Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+2 votes
9.1k views
in Algebra by (48.0k points)
closed by

Determine the nature of the roots for the following quadratic equations 

(i) 15.x2 + 11.x + 2 = 0 

(ii) x2 – x – 1 = 0 

(iii) √2t2 – 3t + 3√2 = 0 

(iv) 9y2 – 6√2y + 2 = 0 

(v) 9a2 b2 x2 – 24abcdx + 16c2 d2 = 0, a ≠ 0, b ≠ 0

by (50 points)
Bro i want ans for the questions i asked pls help me bro i am a slow learner

1 Answer

+2 votes
by (47.6k points)
selected by
 
Best answer

(i) 15x2 + 11x + 2 = 0 comparing with ax2 + bx + c = 0. 

Here a = 15, 6 = 11, c = 2. 

Δ = b2 – 4ac 

= 112 - 4 x 15 x 2 

= 121 – 120 

= 1 > 1. 

∴ The roots are real and unequal.

(ii) x2 – x – 1 = 0, 

Here a = 1, b = -1, c = -1.

Δ = b2 – 4ac 

= (-1)2 – 4 x 1 x -1 

= 1 + 4 = 5 > 0. 

∴ The roots are real and unequal.

(iii) √2t2 – 3t + 3√2 = 0 

Here a = √2, b = -3, c = 3√2

Δ = b2 – 4ac 

= (-3)– 4 x √2 x 3√2 

= 9 – 24 = -15 < 0. 

∴ The roots are not real.

(iv) 9y2 – 6√2y + 2 = 0 

a = 9, b = 6√2, c = 2 

Δ = b2 – 4ac = 

(6√2)2 – 4 x 9 x 2 

= 36 x 2 – 72 

= 72 – 72 = 0 

∴ The roots are real and equal.

(v) 9a2 b2 x2 – 24abcdx + 16c2 d2 = 0

Δ = b2 – 4ac 

= (-24abcd)2 – 4 x 9a2 b2 x 16c2 d

= 576a2 b2 c2 d2 – 576a2 b2 c2 d2 = 0 

∴ The roots are real and equal.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...