(9x-2 -9(r+s)x+9rs=0) को सरल करने के बाद मूल क्या होंगे?
After simplifying (9x-2 -9(r+s)x+9rs=0), what will be the roots?
#quadratic
#simplification
#symbolic
A (x=r,s)
B (x=-r,-s)
C (x=9r,9s)
D (x=r+s,rs)
Explanation opens after your attempt
Correct Answer
A. (x=r,s)
Step 1
Concept
Dividing the whole equation by (9) gives (x-2 -(r+s)x+rs=0). In exams, removing the common factor first makes solving easier.
Step 2
Why this answer is correct
The correct answer is A. (x=r,s). Dividing the whole equation by (9) gives (x-2 -(r+s)x+rs=0). In exams, removing the common factor first makes solving easier.
Step 3
Exam Tip
पूरे समीकरण को (9) से भाग देने पर (x-2 -(r+s)x+rs=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाने से हल आसान होता है।
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यदि \(x^2+px+9=0\) का एक मूल दूसरे का दुगुना है और दोनों ऋणात्मक हैं, तो (p) का सही सरल मान क्या है?
If one root of \(x^2+px+9=0\) is double the other and both are negative, what is the correct simplified value of (p)?
#quadratic
#roots-relation
#simplification
A \( \frac{9\sqrt{2}}{2}\)
B \(3\sqrt{2}\)
C \(9\sqrt{2}\)
D \( \frac{3\sqrt{2}}{2}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{9\sqrt{2}}{2}\)
Step 1
Concept
\(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{9\sqrt{2}}{2}\). \(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
\(\frac{9}{\sqrt{2}}\) को सरल करने पर \(\frac{9\sqrt{2}}{2}\) मिलता है। परीक्षा में हर को परिमेय बनाना न भूलें।
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(4x-2 -4(a+b)x+4ab=0) को सरल करने के बाद मूल क्या होंगे?
After simplifying (4x-2 -4(a+b)x+4ab=0), what will be the roots?
#quadratic
#simplification
#symbolic
A (x=a,b)
B (x=-a,-b)
C (x=4a,4b)
D (x=a+b,ab)
Explanation opens after your attempt
Correct Answer
A. (x=a,b)
Step 1
Concept
Dividing the whole equation by (4) gives (x-2 -(a+b)x+ab=0). In exams, removing the common factor first is easier.
Step 2
Why this answer is correct
The correct answer is A. (x=a,b). Dividing the whole equation by (4) gives (x-2 -(a+b)x+ab=0). In exams, removing the common factor first is easier.
Step 3
Exam Tip
पूरे समीकरण को (4) से भाग देने पर (x-2 -(a+b)x+ab=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाना आसान रहता है।
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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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\(10x^2-90=0\) 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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\(8x^2=72\) 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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\(6x^2-24=0\) 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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\(5x^2=20\) 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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\(2x^2-8=0\) 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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समीकरण (3(x+1)2 +2(x-4)2 =74) का मानक रूप कौन-सा है?
What is the standard form of (3(x+1)2 +2(x-4)2 =74)?
#quadratic-equations
#identity
#simplification
#expert
A \(5x^2-10x-39=0\)
B \(5x^2+10x-39=0\)
C \(5x^2-10x+39=0\)
D \(5x^2-16x-74=0\)
Explanation opens after your attempt
Correct Answer
A. \(5x^2-10x-39=0\)
Step 1
Concept
Expanding gives \(3x^2+6x+3+2x^2-16x+32=74\). Simplifying gives \(5x^2-10x-39=0\).
Step 2
Why this answer is correct
The correct answer is A. \(5x^2-10x-39=0\). Expanding gives \(3x^2+6x+3+2x^2-16x+32=74\). Simplifying gives \(5x^2-10x-39=0\).
Step 3
Exam Tip
विस्तार करने पर \(3x^2+6x+3+2x^2-16x+32=74\) मिलता है। सरल करने पर \(5x^2-10x-39=0\) सही है।
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समीकरण (2(x-1)2 +3(x+2)2 =65) का मानक रूप कौन-सा है?
What is the standard form of (2(x-1)2 +3(x+2)2 =65)?
#quadratic-equations
#identity
#simplification
#expert
A \(5x^2+8x-51=0\)
B \(5x^2-8x-51=0\)
C \(5x^2+8x+51=0\)
D \(5x^2+10x-65=0\)
Explanation opens after your attempt
Correct Answer
A. \(5x^2+8x-51=0\)
Step 1
Concept
Expanding gives \(2x^2-4x+2+3x^2+12x+12=65\). Simplifying gives \(5x^2+8x-51=0\).
Step 2
Why this answer is correct
The correct answer is A. \(5x^2+8x-51=0\). Expanding gives \(2x^2-4x+2+3x^2+12x+12=65\). Simplifying gives \(5x^2+8x-51=0\).
Step 3
Exam Tip
विस्तार करने पर \(2x^2-4x+2+3x^2+12x+12=65\) मिलता है। सरल करने पर \(5x^2+8x-51=0\) सही है।
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समीकरण ((x+2)2 +2(x-3)2 =35) का मानक रूप कौन-सा है?
What is the standard form of ((x+2)2 +2(x-3)2 =35)?
#quadratic-equations
#identity
#simplification
#expert
A \(3x^2-8x-13=0\)
B \(3x^2+8x-13=0\)
C \(3x^2-8x+13=0\)
D \(2x^2-8x-13=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-8x-13=0\)
Step 1
Concept
Expanding gives \(x^2+4x+4+2x^2-12x+18=35\). Hence \(3x^2-8x-13=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-8x-13=0\). Expanding gives \(x^2+4x+4+2x^2-12x+18=35\). Hence \(3x^2-8x-13=0\) is correct.
Step 3
Exam Tip
विस्तार करने पर \(x^2+4x+4+2x^2-12x+18=35\) मिलता है। इसलिए \(3x^2-8x-13=0\) सही है।
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समीकरण ((2x-3)2 +(x+5)2 =34) का मानक रूप कौन-सा है?
What is the standard form of ((2x-3)2 +(x+5)2 =34)?
#quadratic-equations
#identity
#simplification
#hard
A \(5x^2-2x=0\)
B \(5x^2+2x=0\)
C \(5x^2-2x+34=0\)
D \(3x^2-2x=0\)
Explanation opens after your attempt
Correct Answer
A. \(5x^2-2x=0\)
Step 1
Concept
Expanding gives \(4x^2-12x+9+x^2+10x+25=34\). Simplifying gives \(5x^2-2x=0\).
Step 2
Why this answer is correct
The correct answer is A. \(5x^2-2x=0\). Expanding gives \(4x^2-12x+9+x^2+10x+25=34\). Simplifying gives \(5x^2-2x=0\).
Step 3
Exam Tip
विस्तार करने पर \(4x^2-12x+9+x^2+10x+25=34\) मिलता है। सरल करने पर \(5x^2-2x=0\) बनता है।
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समीकरण ((x-2)2 +(2x+1)2 =25) का मानक रूप कौन-सा है?
What is the standard form of ((x-2)2 +(2x+1)2 =25)?
#quadratic-equations
#identity
#simplification
#hard
A \(5x^2+1=25\)
B \(5x^2+1=0\)
C \(5x^2+20=0\)
D \(5x^2-20=0\)
Explanation opens after your attempt
Correct Answer
D. \(5x^2-20=0\)
Step 1
Concept
Expanding gives \(x^2-4x+4+4x^2+4x+1=25\). This gives \(5x^2-20=0\).
Step 2
Why this answer is correct
The correct answer is D. \(5x^2-20=0\). Expanding gives \(x^2-4x+4+4x^2+4x+1=25\). This gives \(5x^2-20=0\).
Step 3
Exam Tip
विस्तार करने पर \(x^2-4x+4+4x^2+4x+1=25\) मिलता है। इससे \(5x^2-20=0\) बनता है।
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यदि \(8x^2-32x+24=0\) को (8) से भाग दें, तो सरल समीकरण कौन-सा होगा?
If \(8x^2-32x+24=0\) is divided by (8), which simplified equation is obtained?
#quadratic-equations
#simplification
#common-factor
#medium
A \(x^2-4x+3=0\)
B \(x^2+4x+3=0\)
C \(8x^2-4x+3=0\)
D \(x^2-32x+24=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x+3=0\)
Step 1
Concept
Dividing every term by (8) gives \(x^2-4x+3=0\). Dividing by a common nonzero factor does not change the roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+3=0\). Dividing every term by (8) gives \(x^2-4x+3=0\). Dividing by a common nonzero factor does not change the roots.
Step 3
Exam Tip
हर पद को (8) से भाग देने पर \(x^2-4x+3=0\) मिलता है। समान अशून्य गुणनखंड से भाग देने पर मूल नहीं बदलते।
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समीकरण ((x+4)2 -3(x+4)-10=0) को सरल करने पर कौन-सा द्विघात समीकरण मिलेगा?
Which quadratic equation is obtained by simplifying ((x+4)2 -3(x+4)-10=0)?
#quadratic-equations
#simplification
#identity
#medium
A \(x^2+5x-6=0\)
B \(x^2+11x-6=0\)
C \(x^2+5x+6=0\)
D \(x^2+8x-10=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x-6=0\)
Step 1
Concept
((x+4)2 =x-2 +8x+16) and (-3(x+4)=-3x-12). Simplifying gives \(x^2+5x-6=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x-6=0\). ((x+4)2 =x-2 +8x+16) and (-3(x+4)=-3x-12). Simplifying gives \(x^2+5x-6=0\).
Step 3
Exam Tip
((x+4)2 =x-2 +8x+16) और (-3(x+4)=-3x-12) है। सरल करने पर \(x^2+5x-6=0\) मिलता है।
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यदि \(6x^2-18x+12=0\) को (6) से भाग दें, तो सरल समीकरण कौन-सा होगा?
If \(6x^2-18x+12=0\) is divided by (6), which simplified equation is obtained?
#quadratic-equations
#simplification
#common-factor
#medium
A \(x^2-3x+2=0\)
B \(x^2+3x+2=0\)
C \(6x^2-3x+2=0\)
D \(x^2-18x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x+2=0\)
Step 1
Concept
Dividing every term by (6) gives \(x^2-3x+2=0\). Dividing by a common nonzero factor does not change the roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x+2=0\). Dividing every term by (6) gives \(x^2-3x+2=0\). Dividing by a common nonzero factor does not change the roots.
Step 3
Exam Tip
हर पद को (6) से भाग देने पर \(x^2-3x+2=0\) मिलता है। समान अशून्य गुणनखंड से भाग देने पर मूल नहीं बदलते।
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समीकरण ((x-3)2 +2(x-3)-8=0) को सरल करने पर कौन-सा द्विघात समीकरण मिलेगा?
Which quadratic equation is obtained by simplifying ((x-3)2 +2(x-3)-8=0)?
#quadratic-equations
#simplification
#identity
#medium
A \(x^2-4x-5=0\)
B \(x^2+4x-5=0\)
C \(x^2-6x+1=0\)
D \(x^2-4x+5=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x-5=0\)
Step 1
Concept
((x-3)2 =x-2 -6x+9) and (2(x-3)=2x-6). Simplifying gives \(x^2-4x-5=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x-5=0\). ((x-3)2 =x-2 -6x+9) and (2(x-3)=2x-6). Simplifying gives \(x^2-4x-5=0\).
Step 3
Exam Tip
((x-3)2 =x-2 -6x+9) है और (2(x-3)=2x-6) है। सरल करने पर \(x^2-4x-5=0\) मिलता है।
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यदि \(3x^2-12x+12=0\) को (3) से भाग दें, तो सरल समीकरण कौन-सा होगा?
If \(3x^2-12x+12=0\) is divided by (3), which simplified equation is obtained?
#quadratic-equations
#simplification
#common-factor
#medium
A \(x^2-4x+4=0\)
B \(x^2+4x+4=0\)
C \(3x^2-4x+4=0\)
D \(x^2-12x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x+4=0\)
Step 1
Concept
Dividing every term by (3) gives \(x^2-4x+4=0\). Dividing by a common nonzero factor does not change the roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+4=0\). Dividing every term by (3) gives \(x^2-4x+4=0\). Dividing by a common nonzero factor does not change the roots.
Step 3
Exam Tip
हर पद को (3) से भाग देने पर \(x^2-4x+4=0\) मिलता है। समान गुणनखंड से भाग देने पर मूल नहीं बदलते।
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समीकरण ((x+2)2 -(x+2)-6=0) को सरल करने पर कौन-सा द्विघात समीकरण मिलेगा?
Which quadratic equation is obtained by simplifying ((x+2)2 -(x+2)-6=0)?
#quadratic-equations
#simplification
#identity
#medium
A \(x^2+3x-4=0\)
B \(x^2+4x-2=0\)
C \(x^2+3x+4=0\)
D \(x^2+5x=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-4=0\)
Step 1
Concept
((x+2)2 =x-2 +4x+4), so the simplified form is \(x^2+3x-4=0\). Write every term while opening brackets.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-4=0\). ((x+2)2 =x-2 +4x+4), so the simplified form is \(x^2+3x-4=0\). Write every term while opening brackets.
Step 3
Exam Tip
((x+2)2 =x-2 +4x+4) है, इसलिए सरल रूप \(x^2+3x-4=0\) है। कोष्ठक खोलते समय हर पद लिखें।
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किस विकल्प में \(\sqrt{3}\) और \(\sqrt{12}\) का योग परिमेय गुणांक वाले सरल अपरिमेय रूप में सही लिखा गया है?
In which option is the sum of \(\sqrt{3}\) and \(\sqrt{12}\) correctly written as a simple irrational form with rational coefficient?
#surd-addition
#simplification
#irrational-form
A \(3\sqrt{3}\)
B \(4\sqrt{3}\)
C \(\sqrt{15}\)
D \(\sqrt{36}\)
Explanation opens after your attempt
Correct Answer
A. \(3\sqrt{3}\)
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\), so \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\). In exams make radicals like terms before adding.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\), so \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\). In exams make radicals like terms before adding.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\), इसलिए \(\sqrt{3}+\sqrt{12}=3\sqrt{3}\) है। परीक्षा में मूलों को जोड़ने से पहले समान मूल बनाएं।
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यदि (x) अपरिमेय है, तो (\(x+\sqrt{2}\)-\(x-\sqrt{2}\)) किसके बराबर है?
If (x) is irrational, what is (\(x+\sqrt{2}\)-\(x-\sqrt{2}\)) equal to?
#algebra
#irrational-expression
#simplification
A \(2\sqrt{2}\)
B (2x)
C (0)
D \(\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{2}\)
Step 1
Concept
The like (x) terms cancel and the value left is \(2\sqrt{2}\). In exams do not be confused by the type of number during algebraic simplification.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{2}\). The like (x) terms cancel and the value left is \(2\sqrt{2}\). In exams do not be confused by the type of number during algebraic simplification.
Step 3
Exam Tip
समान (x) पद कट जाते हैं और मान \(2\sqrt{2}\) बचता है। परीक्षा में बीजीय सरलीकरण में संख्या के प्रकार से भ्रमित न हों।
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यदि \(\frac{35}{2^2\cdot5\cdot7^2}\) को सरलतम रूप में लिखा जाए, तो उसका दशमलव प्रसार कैसा होगा?
If \(\frac{35}{2^2\cdot5\cdot7^2}\) is written in lowest form, what type of decimal expansion will it have?
#decimal-expansion
#simplification
#rational-numbers
A अनवसानी आवर्ती / Non-terminating recurring
B समाप्त / Terminating
C अनवसानी अनावर्ती / Non-terminating non-recurring
D शून्य / Zero
Explanation opens after your attempt
Correct Answer
A. अनवसानी आवर्ती / Non-terminating recurring
Step 1
Concept
After simplification, (7) remains in the denominator, so the decimal is non-terminating recurring. In exams do not decide only from the original denominator.
Step 2
Why this answer is correct
The correct answer is A. अनवसानी आवर्ती / Non-terminating recurring. After simplification, (7) remains in the denominator, so the decimal is non-terminating recurring. In exams do not decide only from the original denominator.
Step 3
Exam Tip
सरलीकरण के बाद हर में (7) बचता है, इसलिए दशमलव अनवसानी आवर्ती होगा। परीक्षा में केवल मूल हर देखकर निर्णय न लें।
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यदि \(x=\sqrt{2}\) और \(y=\sqrt{8}\), तो (x:y) का सरल अनुपात क्या है?
If \(x=\sqrt{2}\) and \(y=\sqrt{8}\), what is the simplified ratio (x:y)?
#ratio
#surds
#simplification
A (1:2)
B (2:1)
C (1:4)
D \(\sqrt{2}:8\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), so \(\sqrt{2}:2\sqrt{2}=1:2\). In exams common radical factors can be cancelled.
Step 2
Why this answer is correct
The correct answer is A. (1:2). \(\sqrt{8}=2\sqrt{2}\), so \(\sqrt{2}:2\sqrt{2}=1:2\). In exams common radical factors can be cancelled.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए \(\sqrt{2}:2\sqrt{2}=1:2\) है। परीक्षा में समान करणी काट सकते हैं।
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यदि \(a=\sqrt{27}-\sqrt{12}\), तो (a) का मान क्या है?
If \(a=\sqrt{27}-\sqrt{12}\), what is the value of (a)?
#surd subtraction
#irrational
#simplification
A \(\sqrt{3}\)
B \(3\sqrt{3}\)
C \(-\sqrt{3}\)
D \(\sqrt{15}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{3}\)
Step 1
Concept
\(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(\sqrt{3}\). Simplify first in exams.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\). \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(\sqrt{3}\). Simplify first in exams.
Step 3
Exam Tip
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\), इसलिए अंतर \(\sqrt{3}\) है। परीक्षा में पहले सरलीकरण करें।
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यदि \(a=\sqrt{2}+\sqrt{8}\), तो (a) का सरल रूप क्या है और वह किस प्रकार की संख्या है?
If \(a=\sqrt{2}+\sqrt{8}\), what is the simplified form and type of (a)?
#like radicals
#simplification
#irrational
A \(3\sqrt{2}\), अपरिमेय / \(3\sqrt{2}\), irrational
B (10), परिमेय / (10), rational
C \(\sqrt{10}\), अपरिमेय / \(\sqrt{10}\), irrational
D \(2\sqrt{2}\), परिमेय / \(2\sqrt{2}\), rational
Explanation opens after your attempt
Correct Answer
A. \(3\sqrt{2}\), अपरिमेय / \(3\sqrt{2}\), irrational
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), so \(a=3\sqrt{2}\), irrational. Combine like radicals in exams.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\), अपरिमेय / \(3\sqrt{2}\), irrational. \(\sqrt{8}=2\sqrt{2}\), so \(a=3\sqrt{2}\), irrational. Combine like radicals in exams.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए \(a=3\sqrt{2}\) अपरिमेय है। परीक्षा में समान करणी वाले पद जोड़ें।
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यदि (p(x)=2x-2 -4x-1) है, तो इसके शून्यकों का सही रूप कौन सा है?
If (p(x)=2x-2 -4x-1), which is the correct form of its zeroes?
#quadratic-formula
#simplification
#irrational
A \(1\pm\frac{\sqrt{6}}{2}\)
B \(1\pm\sqrt{6}\)
C \(2\pm\frac{\sqrt{6}}{2}\)
D \(-1\pm\frac{\sqrt{6}}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(1\pm\frac{\sqrt{6}}{2}\)
Step 1
Concept
By the formula, \(x=\frac{4\pm\sqrt{16+8}}{4}=1\pm\frac{\sqrt{6}}{2}\). Divide the whole expression carefully while simplifying.
Step 2
Why this answer is correct
The correct answer is A. \(1\pm\frac{\sqrt{6}}{2}\). By the formula, \(x=\frac{4\pm\sqrt{16+8}}{4}=1\pm\frac{\sqrt{6}}{2}\). Divide the whole expression carefully while simplifying.
Step 3
Exam Tip
सूत्र से \(x=\frac{4\pm\sqrt{16+8}}{4}=1\pm\frac{\sqrt{6}}{2}\) है। हर को सरल करते समय पूरा पद विभाजित करें।
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यदि (p(x)=\sqrt{3}x-2 -6x+2\sqrt{3}), तो शून्यकों का गुणनफल क्या है?
If (p(x)=\sqrt{3}x-2 -6x+2\sqrt{3}), what is the product of its zeroes?
#product-of-zeroes
#irrational-coefficient
#simplification
A (2)
B \(\sqrt{3}\)
C (6)
D \(2\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
The product is \(\frac{c}{a}=\frac{2\sqrt{3}}{\sqrt{3}}=2\). Like radicals can cancel.
Step 2
Why this answer is correct
The correct answer is A. (2). The product is \(\frac{c}{a}=\frac{2\sqrt{3}}{\sqrt{3}}=2\). Like radicals can cancel.
Step 3
Exam Tip
गुणनफल \(\frac{c}{a}=\frac{2\sqrt{3}}{\sqrt{3}}=2\) है। समान करणी कट सकती है।
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कौन सा विकल्प \(\sqrt{10+\sqrt{96}}\) का सरल रूप है?
Which option is the simplified form of \(\sqrt{10+\sqrt{96}}\)?
#nested-root
#surds
#simplification
A \(\sqrt{6}+\sqrt{4}\)
B \(\sqrt{6}-\sqrt{4}\)
C \(\sqrt{8}+\sqrt{2}\)
D \(\sqrt{10}+\sqrt{96}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{6}+\sqrt{4}\)
Step 1
Concept
(\(\sqrt{6}+2\)2 =6+4+4\sqrt{6}=10+\sqrt{96}). Hence \(\sqrt{6}+\sqrt{4}\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{6}+\sqrt{4}\). (\(\sqrt{6}+2\)2 =6+4+4\sqrt{6}=10+\sqrt{96}). Hence \(\sqrt{6}+\sqrt{4}\) is correct.
Step 3
Exam Tip
(\(\sqrt{6}+2\)2 =6+4+4\sqrt{6}=10+\sqrt{96}) है। इसलिए \(\sqrt{6}+\sqrt{4}\) सही रूप है।
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कौन सा विकल्प \(4\sqrt{7}+2\sqrt{28}-\sqrt{175}\) का सरल रूप है?
Which option is the simplified form of \(4\sqrt{7}+2\sqrt{28}-\sqrt{175}\)?
#surds
#simplification
#like-terms
A \(3\sqrt{7}\)
B \(7\sqrt{7}\)
C \(\sqrt{84}\)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(3\sqrt{7}\)
Step 1
Concept
\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{175}=5\sqrt{7}\). Thus \(4\sqrt{7}+4\sqrt{7}-5\sqrt{7}=3\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{7}\). \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{175}=5\sqrt{7}\). Thus \(4\sqrt{7}+4\sqrt{7}-5\sqrt{7}=3\sqrt{7}\).
Step 3
Exam Tip
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{175}=5\sqrt{7}\) है। इसलिए \(4\sqrt{7}+4\sqrt{7}-5\sqrt{7}=3\sqrt{7}\) है।
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