Ncert Solutions Chapter 4 Quadratic Ncert Solutions Chapter 4 ...

Ncert Solutions Chapter 4 Quadratic Equations Exercise 4.4 Question 1

Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.

( i ) 2 x 2 − 3x + 5 = 0 ( ii ) 3 x 2 − 4 3x + 4 = 0 ( iii ) 2 x 2 − 6 x + 3 = 0 Solution ( i ) 2 x 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) − 4 ( 2 )( 5) = 9 − 40 = −31 2

Discriminant is less than 0 which means equation has no real roots.

Solution ( ii ) 3x 2 − 4 3x + 4 = 0

Comparing this equation with general equation ax 2 + bx + c = 0, we get a = 3, b = −4 3x and c = 4. .

(

Discriminant =b 2 − 4ac = −4 3

)

2

− 4 ( 3 )( 4 ) = 48 − 48 = 0

Discriminant is equal to zero which means equations has equal real roots.

Applying quadratic formula x = x=

−b ± b 2 − 4ac to find roots we get, 2a

4 3± 0 2 3 2 = = 6 3 3

Because, equation has two equal roots, it means x =

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2 2 , 3 3

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Solution ( iii ) 2 x 2 − 6 x + 3 = 0 Comparing equation with general equation ax 2 + bx + c = 0, we get a = 2, b = −6, and c = 3. Discriminant = b 2 − 4ac = ( −6 ) − 4 ( 2 )( 3) = 36 − 24 = 12 2

Value of discriminant is greater than zero. Therefore, equation has distinct and real roots.

Applying quadratic formula x = 6 ± 12 6 ± 2 3 3 ± 3 = = 4 4 2 3+ 3 3− 3 ⇒x= , 2 2 x=

−b ± b 2 − 4ac to find roots we get, 2a

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