Find the roots of the following quadratic equations, if they exist, by the method of completing the
square:
(iii) 4x^2+4 ✓3x+3=0
Answers
Answer:
x=-√3/2
Step-by-step explanation:
Firstly we would check if it has real roots or not by [b²-4ac]
So,
a=4, b=4√3,C=3
[4√3]²-4×4×3
[16×3] - 4×4×3
48-48
0
So, it have real and equal roots
Now by completing the square method
4x²+4√3x+3=0
By dividing 4
x²+√3x+3/4=0
x²+√3x+(√3/2)²-(√3/2)²+3/4=0
(x+√3/2)²-3/4+3/4=0
(x+√3/2)²=0
Now By Transferring the square
x+√3/2=0
x=-√3/2
By Checking that if it have real roots we also found that they are equal. So,
(x+√3/2)(x+√3/2) are the factors.
And X=-√3/2 Ans...
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.