Question

Find the roots of the following quadratic (if they exist) by the method of completing the square.
3x²+11x+10=0

Answer 1

answer : roots of quadratic equation is -2 and -5/3

we have to find the roots of given quadratic equation by the method of completing the square.

3x² + 11x + 10 = 0

⇒3x² + 11x = -10

⇒x² + 11x/3 =- 10/3

⇒x² + 2.(11/6).x = -10/3

adding (11/6)² both sides,

⇒x² + 2.(11/6).x + (11/6)² = -10/3 + (11/6)²

⇒(x + 11/6)² = -10/3 + 121/36 = 1/36

⇒(x + 11/6)² = (1/6)²

taking square root both sides,

⇒x + 11/6 = ±1/6

⇒x = -11/6 ± 1/6

⇒x = -2, -5/3

also read similar questions :Find the roots of the following quadratic equations (if they exist) by the method of completing the square. 3x²+11x+10=0

https://brainly.in/question/8229890

$3 {x }^{2} + 11x + 10 = 0$

by completing square

https://brainly.in/question/3956204

Answer 2

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

SOLUTION :  

Given : 3x² + 11x + 10 = 0

On dividing the whole equation by 3,

(x² + 11x/3 + 10/3) = 0

Shift the constant term on RHS

x² + 11x/3  = - 10/3

Add square of the ½ of the coefficient of x on both sides

On adding (½ of 11/3)² = (11/6)² both sides

x² + 11x/3 + (11/6)² = - 10/3 + (11/6)²

Write the LHS in the form of perfect square

(x + 11/6)² = - 10/3 + 121/36

[a² + 2ab + b² = (a + b)²]

(x + 11/6)² = (-10 × 12 + 121)/36

(x + 11/6)² = (-120 + 121)/16

(x + 11/6)² = 1/36

On taking square root on both sides

(x + 11/6) = √(1/36)

(x + 11/6) = ± 1/6

On shifting constant term (11/6) to RHS

x = ± 1/6 - 11/6

x = 1/6 - 11/6

[Taking +ve sign]

x = (1 - 11)/6

x = - 10/6

x = - 5/3

x = - 1/6 - 11/6

[Taking - ve sign]

x = ( - 1 - 11)/6

x = - 12/6

x = - 2

Hence, the  roots of the given equation are - 2 & - 5/3.