Quadratic Equation

Find the roots of the following Quadratic Equations by factorization. (i) x 2 -3x-10=0 (ii) 2x 2 +x-6=0 (iii) √2 x 2 +7x+5 √2=0 (iv) 2x 2 -x+1/8=0 (v) 100x 2 -20x+1=0   Ans. (i) x 2 -3x-10=0 ⇒ x 2 -5x+2x-10=0   ⇒x(x-5)+2(x-5)=0 ⇒ ( x −5)( x  + 2)=0 ⇒(x-5)=0,(x+2)=0 ⇒ x  = 5, −2 (ii) 2x 2 +x-6=0 ⇒2x 2 +4x-3x-6=0   ⇒ 2 x  ( x  + 2) – 3 ( x  + 2) = 0 ⇒ (2 x  − 3) ( x  + 2) = 0 ⇒(2x-3)=0,(x+2)=0 ⇒ x  = 3/2,-2    (iii) √2 x 2 +7x+5 √2=0 ⇒ √2 x 2 +2x+5x+5 √2=0 ⇒ √2 x(x+ √2 )+5(x+ √2)=0 ⇒ (√2 x+5)(x+ √2)=0 ⇒ (√2 x+5)=0,(x+ √2)=0 ⇒ x=-5/√2,-√2 (iv) 2x 2 -x+1/8=0   ⇒(16x 2 -8x+1)/8=0                                   [L.C.M] ⇒16x 2 -8x+1=0 ⇒16x 2 -8x+1=0 ⇒16x 2 -4x-4x+1=0 ⇒ 4 x (4 x −1)–1(4 x −1)=0 ⇒ (4 x  − 1) (4 x  − 1)=0 ⇒ (4 x −1)=0,(4 x −1)=0   ⇒x=1/4 or 1/4 (v) 100x 2 -20x+1=0 ⇒ 100x 2 -10x-10x+1=0 ⇒ 10x(10x-1)-1(10x-1)=0 ⇒(10x-1)(10x-1)=0 ⇒(10x-1) or (10x-1)=0 ⇒x=1/10 or 1/10 2. Solve the followin

1. Check whether the following are Quadratic Equations. (i)  (x+1) 2 = 2 ( x  − 3) (ii)  x 2 -2x= (−2) (3 −  x ) (iii) ( x  − 2) ( x  + 1) = ( x  − 1) ( x  + 3) (iv) ( x  − 3) (2 x  + 1) =  x  ( x  + 5) (v) (2 x  − 1) ( x  − 3) = ( x  + 5) ( x  − 1) (vi)x 2 +3x+1=(x-2) 2 (vii)(x+2) 3 =2x(x 2 -1) (viii)x 3 -4x 2 -x+1=(x-2) 2   Ans. (i)(x+1) 2 =2(x-3)   {(a+b) 2 =a 2 +2ab+b 2 } ⇒  x 2 +1+2x=2x-6 ⇒  x 2 +7=0 Here, degree of equation = 2. Therefore, it is a Quadratic Equation . (ii)  x 2 -2x=(-2)(3-x) ⇒  x 2 -2x=(-2)(3-x) ⇒ x 2 -2x=-6+2x ⇒ x 2 -2x+6-2x=0 ⇒  x 2 -4x+6=0 Here, degree of equation is=2. Therefore, it is a Quadratic Equation. (iii)  ( x  − 2) ( x  + 1) = ( x  − 1) ( x  + 3) ⇒  x 2 +x-2x-2=x 2 +3x-x-3=0 ⇒  x 2 +x-2x-2-x 2 -3x+x+3=0 ⇒  x-2x-2-3x+x+3=0 ⇒  -3x+1=0 Here, degree of equation = 1. Therefore, it is not a Quadratic Equation. (iv)  ( x  − 3) (2 x  + 1) =  x  ( x  + 5) ⇒  2x 2 +x-6x-3=x 2 +5x ⇒  2x 2 +x-6x-3