\(\frac{x+6}{x}=\frac{49}{x+6}\), \(x\neq0,-6\), के हल क्या हैं?
What are the solutions of \(\frac{x+6}{x}=\frac{49}{x+6}\), \(x\neq0,-6\)?
#quadratic
#rational-equation
#solutions
A (x=1,36)
B (x=-1,-36)
C (x=6,49)
D (x=37,36)
Explanation opens after your attempt
Correct Answer
A. (x=1,36)
Step 1
Concept
(x-2 -37x+36=(x-1)(x-36)), so (x=1) and (x=36). In exams, check solutions against excluded denominator values.
Step 2
Why this answer is correct
The correct answer is A. (x=1,36). (x-2 -37x+36=(x-1)(x-36)), so (x=1) and (x=36). In exams, check solutions against excluded denominator values.
Step 3
Exam Tip
(x-2 -37x+36=(x-1)(x-36)), इसलिए (x=1) और (x=36) हैं। परीक्षा में हर के निषिद्ध मानों से हलों की जांच करें।
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\(\frac{1}{x}+x=\frac{50}{7}\), \(x\neq0\), के हल क्या हैं?
What are the solutions of \(\frac{1}{x}+x=\frac{50}{7}\), \(x\neq0\)?
#quadratic
#reciprocal-equation
#solutions
A \(x=7,\frac{1}{7}\)
B \(x=-7,-\frac{1}{7}\)
C (x=50,7)
D \(x=\frac{50}{7},1\)
Explanation opens after your attempt
Correct Answer
A. \(x=7,\frac{1}{7}\)
Step 1
Concept
(7x-2 -50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=7,\frac{1}{7}\). (7x-2 -50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(7x-2 -50x+7=(7x-1)(x-7)), इसलिए \(x=\frac{1}{7}\) और (7) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
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\(\frac{x+5}{x}=\frac{36}{x+5}\), \(x\neq0,-5\), के हल क्या हैं?
What are the solutions of \(\frac{x+5}{x}=\frac{36}{x+5}\), \(x\neq0,-5\)?
#quadratic
#rational-equation
#solutions
A (x=1,25)
B (x=-1,-25)
C (x=5,36)
D (x=26,25)
Explanation opens after your attempt
Correct Answer
A. (x=1,25)
Step 1
Concept
(x-2 -26x+25=(x-1)(x-25)), so (x=1) and (x=25). In exams, check solutions against excluded denominator values.
Step 2
Why this answer is correct
The correct answer is A. (x=1,25). (x-2 -26x+25=(x-1)(x-25)), so (x=1) and (x=25). In exams, check solutions against excluded denominator values.
Step 3
Exam Tip
(x-2 -26x+25=(x-1)(x-25)), इसलिए (x=1) और (x=25) हैं। परीक्षा में हर के निषिद्ध मानों से हलों की जांच करें।
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\(\frac{1}{x}+x=\frac{37}{6}\), \(x\neq0\), के हल क्या हैं?
What are the solutions of \(\frac{1}{x}+x=\frac{37}{6}\), \(x\neq0\)?
#quadratic
#reciprocal-equation
#solutions
A \(x=6,\frac{1}{6}\)
B \(x=-6,-\frac{1}{6}\)
C (x=37,6)
D \(x=\frac{37}{6},1\)
Explanation opens after your attempt
Correct Answer
A. \(x=6,\frac{1}{6}\)
Step 1
Concept
(6x-2 -37x+6=(6x-1)(x-6)), so \(x=\frac{1}{6}\) and (6). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=6,\frac{1}{6}\). (6x-2 -37x+6=(6x-1)(x-6)), so \(x=\frac{1}{6}\) and (6). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(6x-2 -37x+6=(6x-1)(x-6)), इसलिए \(x=\frac{1}{6}\) और (6) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
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\(\frac{x+4}{x}=\frac{25}{x+4}\), \(x\neq0,-4\), के हल क्या हैं?
What are the solutions of \(\frac{x+4}{x}=\frac{25}{x+4}\), \(x\neq0,-4\)?
#quadratic
#rational-equation
#solutions
A (x=1,16)
B (x=-1,-16)
C (x=4,25)
D (x=17,16)
Explanation opens after your attempt
Correct Answer
A. (x=1,16)
Step 1
Concept
(x-2 -17x+16=(x-1)(x-16)), so (x=1) and (x=16). In exams, check solutions against excluded denominator values.
Step 2
Why this answer is correct
The correct answer is A. (x=1,16). (x-2 -17x+16=(x-1)(x-16)), so (x=1) and (x=16). In exams, check solutions against excluded denominator values.
Step 3
Exam Tip
(x-2 -17x+16=(x-1)(x-16)), इसलिए (x=1) और (x=16) हैं। परीक्षा में हर के निषिद्ध मानों से हलों की जांच करें।
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\(\frac{1}{x}+x=\frac{26}{5}\), \(x\neq0\), के हल क्या हैं?
What are the solutions of \(\frac{1}{x}+x=\frac{26}{5}\), \(x\neq0\)?
#quadratic
#reciprocal-equation
#solutions
A \(x=5,\frac{1}{5}\)
B \(x=-5,-\frac{1}{5}\)
C (x=26,5)
D \(x=\frac{26}{5},1\)
Explanation opens after your attempt
Correct Answer
A. \(x=5,\frac{1}{5}\)
Step 1
Concept
(5x-2 -26x+5=(5x-1)(x-5)), so \(x=\frac{1}{5}\) and (5). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=5,\frac{1}{5}\). (5x-2 -26x+5=(5x-1)(x-5)), so \(x=\frac{1}{5}\) and (5). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(5x-2 -26x+5=(5x-1)(x-5)), इसलिए \(x=\frac{1}{5}\) और (5) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
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\(\frac{x+3}{x}=\frac{16}{x+3}\), \(x\neq0,-3\), के हल क्या हैं?
What are the solutions of \(\frac{x+3}{x}=\frac{16}{x+3}\), \(x\neq0,-3\)?
#quadratic
#rational-equation
#solutions
A (x=1,9)
B (x=-1,-9)
C (x=3,16)
D (x=10,9)
Explanation opens after your attempt
Correct Answer
A. (x=1,9)
Step 1
Concept
(x-2 -10x+9=(x-1)(x-9)), so (x=1) and (x=9). In exams, check solutions against excluded denominator values.
Step 2
Why this answer is correct
The correct answer is A. (x=1,9). (x-2 -10x+9=(x-1)(x-9)), so (x=1) and (x=9). In exams, check solutions against excluded denominator values.
Step 3
Exam Tip
(x-2 -10x+9=(x-1)(x-9)), इसलिए (x=1) और (x=9) हैं। परीक्षा में हर के निषिद्ध मानों से हलों की जांच करें।
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\(\frac{1}{x}+x=\frac{17}{4}\), \(x\neq0\), के हल क्या हैं?
What are the solutions of \(\frac{1}{x}+x=\frac{17}{4}\), \(x\neq0\)?
#quadratic
#reciprocal-equation
#solutions
A \(x=4,\frac{1}{4}\)
B \(x=-4,-\frac{1}{4}\)
C (x=17,4)
D \(x=\frac{17}{4},1\)
Explanation opens after your attempt
Correct Answer
A. \(x=4,\frac{1}{4}\)
Step 1
Concept
(4x-2 -17x+4=(4x-1)(x-4)), so \(x=\frac{1}{4}\) and (4). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=4,\frac{1}{4}\). (4x-2 -17x+4=(4x-1)(x-4)), so \(x=\frac{1}{4}\) and (4). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(4x-2 -17x+4=(4x-1)(x-4)), इसलिए \(x=\frac{1}{4}\) और (4) हैं। परीक्षा में मिले हल मूल समीकरण में मान्य हैं या नहीं जांचें।
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\(\frac{x+2}{x}=\frac{9}{x+2}\), \(x\neq0,-2\), के हल क्या हैं?
What are the solutions of \(\frac{x+2}{x}=\frac{9}{x+2}\), \(x\neq0,-2\)?
#quadratic
#rational-equation
#solutions
A (x=1,4)
B (x=-1,-4)
C (x=2,9)
D (x=5,4)
Explanation opens after your attempt
Correct Answer
A. (x=1,4)
Step 1
Concept
(x-2 -5x+4=(x-1)(x-4)), so (x=1) and (x=4). In exams, check solutions against excluded denominator values.
Step 2
Why this answer is correct
The correct answer is A. (x=1,4). (x-2 -5x+4=(x-1)(x-4)), so (x=1) and (x=4). In exams, check solutions against excluded denominator values.
Step 3
Exam Tip
(x-2 -5x+4=(x-1)(x-4)), इसलिए (x=1) और (x=4) हैं। परीक्षा में हर वाले निषिद्ध मानों से हलों की जांच करें।
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\(\frac{1}{x}+x=\frac{10}{3}\), \(x\neq0\), के हल क्या हैं?
What are the solutions of \(\frac{1}{x}+x=\frac{10}{3}\), \(x\neq0\)?
#quadratic
#reciprocal-equation
#solutions
A \(x=3,\frac{1}{3}\)
B \(x=-3,-\frac{1}{3}\)
C (x=10,3)
D \(x=\frac{10}{3},1\)
Explanation opens after your attempt
Correct Answer
A. \(x=3,\frac{1}{3}\)
Step 1
Concept
(3x-2 -10x+3=(3x-1)(x-3)), so \(x=\frac{1}{3}\) and (3). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=3,\frac{1}{3}\). (3x-2 -10x+3=(3x-1)(x-3)), so \(x=\frac{1}{3}\) and (3). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(3x-2 -10x+3=(3x-1)(x-3)), इसलिए \(x=\frac{1}{3}\) और (3) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
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\(\frac{1}{x}+x=\frac{5}{2}\), \(x\neq0\), के हल क्या हैं?
What are the solutions of \(\frac{1}{x}+x=\frac{5}{2}\), \(x\neq0\)?
#quadratic
#reciprocal-equation
#solutions
A \(x=2,\frac{1}{2}\)
B \(x=-2,-\frac{1}{2}\)
C (x=5,2)
D \(x=\frac{5}{2},1\)
Explanation opens after your attempt
Correct Answer
A. \(x=2,\frac{1}{2}\)
Step 1
Concept
(2x-2 -5x+2=(2x-1)(x-2)), so \(x=\frac{1}{2}\) and (2). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=2,\frac{1}{2}\). (2x-2 -5x+2=(2x-1)(x-2)), so \(x=\frac{1}{2}\) and (2). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(2x-2 -5x+2=(2x-1)(x-2)), इसलिए \(x=\frac{1}{2}\) और (2) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
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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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\(5x^2=80\) 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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\(3x^2=12\) 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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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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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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