Question 23 - CBSE Class 10 Sample Paper for 2018 Boards
Last updated at Sept. 14, 2018 by Teachoo

Transcript

Question 23
Check whether the equation 5x2 6x 2 = 0 has real roots and if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.
5x2 6x 2 = 0
Comparing equation with ax2 + bx + c = 0
a = 5 , b = 6, c = 2
We know that ,
D = b2 4ac
= ( 6)2 4 5 ( 2)
= 36 + 40
= 76
Since D > 0
Hence, there are two real roots
Finding roots by completing the square
5x2 6x 2 = 0
Dividing by 5
(5 2 6 2)/5=0/5
5 2/5 6 /5 2/5=0
x2 6 /5 2/5=0
We know that
(a b)2 = a2 2ab + b2
Here, a = x &
2ab = 6 /5
2xb = 6 /5
b = ( 6 )/5 1/( 2 )
b = 3/5
Now, in our equation
x2 6 /5 2/5=0
Adding and subtracting (3/5)^2
x2 6 /5 2/5+(3/5)^2 (3/5)^2=0
x2 6 /5+(3/5)^2 2/5 (3/5)^2=0
( 3/5)^2 2/5 (3/5)^2=0
( 3/5)^2 2/5 9/25=0
( 3/5)^2=2/5+9/25
( 3/5)^2=(2 5 + 9)/25
( 3/5)^2=(10 + 9)/25
( 3/5)^2=19/25
( 3/5)^2=( 19/5)^2
Cancelling square both sides
( 3/5)= ( 19/5)
x 3/5 = 19/5
x = 19/5+3/5
x = ( 19 + 3)/5
x 3/5 = ( 19)/5
x = ( 19)/5+3/5
x = ( 19 + 3)/5
Now, we will verify zeroes
Verifying x = ( + )/
5x2 6x 2 = 0
5(( 19 + 3)/5)^2 6(( 19 + 3)/5) 2=0
5((( 19 )^2+ 3^(2 )+2 3 19)/25) 6(( 19 + 3)/5) 2=0
(19 + 9 + 6 19)/5 ((6 19 +18)/5) 2=0
(28 + 6 19 6 19 18 2(5))/5=0
(28 18 10)/5=0
(28 28)/5=0
0/5=0
0 =0
Which is true
Hence, ( + )/ is a root of the equation
Verifying x = ( + )/
5x2 6x 2 = 0
5(( 19 + 3)/5)^2 6(( 19 + 3)/5) 2=0
5((( 19 )^2+ 3^(2 )+ 2 3 ( 19))/25) 6(( 19 + 3)/5) 2=0
(19 + 9 6 19)/5 (( 6 19 +18)/5) 2=0
(28 6 19 + 6 19 18 2(5))/5=0
(28 18 10)/5=0
(28 28)/5=0
0/5=0
0 =0
Which is true
Hence, ( + )/ is a root of the equation

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