Question 1/3
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
किस समीकरण के वास्तविक मूल नहीं होंगे?
Which equation will have no real roots?
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x+5=0\)
Step 1
Concept
For the first equation, (D=32-4(1)(5)=-11<0). A negative discriminant gives no real roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x+5=0\). For the first equation, (D=32-4(1)(5)=-11<0). A negative discriminant gives no real roots.
Step 3
Exam Tip
पहले समीकरण में (D=32-4(1)(5)=-11<0) है। ऋणात्मक विविक्तकर वास्तविक मूल नहीं देता।
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Question 2/3
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
किस समीकरण के दो वास्तविक और समान मूल होंगे?
Which equation will have two real and equal roots?
Explanation opens after your attempt
Correct Answer
A. \(x^2+10x+25=0\)
Step 1
Concept
In the first equation, (D=102-4(1)(25)=0). A perfect square form often gives equal roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+10x+25=0\). In the first equation, (D=102-4(1)(25)=0). A perfect square form often gives equal roots.
Step 3
Exam Tip
पहले समीकरण में (D=102-4(1)(25)=0) है। पूर्ण वर्ग रूप अक्सर समान मूल देता है।
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Question 3/3
Easy Mathematics
Quadratic Equations Nature of Roots Class 10 Level 37
किस समीकरण के दो वास्तविक और असमान मूल होंगे?
Which equation will have two real and distinct roots?
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
For the first equation, (D=(-7)2-4(1)(10)=9>0). Hence its roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+10=0\). For the first equation, (D=(-7)2-4(1)(10)=9>0). Hence its roots are real and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=(-7)2-4(1)(10)=9>0) है। इसलिए उसके मूल वास्तविक और असमान हैं।
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