time} \right )$

If train’s speed increased by 5 then speed = (x + 5) km per hour. Time taken by train to cover 360 km at a speed of (x + 5) km per hour=

$\left ( \frac{360}{x+5} \right )\left ( \because speed=\frac{distance}{time} \right )$

Now it is given that the difference between the time taken at a speed of x and at a speed of x + 5 is 48 min.
$\frac{360}{x} - \frac{360}{x+5} =\frac{48}{60}(hour)\\ \frac{360(x+5)-360x}{x(x+5)}=\frac{48}{60}(hour)\\ 5(360x+1800-360x)=4x(x+5)\\ 9000=4x^{2}+20x\\ 4x^{2}+20x-9000=0$

Or

$x^{2}+5x-2250=0$    (divide by 4)

$x^{2}+5x-2250=0\\ x^{2}+50x-45x-2250=0\\ x(x+50)-45(x+50)=0\\ x=-50,45$

We can’t take x = – 50 because speed is always positive.

Hence original speed of train is 45 km per hour.

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#### If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

14

Solution

Let zeba’s current age = x

According to question
$\\(x-5)^{2}=11+5x\\ x^{2}+25-10=11+5x\;\;\;\;\;\;\;\(using \;\;(a-b)^{2}=a^{2}+b^{2}-2ab)\\ x^{2}+25-10x-11-5x=0\\ x^{2}-15x+14=0\\ x^{2}-14x-x+14=0\\ x(x-14)-1(x-14)=0\\ (x-14)(x-1)\\$

x = 14

we can’t take 1 because if we take 5 years younger of 1 then it is equal to – 4 and age cannot be negative.

Hence x = 14 i.e., the age of zeba is 14 years

#### At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

5, 27

Solution

Let Nisha’s present age = x

At present Asha is 2 more than the square of Nisha’s age.

Hence, Asha present age = x2 + 2

According to question

x2 + 2 - x = 10x – 1 - ( x2 + 2 )

x2 + 2 – x = 10x – 1 – x2 - 2

x2 + 2 – x - 10x + x2 + 3=0

2x2-11x+5=0

2x(x-5)-1(x-5)=0

(x-5)(2x-1)=0

$x=5,\frac{1}{2}$?

If Nisha’s age is 5

Asha age = (5)2+2=27

If Nisha’s age is 1/2  then Asha’s age is 2.25  which is not possible.

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#### In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2 [see Fig. 4.1]. Find the length and breadth of the pond.

length = 34m, breadth = 24m

Solution

Let the distance between the pond and loan boundary is x.

Hence the length of pond =50-x-x

From the figure

Area of lawn = Area of pond + Area of grass

50 × 40=(50-2x)(40-2x)+1184     (using Area of rectangle = l × B)

2000-1184=2000-100x-80x+4x2

4x2-180x+2000-2000+1184=0

4x2-180x+1184=0

Divide by 4

x2-45x+296=0

x2 - 37x - 8x + 296 = 0

x(x-37)-8(x-37)=0

(x-37)(x-8)=0

x = 37, 8

If x = 37

Length of pond =50-2(37)

=50-74

=-24

Which is not possible because length can’t be negative.

If x = 8

Length of pond =50-2(8)

=50-16=34m

=40-16=24m

#### At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than $\frac{t^{2}}{4}$ minutes. Find t.

14

Solution

At t minutes past 2 pm, the time need by minute hand to show 3 pm = 60 – t

According to question, it is given that (60 – t) is equal to 3 minutes less than $\frac{t^{2}}{4}$
$60-t=\frac{t^{2}}{4}-3\\ 60-t=\frac{t^{2}-12}{4}\\ t^{2}-12=4(60-t)\\ t^{2}-12=240-4t\\ t^{2}+4t-12-240=0\\ t^{2}+4t-252=0\\ t^{2}+18t-14t-252=0\\ t(t+18)-14(t+18)=0\\ (t+18)(t-14)=0\\ t=-18,14\\$

Time can’t equal to negative term.

Hence t = 14 minutes

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#### The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is 5 m/sec2.The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is 6 m/sec2.

It is given that force is directly proportional to acceleration

i.e.,$F \propto a$

F = ma {Here m is proportionality constant and also mass of body}

It is also given that m = 6

$\therefore F=6a$

 F 6 12 a 1 2

The given equation is F = 6a

When acceleration is 5 m/sec2

Put a = 5

$F = 6 \times 5 = 30$ kgm/sec2

Hence the required answer is 30 kgm/sec2

It is given that force is directly proportional to acceleration

i.e.,$F \propto a$

F = ma {Here m is proportionality constant and also mass of body}

It is also given that m = 6

$\therefore F=6a$

 F 6 12 a 1 2

The given equation is

F = 6a

When acceleration is 6 m/sec2

Put a = 6

$F = 6 \times 6 = 36$ kgm/sec2

Hence the required answer is 36 kgm/sec2

#### If the temperature of a liquid can be measured in Kelvin units as $x^{\circ}K$ or in Fahrenheit units as $y^{\circ}F$, the relation between the two systems of measurement of temperature is given by the linear equation $y =\frac{9}{5}(x-273)+32$.Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is $313^{\circ}K$.If the temperature of a liquid can be measured in Kelvin units as $x^{\circ}K$ or in Fahrenheit units as $y^{\circ}F$, the relation between the two systems of measurement of temperature is given by the linear equation $y =\frac{9}{5}(x-273)+32$.If the temperature is $158^{\circ}F$, then find the temperature in Kelvin.

$104^{\circ}F$

Solution:

Given that,

$y =\frac{9}{5}(x-273)+32$

Put the value of x = $313^{\circ}K$ in equation (1), we get

$y =\frac{9}{5}(313-273)+32$

$y =\frac{9}{5}(40)+32$

$y =\frac{360}{5}+32$

y = 72 + 32

$y =104^{\circ}F$

343K

Solution:

Given that, $y =\frac{9}{5}(x-273)+32$

Put $y=158^{\circ}$ in equation (1) we get

$158^{\circ} =\frac{9}{5}(x-273)+32$

$158^{\circ} =\frac{9(x-273)+160}{5}$

790 – 160 = 9x – 2457

630 + 2457 = 9x

3087 = 9x

343 = x