यदि किसी छात्र ने \(x^2=81\) से केवल (x=9) लिखा, तो सही सुधार क्या है?
If a student wrote only (x=9) from \(x^2=81\), what is the correct correction?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm9\) लिखना चाहिए / One should write \(x=\pm9\)
B (x=81) लिखना चाहिए / One should write (x=81)
C (x=-81) लिखना चाहिए / One should write (x=-81)
D \(x=\pm81\) लिखना चाहिए / One should write \(x=\pm81\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm9\) लिखना चाहिए / One should write \(x=\pm9\)
Step 1
Concept
From \(x^2=81\), \(x=\pm\sqrt{81}=\pm9\). In exams, both signs are necessary in the square root method.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm9\) लिखना चाहिए / One should write \(x=\pm9\). From \(x^2=81\), \(x=\pm\sqrt{81}=\pm9\). In exams, both signs are necessary in the square root method.
Step 3
Exam Tip
\(x^2=81\) से \(x=\pm\sqrt{81}=\pm9\) मिलता है। परीक्षा में वर्गमूल विधि में दोनों चिन्ह अनिवार्य हैं।
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((x-7)2 =11) को हल करने पर (x) का मान क्या होगा?
Solving ((x-7)2 =11), what will be the value of (x)?
#quadratic
#square-root-method
#irrational-roots
A \(x=7\pm\sqrt{11}\)
B \(x=-7\pm\sqrt{11}\)
C \(x=7\pm11\)
D \(x=\sqrt{7}\pm11\)
Explanation opens after your attempt
Correct Answer
A. \(x=7\pm\sqrt{11}\)
Step 1
Concept
\(x-7=\pm\sqrt{11}\), so \(x=7\pm\sqrt{11}\). In exams, write \(\pm\) with the whole square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=7\pm\sqrt{11}\). \(x-7=\pm\sqrt{11}\), so \(x=7\pm\sqrt{11}\). In exams, write \(\pm\) with the whole square root.
Step 3
Exam Tip
\(x-7=\pm\sqrt{11}\), इसलिए \(x=7\pm\sqrt{11}\) है। परीक्षा में \(\pm\) को पूरे वर्गमूल के साथ लिखें।
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\(7x^2=175\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(7x^2=175\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm5\)
B (x=5)
C (x=-5)
D \(x=\pm25\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm5\)
Step 1
Concept
First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm5\). First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=25\) मिलता है, इसलिए \(x=\pm5\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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यदि किसी छात्र ने \(x^2=49\) से केवल (x=7) लिखा, तो सही सुधार क्या है?
If a student wrote only (x=7) from \(x^2=49\), what is the correct correction?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm7\) लिखना चाहिए / One should write \(x=\pm7\)
B (x=49) लिखना चाहिए / One should write (x=49)
C (x=-49) लिखना चाहिए / One should write (x=-49)
D \(x=\pm49\) लिखना चाहिए / One should write \(x=\pm49\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm7\) लिखना चाहिए / One should write \(x=\pm7\)
Step 1
Concept
From \(x^2=49\), \(x=\pm\sqrt{49}=\pm7\). In exams, both signs are necessary in the square root method.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm7\) लिखना चाहिए / One should write \(x=\pm7\). From \(x^2=49\), \(x=\pm\sqrt{49}=\pm7\). In exams, both signs are necessary in the square root method.
Step 3
Exam Tip
\(x^2=49\) से \(x=\pm\sqrt{49}=\pm7\) मिलता है। परीक्षा में वर्गमूल विधि में दोनों चिन्ह अनिवार्य हैं।
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((x+6)2 =5) को हल करने पर (x) का मान क्या होगा?
Solving ((x+6)2 =5), what will be the value of (x)?
#quadratic
#square-root-method
#irrational-roots
A \(x=-6\pm\sqrt{5}\)
B \(x=6\pm\sqrt{5}\)
C \(x=-6\pm5\)
D \(x=\sqrt{6}\pm5\)
Explanation opens after your attempt
Correct Answer
A. \(x=-6\pm\sqrt{5}\)
Step 1
Concept
\(x+6=\pm\sqrt{5}\), so \(x=-6\pm\sqrt{5}\). In exams, write \(\pm\) with the whole square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=-6\pm\sqrt{5}\). \(x+6=\pm\sqrt{5}\), so \(x=-6\pm\sqrt{5}\). In exams, write \(\pm\) with the whole square root.
Step 3
Exam Tip
\(x+6=\pm\sqrt{5}\), इसलिए \(x=-6\pm\sqrt{5}\) है। परीक्षा में \(\pm\) को पूरे वर्गमूल के साथ लिखें।
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\(5x^2=80\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(5x^2=80\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm4\)
B (x=4)
C (x=-4)
D \(x=\pm16\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm4\)
Step 1
Concept
First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm4\). First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=16\) मिलता है, इसलिए \(x=\pm4\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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यदि किसी छात्र ने \(x^2=25\) से केवल (x=5) लिखा, तो सुधार क्या है?
If a student wrote only (x=5) from \(x^2=25\), what is the correction?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm5\) लिखना चाहिए / One should write \(x=\pm5\)
B (x=25) लिखना चाहिए / One should write (x=25)
C (x=-25) लिखना चाहिए / One should write (x=-25)
D \(x=\pm25\) लिखना चाहिए / One should write \(x=\pm25\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm5\) लिखना चाहिए / One should write \(x=\pm5\)
Step 1
Concept
From \(x^2=25\), \(x=\pm\sqrt{25}=\pm5\). In exams, both signs are necessary in the square root method.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm5\) लिखना चाहिए / One should write \(x=\pm5\). From \(x^2=25\), \(x=\pm\sqrt{25}=\pm5\). In exams, both signs are necessary in the square root method.
Step 3
Exam Tip
\(x^2=25\) से \(x=\pm\sqrt{25}=\pm5\) मिलता है। परीक्षा में वर्गमूल विधि में दोनों चिन्ह अनिवार्य हैं।
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((x-5)2 =3) को हल करने पर (x) का मान क्या होगा?
Solving ((x-5)2 =3), what will be the value of (x)?
#quadratic
#square-root-method
#irrational-roots
A \(x=5\pm\sqrt{3}\)
B \(x=-5\pm\sqrt{3}\)
C \(x=5\pm3\)
D \(x=\sqrt{5}\pm3\)
Explanation opens after your attempt
Correct Answer
A. \(x=5\pm\sqrt{3}\)
Step 1
Concept
\(x-5=\pm\sqrt{3}\), so \(x=5\pm\sqrt{3}\). In exams, write \(\pm\) with the whole square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=5\pm\sqrt{3}\). \(x-5=\pm\sqrt{3}\), so \(x=5\pm\sqrt{3}\). In exams, write \(\pm\) with the whole square root.
Step 3
Exam Tip
\(x-5=\pm\sqrt{3}\), इसलिए \(x=5\pm\sqrt{3}\) है। परीक्षा में \(\pm\) को पूरे वर्गमूल के साथ लिखें।
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\(3x^2=12\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(3x^2=12\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm2\)
B (x=2)
C (x=-2)
D \(x=\pm4\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm2\)
Step 1
Concept
First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm2\). First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=4\) मिलता है, इसलिए \(x=\pm2\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(12x^2=108\) के हल क्या हैं?
What are the solutions of \(12x^2=108\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm3\)
B (x=3)
C (x=-3)
D \(x=\pm9\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm3\)
Step 1
Concept
From \(12x^2=108\), \(x^2=9\), so \(x=\pm3\). In exams, write both values in the final answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm3\). From \(12x^2=108\), \(x^2=9\), so \(x=\pm3\). In exams, write both values in the final answer.
Step 3
Exam Tip
\(12x^2=108\) से \(x^2=9\), इसलिए \(x=\pm3\) है। परीक्षा में अंतिम उत्तर में दोनों मान लिखें।
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\(12x^2=108\) को हल करने का सही सरल रूप कौनसा है?
What is the correct simplified form to solve \(12x^2=108\)?
#quadratic
#simplification
#square-root-method
A \(x^2=9\)
B \(x^2=108\)
C \(x^2=12\)
D \(x^2=96\)
Explanation opens after your attempt
Correct Answer
A. \(x^2=9\)
Step 1
Concept
Dividing both sides by (12) gives \(x^2=9\). In exams, remove the coefficient first.
Step 2
Why this answer is correct
The correct answer is A. \(x^2=9\). Dividing both sides by (12) gives \(x^2=9\). In exams, remove the coefficient first.
Step 3
Exam Tip
दोनों पक्षों को (12) से भाग देने पर \(x^2=9\) मिलता है। परीक्षा में पहले गुणांक हटाएं।
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\(x^2+36=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?
What is the correct statement about the real roots of \(x^2+36=0\)?
#quadratic
#no-real-roots
#square-root-method
A कोई वास्तविक मूल नहीं / No real roots
B दो वास्तविक मूल (6) और (-6) / Two real roots (6) and (-6)
C एक वास्तविक मूल (36) / One real root (36)
D दो समान वास्तविक मूल (0) / Two equal real roots (0)
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक मूल नहीं / No real roots
Step 1
Concept
\(x^2=-36\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. \(x^2=-36\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).
Step 3
Exam Tip
\(x^2=-36\) वास्तविक संख्याओं में संभव नहीं है। परीक्षा में \(x^2\) का मान वास्तविक (x) के लिए ऋणात्मक नहीं होता।
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((x-4)2 =25) को हल करने पर कौनसे मान मिलते हैं?
Solving ((x-4)2 =25) gives which values?
#quadratic
#square-root-method
#roots
A (x=9,-1)
B (x=5,-5)
C (x=4,25)
D (x=1,9)
Explanation opens after your attempt
Correct Answer
A. (x=9,-1)
Step 1
Concept
\(x-4=\pm5\), so (x=9) or (x=-1). In exams, write both \(\pm\) cases.
Step 2
Why this answer is correct
The correct answer is A. (x=9,-1). \(x-4=\pm5\), so (x=9) or (x=-1). In exams, write both \(\pm\) cases.
Step 3
Exam Tip
\(x-4=\pm5\), इसलिए (x=9) या (x=-1) है। परीक्षा में दोनों \(\pm\) स्थितियां लिखें।
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\(x^2=169\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=169\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm13\)
B (x=13)
C (x=-13)
D \(x=\pm169\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm13\)
Step 1
Concept
\(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm13\). \(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{169}=\pm13\) होता है। परीक्षा में केवल (13) लिखना अधूरा उत्तर है।
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\(10x^2-90=0\) को पहले सरल करने पर क्या मिलेगा?
What will be obtained first after simplifying \(10x^2-90=0\)?
#quadratic
#simplification
#square-root-method
A \(x^2-9=0\)
B \(x^2+9=0\)
C \(10x^2=0\)
D \(x^2-90=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-9=0\)
Step 1
Concept
Dividing both sides by (10) gives \(x^2-9=0\). In exams, simplifying the equation first saves time.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-9=0\). Dividing both sides by (10) gives \(x^2-9=0\). In exams, simplifying the equation first saves time.
Step 3
Exam Tip
दोनों पक्षों को (10) से भाग देने पर \(x^2-9=0\) मिलता है। परीक्षा में पहले समीकरण को सरल करना समय बचाता है।
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वर्गमूल विधि से \(x^2=144\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=144\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm12\)
B (x=12)
C (x=-12)
D \(x=\pm72\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm12\)
Step 1
Concept
\(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm12\). \(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{144}=\pm12\) होता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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\(8x^2=72\) के हल क्या हैं?
What are the solutions of \(8x^2=72\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm3\)
B (x=3)
C (x=-3)
D \(x=\pm9\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm3\)
Step 1
Concept
From \(8x^2=72\), \(x^2=9\), so \(x=\pm3\). In exams, write both values in the final answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm3\). From \(8x^2=72\), \(x^2=9\), so \(x=\pm3\). In exams, write both values in the final answer.
Step 3
Exam Tip
\(8x^2=72\) से \(x^2=9\), इसलिए \(x=\pm3\) है। परीक्षा में अंतिम उत्तर में दोनों मान लिखें।
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\(8x^2=72\) को हल करने का सही सरल रूप कौनसा है?
What is the correct simplified form to solve \(8x^2=72\)?
#quadratic
#simplification
#square-root-method
A \(x^2=9\)
B \(x^2=72\)
C \(x^2=64\)
D \(x^2=8\)
Explanation opens after your attempt
Correct Answer
A. \(x^2=9\)
Step 1
Concept
Dividing both sides by (8) gives \(x^2=9\). In exams, remove the coefficient first.
Step 2
Why this answer is correct
The correct answer is A. \(x^2=9\). Dividing both sides by (8) gives \(x^2=9\). In exams, remove the coefficient first.
Step 3
Exam Tip
दोनों पक्षों को (8) से भाग देने पर \(x^2=9\) मिलता है। परीक्षा में पहले गुणांक हटाएं।
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\(x^2+25=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?
What is the correct statement about the real roots of \(x^2+25=0\)?
#quadratic
#no-real-roots
#square-root-method
A कोई वास्तविक मूल नहीं / No real roots
B दो वास्तविक मूल (5) और (-5) / Two real roots (5) and (-5)
C एक वास्तविक मूल (25) / One real root (25)
D दो समान वास्तविक मूल (0) / Two equal real roots (0)
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक मूल नहीं / No real roots
Step 1
Concept
\(x^2=-25\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. \(x^2=-25\) is not possible in real numbers. In exams, \(x^2\) is not negative for real (x).
Step 3
Exam Tip
\(x^2=-25\) वास्तविक संख्याओं में संभव नहीं है। परीक्षा में \(x^2\) का मान वास्तविक (x) के लिए ऋणात्मक नहीं होता।
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((x+3)2 =16) को हल करने पर कौनसे मान मिलते हैं?
Solving ((x+3)2 =16) gives which values?
#quadratic
#square-root-method
#roots
A (x=1,-7)
B (x=4,-4)
C (x=7,-1)
D (x=3,16)
Explanation opens after your attempt
Correct Answer
A. (x=1,-7)
Step 1
Concept
\(x+3=\pm4\), so (x=1) or (x=-7). In exams, write both \(\pm\) cases.
Step 2
Why this answer is correct
The correct answer is A. (x=1,-7). \(x+3=\pm4\), so (x=1) or (x=-7). In exams, write both \(\pm\) cases.
Step 3
Exam Tip
\(x+3=\pm4\), इसलिए (x=1) या (x=-7) है। परीक्षा में दोनों \(\pm\) स्थितियां लिखें।
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\(x^2=121\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=121\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm11\)
B (x=11)
C (x=-11)
D \(x=\pm121\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm11\)
Step 1
Concept
\(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm11\). \(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{121}=\pm11\) होता है। परीक्षा में केवल (11) लिखना अधूरा उत्तर है।
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\(6x^2-24=0\) को पहले सरल करने पर क्या मिलेगा?
What will be obtained first after simplifying \(6x^2-24=0\)?
#quadratic
#simplification
#square-root-method
A \(x^2-4=0\)
B \(x^2+4=0\)
C \(6x^2=0\)
D \(x^2-24=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4=0\)
Step 1
Concept
Dividing both sides by (6) gives \(x^2-4=0\). In exams, simplify the equation first.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4=0\). Dividing both sides by (6) gives \(x^2-4=0\). In exams, simplify the equation first.
Step 3
Exam Tip
दोनों पक्षों को (6) से भाग देने पर \(x^2-4=0\) मिलता है। परीक्षा में पहले समीकरण सरल करें।
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वर्गमूल विधि से \(x^2=64\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=64\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm8\)
B (x=8)
C (x=-8)
D \(x=\pm32\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm8\)
Step 1
Concept
\(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm8\). \(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{64}=\pm8\) होता है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(5x^2=20\) के हल क्या हैं?
What are the solutions of \(5x^2=20\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm2\)
B (x=2)
C (x=-2)
D \(x=\pm4\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm2\)
Step 1
Concept
From \(5x^2=20\), \(x^2=4\), so \(x=\pm2\). In exams, write both values in the final answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm2\). From \(5x^2=20\), \(x^2=4\), so \(x=\pm2\). In exams, write both values in the final answer.
Step 3
Exam Tip
\(5x^2=20\) से \(x^2=4\), इसलिए \(x=\pm2\) है। परीक्षा में अंतिम उत्तर में दोनों मान लिखें।
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\(5x^2=20\) को हल करने का सही सरल रूप कौनसा है?
What is the correct simplified form to solve \(5x^2=20\)?
#quadratic
#simplification
#square-root-method
A \(x^2=4\)
B \(x^2=20\)
C \(x^2=25\)
D \(x^2=100\)
Explanation opens after your attempt
Correct Answer
A. \(x^2=4\)
Step 1
Concept
Dividing both sides by (5) gives \(x^2=4\). In exams, remove the coefficient first and then take square root.
Step 2
Why this answer is correct
The correct answer is A. \(x^2=4\). Dividing both sides by (5) gives \(x^2=4\). In exams, remove the coefficient first and then take square root.
Step 3
Exam Tip
दोनों पक्षों को (5) से भाग देने पर \(x^2=4\) मिलता है। परीक्षा में पहले गुणांक हटाकर वर्गमूल लें।
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\(x^2+9=0\) के वास्तविक मूलों के बारे में सही कथन क्या है?
What is the correct statement about the real roots of \(x^2+9=0\)?
#quadratic
#no-real-roots
#square-root-method
A कोई वास्तविक मूल नहीं / No real roots
B दो वास्तविक मूल (3) और (-3) / Two real roots (3) and (-3)
C एक वास्तविक मूल (9) / One real root (9)
D दो समान वास्तविक मूल (0) / Two equal real roots (0)
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक मूल नहीं / No real roots
Step 1
Concept
\(x^2=-9\) gives no solution in real numbers. In exams, remember that \(x^2\) cannot be negative for real (x).
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. \(x^2=-9\) gives no solution in real numbers. In exams, remember that \(x^2\) cannot be negative for real (x).
Step 3
Exam Tip
\(x^2=-9\) से वास्तविक संख्या में हल नहीं मिलता। परीक्षा में \(x^2\) ऋणात्मक नहीं हो सकता यह याद रखें।
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((x-2)2 =9) को हल करने पर कौनसे मान मिलते हैं?
Solving ((x-2)2 =9) gives which values?
#quadratic
#square-root-method
#roots
A (x=5,-1)
B (x=2,9)
C (x=3,-3)
D (x=5,1)
Explanation opens after your attempt
Correct Answer
A. (x=5,-1)
Step 1
Concept
\(x-2=\pm3\), so (x=5) or (x=-1). In exams, write both \(\pm\) cases.
Step 2
Why this answer is correct
The correct answer is A. (x=5,-1). \(x-2=\pm3\), so (x=5) or (x=-1). In exams, write both \(\pm\) cases.
Step 3
Exam Tip
\(x-2=\pm3\), इसलिए (x=5) या (x=-1) है। परीक्षा में \(\pm\) के दोनों केस लिखें।
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किस समीकरण को वर्गमूल विधि से सीधे हल किया जा सकता है?
Which equation can be solved directly by square root method?
#quadratic
#square-root-method
#method-selection
A ((x-2)2 =9)
B \(x^2+5x+6=0\)
C \(x^2+2x+1=0\)
D \(2x^2+3x+1=0\)
Explanation opens after your attempt
Correct Answer
A. ((x-2)2 =9)
Step 1
Concept
\(In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\)expression\()^2=k).\)
Step 2
Why this answer is correct
\(The correct answer is A. ((x-2)^2=9). In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\)expression\()^2=k).\)
Step 3
Exam Tip
((x-2)2 =9) में सीधे वर्गमूल लिया जा सकता है। परीक्षा में ((expression\()^2=k) रूप को पहचानें\)।
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\(x^2=49\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=49\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm7\)
B (x=7)
C (x=-7)
D \(x=\pm49\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm7\)
Step 1
Concept
\(x=\pm\sqrt{49}=\pm7\). In exams, writing only the positive root is a common mistake.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm7\). \(x=\pm\sqrt{49}=\pm7\). In exams, writing only the positive root is a common mistake.
Step 3
Exam Tip
\(x=\pm\sqrt{49}=\pm7\) होता है। परीक्षा में केवल धनात्मक मूल लिखना सामान्य गलती है।
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\(2x^2-8=0\) को पहले सरल करने पर क्या मिलेगा?
What do we get first after simplifying \(2x^2-8=0\)?
#quadratic
#simplification
#square-root-method
A \(x^2-4=0\)
B \(x^2+4=0\)
C \(2x^2=0\)
D \(x^2-8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4=0\)
Step 1
Concept
Dividing both sides by (2) gives \(x^2-4=0\). In exams, simplifying the equation first saves time.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4=0\). Dividing both sides by (2) gives \(x^2-4=0\). In exams, simplifying the equation first saves time.
Step 3
Exam Tip
दोनों पक्षों को (2) से भाग देने पर \(x^2-4=0\) मिलता है। परीक्षा में समीकरण को पहले सरल करना समय बचाता है।
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