Question

Que.2 find the value of (x) in the following:-
1.2(x+3)+3(x+1)=4(2x-3)+3​

Answer 1

Step-by-step explanation:

Main concept used: For real roots of quadratic equationax

2

+bx+c=0,b

2

−4ac>0

(iv)

(2x−3)

1

+

(x−5)

1

=1,x



=

2

3

,5

(2x−3)

1

+

(x−5)

1

=1,x



=

2

3

,5

⇒

(2x−3)(x−5)

(x−5)+(2x−3)

=1

⇒x−5+2x−3=2x

2

−3x−10x+15

⇒2x

2

−13x+15=3x−8

⇒2x

2

−13x+15−3x+8=0

⇒2x

2

−16x+23=0

Now, D=b

2

−4ac

=(−16)

2

−4(2)(23)(a=2,b=−16,c=23)

D=256−184=72>0

⇒

D

=

72

⇒

D

=6

2

Now, x=

2a

−b±

D

=

2×2

16±6

2

=

4

16

±

4

6

2

⇒x

1

=4+

2

3

2

and x

2

=4−

2

3

2

Hence, the root of the given quadratic equation are (4+

2

3

2

) and (4−

2

3

)

2

Answer 2

Answer:

2x+6+3x+3=8x-12+3

=2x+3x+6+3=8x-9

=5x+9=8x-9

=5x-8x=-9-9

= -3x= -18

=x= -18/-3

= x= 6