Find the roots of the following quadratic equations (if they exist) by the method of completing the square. √3x²+10x+7√3=0
Answers
SOLUTION :
Given : √3x² + 10x + 7√3 = 0
On dividing the whole equation by √3,
(x² + 10 x/√3 + 7) = 0
Shift the constant term on RHS
x² + 10 x/√3 = - 7
Add square of the ½ of the coefficient of x on both sides
On adding (½ of 10/√3)² = (10/2√3)² both sides
x² + 10 x/√3 + (10/2√3)² = - 7 + (10/2√3)²
Write the LHS in the form of perfect square
(x + 10/2√3)² = - 7 + 100/12
[a² + 2ab + b² = (a + b)²]
(x + 10/2√3)² = (- 7 × 12 + 100)/12
(x + 10/2√3)² = (- 84 + 100)/12
(x + 10/2√3)² = 16/12
On taking square root on both sides
(x + 10/2√3) = √(16/12)
(x + 10/2√3) = ± 4/√12
(x + 10/2√3) = ± 4/2√3
[√12 = √4 × 3 = 2√3]
On shifting constant term (10/2√3) to RHS
x = ± 4/2√3 - 10/2√3
x = 4/2√3 - 10/2√3
[Taking +ve sign]
x = (4 - 10)/2√3
x = -6/2√3 = - 3/√3
On rationalising the denominator
x = (-3 × √3) / (√3 ×√3)
x = -3√3/3
x = - √3
x = - 4/2√3 - 10/2√3
[Taking - ve sign]
x = (- 4 - 10)/2√3
x = - 14/2√3 = - 7/√3
x = - 7/√3
Hence, the roots of the given equation are - √3 & - 7/√3
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Solution:
Compare given Quadratic
equation √3x²+10x+7√3=0
with ax²+bx+c=0, we get
a = √3 , b = 10 , c = 7√3,
i ) Discriminant (D)
= b² - 4ac
= 10² - 4×√3×7√3
= 100 - 84
= 16
= 4²
D > 0
Therefore ,
Roots are real and distinct.
ii ) Finding roots by
completing square method:
√3x²+10x+7√3 = 0
Divide each term by √3 ,
we get
x² + (10/√3)x + 7 = 0
=> x² + 2•x•(5/√3)= -7
=> x²+2•x•(5/√3)+(5/√3)²=-7+(5/√3)²
=> (x+5/√3)² = -7 + 25/3
=> ( x + 5/√3 )² = (-21+25)/3
=> ( x + 5/√3 )² = 4/3
=> x + 5/√3 = ± (√4/√3 )
=> x = -5/√3 ± 2/√3
=> x = ( -5 ± 2 )/√3
Therefore ,
x = (-5+2)/√3 or x = (-5-2)/√3
=> x = -3/√3 Or x = -7/√3
=> x = -√3 Or x = -7/√3
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