Quadratic Equations : A quadratic equation is an equation in the form of

$$ax^{2}$$ + bx + c  = 0  where a,b,c  are real numbers and a $$\neq$$ 0. The General form of the quadratic equation is $$ax^{2}$$ + bx +c = 0 ., where the roots of the  above equation is given by $$x = \frac{-b\pm \sqrt{b^{2}- 4ac}}{2a}$$ Sum of the roots = $$\frac{-b}{a}$$ and Product of the roots  = $$\frac{c}{a}$$ . Behaviour of the roots :

1. If $$b^{2}$$ -4ac=0 , then the roots of the equation are real and equal $$,x_{1} = x_{2}$$ .
2. If $$b^{2}$$ – 4ac > 0 , then the roots are real and distinct , $$x_{1} \neq x_{2}$$ .
3. If $$b^{2}$$ – ac < 0 , then the roots are imaginary , $$x_{1}$$ and  $$x_{2}$$ are complex numbers.

Word Problem : Problem :

$$(x+1)^{2}$$ = 2 (x-3) Soution : $$x^{2}$$ + 2x +1 = 2x – 6 $$x^{2}$$ + 2x +1 = 2x -6 $$x^{2}$$ + 2x + 1 -2x + 6 = 0 $$\Rightarrow  x^{2}$$ + 7 = 0. a = 1 , b = 0 , c = 7 , where a $$\neq$$ 0 . The given equation is a quadratic  equation . 1. $$x^{2}$$ – 2x = (-2) (3-x). $$x^{2}$$ -2x = -6 +2x $$x^{2}$$ -4x + 6 = 0 a $$x^{2}$$ + bx + c = 0 Then a = 1 , b = -4 and c = 6 , where a $$\neq$$ 0 . This equation is also a Quadratic equation.

1. The area of a rectangular plot is 528 $$m^{2}$$ .The length of the plot is  twice more than its breadth .

Solution : Area = 528

$$m^{2}$$ , length = x , braeadth = $$\frac{x} {2}$$ , Length $$\times$$ breadth = Area $$x\times \frac{x}{2}$$ = 528 $$x^{2} = 528 \times 2  \Rightarrow = 528 \times$$ 2 = 1056. $$\Rightarrow  x^{2}$$ -1056 =0 Factorise : $$x^{2}$$ -3x – 10 = 0 Solution : $$x^{2}$$ -5x +2x – 10 =0    $$\Rightarrow$$ x(x-5)+2(x-5) = 0 (x+2)(x-5) = 0 $$\Rightarrow$$ x + 2 = 0    $$\Rightarrow$$ x = -2 $$\Rightarrow$$ x – 5 = 0 $$\Rightarrow$$ x = 5. (ii) $$2x^{2}$$ + x – 6 = 0 $$2x^{2}$$ + 4x – 3x – 6 = 0 . 2x(x+2) – 3 (x+2) = 0 (2x-3) (x+2) = 0 2x-3 = 0 or  x = $$\frac{3}{2}$$ X + 2 = 0 $$\Rightarrow$$   x = -2

1. John and Jivan together have 45 marbles .Both of them lost 5 marbles each and the product of number of marbles they have now is 124. Find out how many marbles did they start with.

Solution : Let us assume  , Number of marbles John has be ‘x’. Jivan has 45 – x . John = x – 5 Jivan = 45 –(x-5) = 40 – x. (x-5)(40-x) = 0 40x – 200 –

$$x^{2}$$ + 5x = 0 . 45x – $$x^{2}$$ – 200 = 0 . $$x^{2}$$ – 40 x – 5(x-40) = 0 x(x-40) – 5 (x-40) = 0 . (x-5) (x-40) = 0 . X = 5 0r  x = 40 .

2. John and Jivan together have 45 marbles .Both of them lost 5 marbles each and the product of number of marbles they have now is 124. Find out how many marbles did they start with.

Solution : Let us assume  , Number of marbles John has be ‘x’. Jivan has 45 – x . John = x – 5 Jivan = 45 –(x-5) = 40 – x. 40x – 200 –

$$x^{2}$$ + 5x = 124. 45x – $$x^{2}$$ – 200 = 124 $$x^{2}$$ – 45 x + 200 + 124 = 0    $$\Rightarrow  x^{2}$$ – 45 x + 324 = 0 $$x^{2}$$ -36 x – 9x + 324 = 0 $$\Rightarrow$$ x (x-36 ) – 9 (x-36) = 0 (x- 9 ) (x- 36 ) = 0  $$\Rightarrow$$   x = 9 . Condition (1) , x = 9 John x  i.e x = 9  , Jivan 45 –x i.e 45 -9 = 36 John = 9 marbles and Jivan = 36 marbles. Condition (ii)  , x = 36 John has 36 marbles , Jivan has 45 -36 = 9 marbles . John = 36 marbles , Jivan = 9 marbles. Problem :

• Find the two numbers whose sum is 27 and the product is 182.

Solution : Let us assume one number as ‘x’ , Other number is 27 – x . x(27-x)  = 182

$$\Rightarrow$$ 27 x – $$x ^{2}$$ = 182. $$x^{2}$$ – 27x +182 = 0 $$\Rightarrow$$   $$x^{2}$$ – 14x – 13x + 182 = 0. x(x-14) – 13 (x-14) = 0 (x-13)(x-14) = 0 $$\Rightarrow$$   x = 13 or x = 14 x = 13 27-13 = 14 . Two numbers  are 13 and 14 . Problem (2) : Find two consecutive positive numbers or integers , sum of whose squares is 365 . Solution :Let us assume one integer as  x and the other is x + 1 . $$(x)^{2} + (x+1)^{2}$$ = 365 $$x^{2} + x^{2}$$ +2x + 1 = 365 $$2x^{2}$$ + 2x + 1 – 365 = 0 $$\Rightarrow 2x^{2}$$ + 2x – 364 = 0. 2 $$[x^{2}+x – 182]$$ = 0 $$x^{2}$$ + x – 182 = 0 $$\Rightarrow  x^{2}$$ + 14 x -  13x -182 = 0 x(x + 14 ) -13 (x+14 ) = 0 (x – 13) (x + 14 ) = 0 . x – 13 = 0 , $$\Rightarrow$$ = 13 x+ 1 = 13 + 1 = 14 . x + 14 = 0 $$\Rightarrow$$   x = -14 . x +1 = -14 +1 = -13 therefore  Two numbers are  -13 and -14 . Problem (3 ) : The altitude of a right angle triangle  is 7cms less  than its base . If hypotenuse is 13 cms . Find the other two sides ? Solution : Let the base of the triangle be ‘x’ cms , Altitude  = (x-7) cms . Base $$^{2}$$ + Altitude $$^{2}$$ = Hyptoneuse $$^{2}$$ . $$(x) ^{2} + (x-7)^{2} = 13 ^{2}$$ $$x^{2} + x^{2}$$ – 14 x + 49 = 169 . $$x^{2} + x^{2}$$ -14x +49 -169 = 0 . $$2x^{2}$$ – 14 x – 120 = 0 $$2[x ^{2} – 7x – 60 ] = 0 \Rightarrow   x^{2}$$ – 7x – 60 = 0 . $$x^{2}$$ – 12 x + 5x -60 = 0 . x (x-12) + 5(x-12) = 0 x + 5 = 0  $$\Rightarrow$$ x = -5 x -12 = 0 $$\Rightarrow$$   x = 12 . One side of the triangle = 12  cms . Other side  (12-7) = 5 cms. Therefore the two sides of the triangle are 12 and 5 cms .  

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