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Introduction to Quadratic Equations- 1

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 ax^{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 .

  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. 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 .