Quadratic Equations-Exercise 4.2-Solution-Class 10-Mathematics

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Exercise:4.2

1.Find the roots of the following quadratic equations by factorization:

(i) $Find the roots of the following quadratic equations by factorization:$

Solution:

⇨ $x^2-3x-10=0$
⇨ $x^2-(5-2)x-10=0$
⇨ $x^2-5x+2x-10=0$
⇨ $x(x-5)+2(x-5)=0$
⇨ $(x+2)(x-5)=0$
⇨ $x+2=0, x-5=0$
⇨ $x=-2, x=5$

(ii) $2x^2+x-6=0$
Solution:
⇨ $2x^2+x-6=0$
$\Rightarrow 2x^2+(4-3)x-6=0$
$\Rightarrow 2x^2+4x-3x-6=0$
$\Rightarrow 2x(x+2)-3(x+2)=0$
$\Rightarrow (2x-3)(x+2)=0$
$\Rightarrow 2x-3=0,x+2=0$
$\Rightarrow x=\frac{3}{2},x=-2$

(iii) $\sqrt{2}x^2+7x+5\sqrt{2}=0$
Solution:
⇨ $\sqrt{2}x^2+7x+5\sqrt{2}=0$
⇨ Divided by $\sqrt{2}$ on both side
$\Rightarrow x^2+\frac{7x}{\sqrt{2}}+5=0$
$\Rightarrow x^2+(\sqrt{2}+\frac{5}{\sqrt{2}})x+5=0$
$\Rightarrow x^2+\sqrt{2}x+\frac{5}{\sqrt{2}}x+5=0$
$\Rightarrow x(x+\sqrt{2})+\frac{5}{\sqrt{2}}(x+\sqrt{2})=0$
$\Rightarrow (x+\sqrt{2})(x+\frac{5}{\sqrt{2}})=0$
$\Rightarrow x+\sqrt{2}=0, x+\frac{5}{\sqrt{2}}=0$
$\Rightarrow x=-\sqrt{2}, x=-\frac{5}{\sqrt{2}}$

(iv) $2x^2-x+\frac{1}{8}=0$

Solution:

⇨ $2x^2-x+\frac{1}{8}=0$

⇨Multiply by 2 on both side

$\Rightarrow 4x^2-2x+\frac{1}{4}=0$

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

$\Rightarrow 4x^2-x-x+\frac{1}{4}=0$

$\Rightarrow 4x(x-\frac{1}{4})-1(x-\frac{1}{4})=0$

$\Rightarrow (4x-1)(x-\frac{1}{4})=0$

$\Rightarrow 4x-1=0,x-\frac{1}{4}=0$

$\Rightarrow x=\frac{1}{4},x=\frac{1}{4}$

(v) $100x^2-20x+1=0$
Solution:
⇨ $100x^2-20x+1=0$
⇨Divided by 10 on both side

$\large \Rightarrow 10x^2-2x+\frac{1}{10}=0$

$\large \Rightarrow 10x^2-(1+1)x+\frac{1}{10}=0$

$\large \Rightarrow 10x^2-x-x+\frac{1}{10}=0$

$\large \Rightarrow 10x(x-\frac{1}{10})-1(x-\frac{1}{10})=0$

$\large \Rightarrow (10x-1)(x-\frac{1}{10})=0$

$\large \Rightarrow 10x-1=0,x-\frac{1}{10}=0$

$\large \Rightarrow x=\frac{1}{10},x=\frac{1}{10}$

3. Find two numbers whose sum is 27 and product is 182.
Solution:

⇨Let one of the number be x

⇨ Then another number is = 27-x

A/Q

x(27-x)=182

$\large \Rightarrow 27x-x^{2}= 182$

$\large \Rightarrow x^{2}-27+182= 0$

$\large \Rightarrow x^{2}-27x+182= 0$

$\large \Rightarrow x^{2}-(14+13)x+182= 0$

$\large \Rightarrow x^{2}-14x-13x+182= 0$

$\large \Rightarrow x(x-14)-13(x-14)=0$

$\large \Rightarrow (x-13)(x-14)=0$

$\large \Rightarrow x-13=0,x-14=0$

$\large \Rightarrow x=13,x=14$

Therefore the two numbers are = 13, 14

4. Find two consecutive positive integers, sum of whose squares is 365.
Solution:

⇨ Let the two consecutive positive integers are x, x+1

A/Q

$\large x^2+(x+1)^2=365$

$\large \Rightarrow x^2+x^2+2x+1=365$

$\large \Rightarrow 2x^2+2x+1-365=0$

$\large \Rightarrow 2x^2+2x-364=0$

⇨Divided by 2 on both side

$\large \Rightarrow x^2+x-182=0$

$\large \Rightarrow x^2+(14-13)x-182=0$

$\large \Rightarrow x^2+14x-13x-182=0$

$\large \Rightarrow x(x+14)-13(x+14)=0$

$\large \Rightarrow (x-13)(x+14)=0$

$\large \Rightarrow x-13=0,x+14=0$

$\large \Rightarrow x=13,x=-14$

⇨x = -14, We only consider positive integers, so -14 is rejected.

⇨Therefore, Two consecutive positive integers are = x , x+1

⇨13 , 13+1

⇨13, 14

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

⇨Let the base of the right angle triangle be x cm

⇨Than Altitude/Perpendicular is = x-7 cm

⇨Given, Hypotenuse is = 13 cm

From Pythagoras Formula

$\large h^2=p^2+b^2$

$\large \Rightarrow (13)^2=x^2+(x-7)^2$

$\large \Rightarrow 169=x^2+x^2-2.x.7+7^2$

$\large \Rightarrow 169=2x^2-2.x.7+49$

$\large \Rightarrow 169=2x^2-14x+49$

$\large \Rightarrow 2x^2-14x+49-169=0$

$\large \Rightarrow 2x^2-14x-120=0$

$\large \Rightarrow x^2-7x-60=0$

$\large \Rightarrow x^2-(12-5)x-60=0$

$\large \Rightarrow x^2-12x+5x-60=0$

$\large \Rightarrow x(x-12)+5(x-12)=0$

$\large \Rightarrow (x+5)(x-12)=0$

$\large \Rightarrow x+5=0,x-12=0$

$\large \Rightarrow x=-5,x=12$

⇨x=-5, Length can not be negative, Hence it rejected.

⇨Therefore Length of the Base = 12 cm

⇨Length of Altitude/ Perpendicular = x-7 = 12-7 = 5 cm

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90, find the number of articles produced and the cost of each article.

Solution:

⇨Let the number of pottery articles produced be x

⇨Than price of pottery articles is = 2x+3

A/Q

x(2x+3)= 90

$\large \Rightarrow 2x^2+3x=90$
$\large \Rightarrow 2x^2+3x-90=0$

$\large \Rightarrow 2x^2+(15-12)x-90=0$

$\large \Rightarrow 2x^2+15x-12x-90=0$

$\large \Rightarrow 2x^2-12x+15x-90=0$

$\large \Rightarrow 2x(x-6)+15(x-6)=0$

$\large \Rightarrow (x-6)(2x+15)=0$

$\large \Rightarrow x-6=0,2x+15=0$

$\large \Rightarrow x=6,x=-\frac{15}{2}$

⇨x=-15/2, Number of articles never be negative or in fraction.

⇨Therefore number of articles = 6

⇨Price of articles = 2x+3 =2x6+3 = 15 Rs

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