√3x²+x-4√3=0 solve the quadratic equations
Answers
Answer:
3x
2
−4
3
x+4=0
a=3,b=−4
3
,c=4
Discriminant D=b
2
−4ac
=(−4
3
)
2
−4(3)(4)=48−48=0
⇒D=0
⇒ Two roots are equal.
The roots are
=
2a
−b±
D
=
2×3
4
3
±0
=
3
2
,
3
2
Hence, the roots are
3
2
and
3
2
hi dear
Divide both sides of the equation by 3 to have 1 as the coefficient of the first term :
x2-(1/3)x-(4/3) = 0
Add 4/3 to both side of the equation :
x2-(1/3)x = 4/3
Now the clever bit: Take the coefficient of x , which is 1/3 , divide by two, giving 1/6 , and finally square it giving 1/36
Add 1/36 to both sides of the equation :
On the right hand side we have :
4/3 + 1/36 The common denominator of the two fractions is 36 Adding (48/36)+(1/36) gives 49/36
So adding to both sides we finally get :
x2-(1/3)x+(1/36) = 49/36
Adding 1/36 has completed the left hand side into a perfect square :
x2-(1/3)x+(1/36) =
(x-(1/6)) • (x-(1/6)) =
(x-(1/6))2
Things which are equal to the same thing are also equal to one another. Since
x2-(1/3)x+(1/36) = 49/36 and
x2-(1/3)x+(1/36) = (x-(1/6))2
then, according to the law of transitivity,
(x-(1/6))2 = 49/36
We'll refer to this Equation as Eq. #4.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-(1/6))2 is
(x-(1/6))2/2 =
(x-(1/6))1 =
x-(1/6)
Now, applying the Square Root Principle to Eq. #4.2.1 we get:
x-(1/6) = √ 49/36
Add 1/6 to both sides to obtain:
x = 1/6 + √ 49/36
Since a square root has two values, one positive and the other negative
x2 - (1/3)x - (4/3) = 0
has two solutions:
x = 1/6 + √ 49/36
or
x = 1/6 - √ 49/36
Note that √ 49/36 can be written as
√ 49 / √ 36 which is 7 / 6
hopes its helps you