2x² + x +4= 0
find the roots
Answers
Answer:
2x² + x - 4 = 0,
=> x^2 + \frac{x}{2} - 2 = 0x
2
+
2
x
−2=0 (Dividing both the sides by \frac{1}{2}
2
1
),
=> x^2 + 2( \frac{1}{4} ) x - 2 = 0x
2
+2(
4
1
)x−2=0
(Multiplying & dividing the middle term by 2 to get the term 2ab (here a = x, b = \frac{1}{2}b=
2
1
))
So, we have the form,
a² + 2ab + c (for some value c, here it is - 4), we need b² to bring this equation to that form,
By adding ( \frac{1}{4}) ^2(
4
1
)
2
both the sides ,
we get,
=> x^2 + 2\frac{1}{4}x + ( \frac{1}{4} ) ^2 - 2 = ( \frac{1}{4} )^2x
2
+2
4
1
x+(
4
1
)
2
−2=(
4
1
)
2
It can be, reduced to the form (a + b)²
=> (x + \frac{1}{4} )^2 - 2 = \frac{1}{16}(x+
4
1
)
2
−2=
16
1
=> (x + \frac{1}{4} )^2 = \frac{1}{16} + 2(x+
4
1
)
2
=
16
1
+2
=> (x + \frac{1}{4} )^2 = \frac{1+32}{16}(x+
4
1
)
2
=
16
1+32
=> (x + \frac{1}{4} )^2 = \frac{33}{16}(x+
4
1
)
2
=
16
33
By moving the square from LHS to RHS, we get,
=> (x + \frac{1}{4} ) = \sqrt{ \frac{33}{16} }(x+
4
1
)=
16
33
=> (x + \frac{1}{4} ) = \frac{ \sqrt{33} }{4}(x+
4
1
)=
4
33
(or) (x + \frac{1}{4} ) = \frac{ - \sqrt{33} }{4}(x+
4
1
)=
4
−
33
=> x = \frac{ \sqrt{33} }{4} - \frac{1}{4}x=
4
33
−
4
1
(or) x = \frac{ - \sqrt{33} }{4} - \frac{1}{4}x=
4
−
33
−
4
1
=> x = \frac{ \sqrt{33} - 1 }{4}x=
4
33
−1
(or) x = \frac{ -\sqrt{33} - 1}{4}x=
4
−
33
−1