identity se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
341 questions tagged with identity.
If \(\alpha\) and \(\beta\) are zeroes of \(x^2-7x+10\), what is \(\alpha^2+\beta^2\)?
A. (29)
(\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) होता है। (72-2(10)=29) है। / (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta). Thus (72-2(10)=29).
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If (p(x)=x-3-3x-2+3x-1), what is (p(2))?
B. (1)
यह ((x-1)3) है, इसलिए (p(2)=(2-1)3=1)। पहचान पहचानने से गणना कम होती है। / It is ((x-1)3), so (p(2)=(2-1)3=1). Recognizing identities reduces calculation.
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Which polynomial is a perfect square?
A. \(x^2+10x+25\)
(x-2+10x+25=(x+5)2) है। पहले और अंतिम पद के वर्ग तथा मध्य पद (2ab) को जांचें। / (x-2+10x+25=(x+5)2). Check the squares of first and last terms and the middle term (2ab).
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What is the expanded form of (\(x^2-1\)\(x^2+1\))?
A. \(x^4-1\)
यह \(a^2-b^2\) की पहचान है, इसलिए (\(x^2\)2-12=x-4-1)। पहचान से विस्तार तेज होता है। / This is the identity \(a^2-b^2\), so (\(x^2\)2-12=x-4-1). Identities make expansion faster.
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What is the simplified polynomial form of ((x+4)2-(x-4)2)?
B. (16x)
((x+4)2-(x-4)2=16x)। पहचान (a-2-b-2=(a-b)(a+b)) से जल्दी हल करें। / ((x+4)2-(x-4)2=16x). Use the identity (a-2-b-2=(a-b)(a+b)) for a quick solution.
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What is the expanded polynomial form of ((2x+3)2)?
A. \(4x^2+12x+9\)
((2x+3)2=4x-2+12x+9)। वर्ग करते समय मध्य पद (2ab) को न छोड़ें। / ((2x+3)2=4x-2+12x+9). Do not skip the middle term (2ab) while squaring.
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If \(x^{2}-\frac{1}{x^{2}}=60\) and \(x-\frac{1}{x}=6\), what is the value of \(x+\frac{1}{x}\)?
C. (10)
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)) है। इसलिए (60=6\left\(x+\frac{1}{x}\right\)) और मान (10) है। / We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (60=6\left\(x+\frac{1}{x}\right\)), so the value is (10).
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If \(s=4+\sqrt{17}\), what is the value of \(s^{2}-\frac{1}{s^{2}}\)?
A. \(16\sqrt{17}\)
\(\frac{1}{s}=\sqrt{17}-4\), इसलिए \(s-\frac{1}{s}=8\) और \(s+\frac{1}{s}=2\sqrt{17}\)। अतः \(s^{2}-\frac{1}{s^{2}}=16\sqrt{17}\)। / Here \(\frac{1}{s}=\sqrt{17}-4\), so \(s-\frac{1}{s}=8\) and \(s+\frac{1}{s}=2\sqrt{17}\). Thus \(s^{2}-\frac{1}{s^{2}}=16\sqrt{17}\).
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If \(x=\sqrt{11}-\sqrt{6}\), what is the value of \(x^{2}+2\sqrt{66}\)?
C. (17)
\(x^{2}=11+6-2\sqrt{66}=17-2\sqrt{66}\)। इसलिए \(x^{2}+2\sqrt{66}=17\)। / Since \(x^{2}=11+6-2\sqrt{66}=17-2\sqrt{66}\), \(x^{2}+2\sqrt{66}=17\).
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Which option is the simplified form of \(\frac{x^{12}-4096}{x^{6}-64}\), where \(x^{6}\neq64\)?
B. \(x^{6}+64\)
(x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\))। समान गुणनखंड कटने पर \(x^{6}+64\) मिलता है। / Since (x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\)), cancelling the common factor gives \(x^{6}+64\).
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Which option gives the correct expansion of (\(4\sqrt{3}-3\sqrt{5}\)^{2})?
A. \(93-24\sqrt{15}\)
(\(4\sqrt{3}\)^{2}=48), (\(3\sqrt{5}\)^{2}=45), और मध्य पद \(24\sqrt{15}\) है। इसलिए विस्तार \(93-24\sqrt{15}\) है। / Here (\(4\sqrt{3}\)^{2}=48), (\(3\sqrt{5}\)^{2}=45), and the middle term is \(24\sqrt{15}\). Therefore, the expansion is \(93-24\sqrt{15}\).
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If \(r=\sqrt{21}+\sqrt{14}\), what is the value of \(r^{2}-14\sqrt{6}\)?
C. (35)
\(r^{2}=21+14+2\sqrt{294}=35+14\sqrt{6}\)। इसलिए \(r^{2}-14\sqrt{6}=35\)। / Since \(r^{2}=21+14+2\sqrt{294}=35+14\sqrt{6}\), \(r^{2}-14\sqrt{6}=35\).
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If \(A=19+6\sqrt{10}\), what is the simplified form of \(\sqrt{A}\)?
A. \(3+\sqrt{10}\)
क्योंकि (\(3+\sqrt{10}\)^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), इसलिए \(\sqrt{A}=3+\sqrt{10}\)। परीक्षा में पूर्ण वर्ग करणी पहचानें। / Because (\(3+\sqrt{10}\)^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), \(\sqrt{A}=3+\sqrt{10}\). In exams, identify perfect-square surd forms.
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A. \(\frac{x^{2}+y^{2}}{x^{2}y^{2}}\)
मान लें \(A=x^{-2}\) और \(B=y^{-2}\), तो \(\frac{A^{2}-B^{2}}{A-B}=A+B\)। इसलिए उत्तर \(x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}\) है। / Let \(A=x^{-2}\) and \(B=y^{-2}\). Then \(\frac{A^{2}-B^{2}}{A-B}=A+B\), so the answer is \(x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}\).
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If \(x+\frac{1}{x}=7\), what is the value of \(x^{2}+\frac{1}{x^{2}}\)?
B. (47)
(\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2) होता है। इसलिए \(49=x^{2}+\frac{1}{x^{2}}+2\) और मान (47) है। / We use (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).
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C. \(\frac{8\sqrt{34}}{9}\)
(u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}) और (uv=9) है। इसलिए मान \(\frac{8\sqrt{34}}{9}\) है। / Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}), and (uv=9). Hence the value is \(\frac{8\sqrt{34}}{9}\).
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If \(x^{2}-\frac{1}{x^{2}}=40\) and \(x-\frac{1}{x}=5\), what is the value of \(x+\frac{1}{x}\)?
C. (8)
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)) है। इसलिए (40=5\left\(x+\frac{1}{x}\right\)), और मान (8) है। / We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (40=5\left\(x+\frac{1}{x}\right\)), so the value is (8).
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If \(s=3+\sqrt{10}\), what is the value of \(s^{2}-\frac{1}{s^{2}}\)?
A. \(12\sqrt{10}\)
\(\frac{1}{s}=\sqrt{10}-3\), इसलिए \(s-\frac{1}{s}=6\) और \(s+\frac{1}{s}=2\sqrt{10}\)। अतः \(s^{2}-\frac{1}{s^{2}}=12\sqrt{10}\)। / Here \(\frac{1}{s}=\sqrt{10}-3\), so \(s-\frac{1}{s}=6\) and \(s+\frac{1}{s}=2\sqrt{10}\). Thus \(s^{2}-\frac{1}{s^{2}}=12\sqrt{10}\).
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If \(x=\sqrt{7}-\sqrt{3}\), what is the value of \(x^{2}+2\sqrt{21}\)?
C. (10)
\(x^{2}=7+3-2\sqrt{21}=10-2\sqrt{21}\)। इसलिए \(x^{2}+2\sqrt{21}=10\)। / Since \(x^{2}=7+3-2\sqrt{21}=10-2\sqrt{21}\), \(x^{2}+2\sqrt{21}=10\).
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Which option is the simplified form of \(\frac{x^{8}-81}{x^{4}-9}\), where \(x^{4}\neq9\)?
B. \(x^{4}+9\)
(x^{8}-81=\(x^{4}\)^{2}-9^{2}=\(x^{4}-9\)\(x^{4}+9\))। समान गुणनखंड कटने पर \(x^{4}+9\) मिलता है। / Since (x^{8}-81=\(x^{4}\)^{2}-9^{2}=\(x^{4}-9\)\(x^{4}+9\)), cancelling the common factor gives \(x^{4}+9\).
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Which option gives the correct expansion of (\(3\sqrt{5}-2\sqrt{7}\)^{2})?
A. \(73-12\sqrt{35}\)
(\(3\sqrt{5}\)^{2}=45), (\(2\sqrt{7}\)^{2}=28), और मध्य पद \(12\sqrt{35}\) है। इसलिए विस्तार \(73-12\sqrt{35}\) है। / Here (\(3\sqrt{5}\)^{2}=45), (\(2\sqrt{7}\)^{2}=28), and the middle term is \(12\sqrt{35}\). Therefore, the expansion is \(73-12\sqrt{35}\).
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If \(r=\sqrt{15}+\sqrt{6}\), what is the value of \(r^{2}-6\sqrt{10}\)?
C. (21)
\(r^{2}=15+6+2\sqrt{90}=21+6\sqrt{10}\)। इसलिए \(r^{2}-6\sqrt{10}=21\)। / Since \(r^{2}=15+6+2\sqrt{90}=21+6\sqrt{10}\), \(r^{2}-6\sqrt{10}=21\).
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If \(A=14+6\sqrt{5}\), what is the simplified form of \(\sqrt{A}\)?
A. \(3+\sqrt{5}\)
क्योंकि (\(3+\sqrt{5}\)^{2}=9+5+6\sqrt{5}=14+6\sqrt{5}), इसलिए \(\sqrt{A}=3+\sqrt{5}\)। परीक्षा में पूर्ण वर्ग करणी पहचानें। / Because (\(3+\sqrt{5}\)^{2}=9+5+6\sqrt{5}=14+6\sqrt{5}), \(\sqrt{A}=3+\sqrt{5}\). In exams, identify perfect-square surd forms.
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If \(x-\frac{1}{x}=6\), what is the value of \(x^{2}+\frac{1}{x^{2}}\)?
C. (38)
(\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2) होता है। इसलिए \(36=x^{2}+\frac{1}{x^{2}}-2\) और मान (38) है। / We use (\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2). Thus \(36=x^{2}+\frac{1}{x^{2}}-2\), so the value is (38).
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B. \(2\sqrt{65}\)
(u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{5}\cdot2\sqrt{13}=4\sqrt{65}) और (uv=8)। इसलिए मान \(\frac{\sqrt{65}}{2}\) है। / Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{5}\cdot2\sqrt{13}=4\sqrt{65}) and (uv=8). Hence the value is \(\frac{\sqrt{65}}{2}\).
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If \(x^{2}-\frac{1}{x^{2}}=24\) and \(x-\frac{1}{x}=4\), what is the value of \(x+\frac{1}{x}\)?
A. (6)
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)), इसलिए (24=4\left\(x+\frac{1}{x}\right\))। परीक्षा में वर्गों के अंतर की पहचान लगाएं। / Since (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)), (24=4\left\(x+\frac{1}{x}\right\)). In exams, use the difference of squares identity.
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If \(s=2+\sqrt{7}\), what is the value of \(s^{2}-\frac{1}{s^{2}}\)?
A. \(8\sqrt{7}\)
\(\frac{1}{s}=\sqrt{7}-2\), इसलिए \(s-\frac{1}{s}=4\) और \(s+\frac{1}{s}=2\sqrt{7}\)। अतः \(s^{2}-\frac{1}{s^{2}}=8\sqrt{7}\)। / Here \(\frac{1}{s}=\sqrt{7}-2\), so \(s-\frac{1}{s}=4\) and \(s+\frac{1}{s}=2\sqrt{7}\). Thus \(s^{2}-\frac{1}{s^{2}}=8\sqrt{7}\).
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If \(x=\sqrt{5}-\sqrt{2}\), what is the value of \(x^{2}+2\sqrt{10}\)?
A. (7)
\(x^{2}=5+2-2\sqrt{10}=7-2\sqrt{10}\), इसलिए \(x^{2}+2\sqrt{10}=7\)। परीक्षा में ((a-b)^{2}) का मध्य पद ध्यान से लिखें। / Since \(x^{2}=5+2-2\sqrt{10}=7-2\sqrt{10}\), \(x^{2}+2\sqrt{10}=7\). In exams, write the middle term of ((a-b)^{2}) carefully.
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Which option is the simplified form of \(\frac{x^{6}-1}{x^{3}-1}\), where \(x^{3}\neq1\)?
A. \(x^{3}+1\)
(x^{6}-1=\(x^{3}-1\)\(x^{3}+1\)), इसलिए समान गुणनखंड कटने पर \(x^{3}+1\) मिलता है। परीक्षा में \(A^{2}-B^{2}\) रूप पहचानें। / Since (x^{6}-1=\(x^{3}-1\)\(x^{3}+1\)), cancelling the common factor gives \(x^{3}+1\). In exams, recognize the \(A^{2}-B^{2}\) form.
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Which option gives the correct expansion of (\(2\sqrt{3}-3\sqrt{2}\)^{2})?
A. \(30-12\sqrt{6}\)
(\(2\sqrt{3}\)^{2}=12), (\(3\sqrt{2}\)^{2}=18), और मध्य पद \(2\cdot2\sqrt{3}\cdot3\sqrt{2}=12\sqrt{6}\) है। इसलिए उत्तर \(30-12\sqrt{6}\) है। / Here (\(2\sqrt{3}\)^{2}=12), (\(3\sqrt{2}\)^{2}=18), and the middle term is \(2\cdot2\sqrt{3}\cdot3\sqrt{2}=12\sqrt{6}\). Therefore, the answer is \(30-12\sqrt{6}\).
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