Math, asked by neetusinghak2512, 1 month ago

Find roots of quadratic equation using completing square methods if roots exist.2x^2-7x-6=0​

Answers

Answered by nidhijain352
0

Answer:

the question is not clear enough

Answered by iamgourichakraborty
0

Answer:

Answer

a). 2x

2

−7x+3=0

⇒x

2

2

7

x=−

2

3

Adding (

4

7

)

2

on both sides

⇒x

2

2

7

x+(

4

7

)

2

=

2

−3

+(

4

7

)

2

⇒(x−

4

7

)

2

=

2

−3

+

16

49

⇒(x−

4

7

)

2

=

16

25

⇒(x−

4

7

)

2

=(

4

5

)

2

Taking square root on both sides

⇒(x−

4

7

)=±

4

5

⇒x−

4

7

=

4

5

, x−

4

7

=

4

−5

x=

4

5

+

4

7

x=

4

−5

+

4

7

x=3 x=

2

1

b). 2x

2

+x−4=0

x

2

+

2

x

=2

Adding (

4

1

)

2

on both sides

⇒x

2

+

2

x

+(

4

1

)

2

=2+(

4

1

)

2

(x+

4

1

)

2

=2+

16

1

(x+

4

1

)

2

=

16

33

Taking square root on both sides

⇒x+

4

1

4

33

⇒x=±

4

33

4

1

, x=

4

33

4

1

⇒x=±

4

33−1

, x=

4

33−1

c). 4x

2

+4

32

+3=0

⇒x

2

+

3

x+

4

3

=0

x

2

+

3

x=

4

−3

Adding (

2

3

)62 on both sides

⇒x

2

+

3

x+(

2

3

)

2

=

4

−3

+(

2

3

)

2

⇒(x+

2

3

)

2

=

4

−3

+

4

3

⇒(x+

2

3

)

2

=0

x=

2

3

,

2

3

same roots.

d). 2x

2

+x+4=0

⇒x

2

+

2

x

+2=0

x

2

+

2

x

=−2

Adding (

4

1

)

2

on both sides

⇒x

2

+

2

x

+(

4

1

)

2

=−2+(

4

1

)62

⇒(x+

4

1

)

2

=−2+

16

1

⇒(x+

4

1

)

2

=−

16

−31

Hence, solved.

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