Solve using formula y 2 + 1 3 y = 2

Given equation is y 2 + 1 3 y = 2 ⇒ y 2 + 1 3 y - 2 = 0 ⇒ 3 y 2 + y - 6 3 = 0 ⇒ 3 y 2 + y - 6 = 0 Comparing with a x 2 + b x + c = 0 , we get a = 3 , b = 1 , c = - 6 We know that y = - b ± b 2 - 4 a c 2 a y = - ( 1 ) ± ( 1 ) 2 - 4 . 3 . ( - 6 ) 2 . 3 y = - 1 ± 73 6 ∴ y = - 1 + 73 6 , - 1 - 73 6

Solve the equation x 2 + 2 3 x + 3 = 0 using the formula.

Given equation is x 2 + 2 3 x + 3 = 0 Comparing with a x 2 + b x + c = 0 , we get a = 1 , b = 2 3 , c = 3 D = b 2 - 4 a c = 2 3 2 - 4 × 1 × 3 = 12 - 12 = 0 We know that x = - b ± b 2 - 4 a c 2 a ⇒ x = - 2 3 ± 0 2 × 1 ⇒ x = - 2 3 2 ∴ x = - 3 , - 3

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Algebra Standard X>Chapter 2 - Quadratic Equations>Practice Set

Maharashtra Board Solutions for Chapter: Quadratic Equations, Exercise 4: Practice Set

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Quadratic Equations, Exercise 4: Practice Set

Attempt the practice questions on Chapter 2: Quadratic Equations, Exercise 4: Practice Set with hints and solutions to strengthen your understanding. Algebra Standard X solutions are prepared by Experienced Embibe Experts.