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NCERT-class-10th - Quadratic equation - Exercise 4.4

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Sum No.1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them. (i) 2x 2 -3x+5=0  (ii) 3x 2 -4 √ 3x +4=0 (iii) 2x 2 -6x+3=0      Ans. (i) 2x 2 -3x+5=0  Comparing this equation with general equation ax 2 +bx+c=0, We get a=2, b=-3 and c=5 Discriminant = b 2 -4ac=(-3) 2 -4*2*5   =9-40=-31 Discriminant is less than 0 which means equation has no real roots.   (ii) 3x 2 -4 √ 3x +4=0 Comparing this equation with general equation ax 2 +bx+c=0, We get a=3, b=-4 √ 3 and c=4 Discriminant = b 2 -4ac=(-4 √ 3) 2 -4*3*4 =48-48=0 Discriminant is equal to 0 which means equation has equal real roots. Applying quadratic   x= [-b 士 √b 2 -4ac]/ 2a to find roots ⇒ x= [ 4 √ 3 士 √0 ]/ 2*3  =( 2 √ 3)  /3 Because, equation has two equal roots, it means  x= (2 √ 3)/3 , (2 √ 3)/3   (iii) 2x 2 -6x+3=0   Comparing this equation with general equation ax 2 +bx+c=0, We get a=2, b=-6 and c=3 Discriminant=b 2 -4ac=(-6) 2 -4*2*3  =
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Class 10th-Quadratic Equation- Exercise 4.3.Full notes

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1. Find the roots of the following quadratic equations if they exist by the method of completing square. (i) 2x 2 -7x+3=0 (ii) 2x 2 +x-4=0 (iii)4x 2 +4 √ 3x+3=0 (iv) 2x 2 +x+4=0  Ans. (i) 2x 2 -7x+3=0 First we divide equation by 2 to make co-efficient of  x 2  equal to 1, x 2 -(7/2)x+3/2=0 Divide middle term of the equation by 2 x , we get (7/2)x*(1/2x)=7/4  Add and subtract square of 7/4  from the equation  x 2 -(7/2)x+3/2=0 x 2 -(7/2)x+3/2+(7/4) 2 -(7/4) 2 =0 x 2 +(7/4) 2 -7/2 x+3/2+(-7/4) 2 =0 {(a-b) 2 =a 2 +b 2 -2ab} (x-7/4) 2 -(24-49)/16=0 (x-7/4) 2 =(49-24)/16 Taking Square root on both sides, x-7/4= 士 5/4 X=5/4+7/4=12/4=3  or  x=-5/4+7/4=2/4=1/2 Therefore, x=1/2 or 3   (ii) 2x 2 +x-4=0 First we divide equation by 2 to make co-efficient of  x 2  equal to 1, x 2 +x/2-2=0 Divide middle term of the equation by 2 x , we get x/2*1/2x=1/4 Add and subtract square of 1/4  from the equation  x 2 +x/2-2=0 x 2 -x/2-2+(1/4) 2 -(1/4) 2 =0
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