Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
683 views
in Quadratic Equations by (115k points)
closed by
If (a2 + b2) x2 + 2(ac + bd) x + c2 + d2 = 0 has no real roots then
1. ad = bc
2. ab = cd
3. ac = bd
4. ad ≠ bc

1 Answer

0 votes
by (113k points)
selected by
 
Best answer
Correct Answer - Option 4 : ad ≠ bc

Concept used:

We know, for no real roots, \(D = b^2 - 4ac < 0\)

Calculation:

Given equation is (a2 + b2) x2 + 2(ab + bd) x + c2 + d2 = 0

Here, a = (a2 + b2), b = 2(ab + bd), c = c2 + d2

\([(ac + bd)^2] - 4(a^2 + b^2) (c^2 + d^2) < 0\)

\(\Rightarrow [4(a^2 c^2 + b^2 d^2 + 2abcd)] - 4 (a^2 + b^2) (c^2 + d^2) < 0\)

\(\Rightarrow 4[(a^2 c^2 + b^2 d^2 + 2abcd) - (a^2 c^2 + a^2 d^2 + b^2 c^2 +b^2)] < 0\)

\(\Rightarrow 4[a^2 c^2 + b^2 d^2 + 2abcd - a^2 c^2 - a^2 d^2 - b^2 c^2 - b^2] < 0\)

\(\Rightarrow 4 [2abcd - b^2 c^2 - a^2 d^2] < 0\)

\(\Rightarrow -4 [a^2 d^2 + b^2 c^2 -2 abcd] < 0\)

\(\Rightarrow -4 [ad - bc]^2 < 0\)

Therefore,

(ad - bc) < 0

or ad < bc

∴ ad ≠ bc

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.

...