difference of squares se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
It is ((x-d)2-36), so \(x-d=\pm6\) and the zeroes are (d-6), (d+6). Tip: use difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (d-6) और (d+6) / (d-6) and (d+6). It is ((x-d)2-36), so \(x-d=\pm6\) and the zeroes are (d-6), (d+6). Tip: use difference of squares.
Step 3
Exam Tip
यह ((x-d)2-36) है, इसलिए \(x-d=\pm6\) और शून्यक (d-6), (d+6) हैं। टिप: वर्गों के अंतर का उपयोग करें।
A. (\left\(\frac{7}{6},0\right\)) और (\left\(-\frac{7}{6},0\right\))/(\left\(\frac{7}{6},0\right\)) and (\left\(-\frac{7}{6},0\right\))
Step 1
Concept
From \(36x^2-49=0\), \(x=\pm\frac{7}{6}\). Tip: treat \(36x^2\) as ((6x)2).
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{7}{6},0\right\)) और (\left\(-\frac{7}{6},0\right\)) / (\left\(\frac{7}{6},0\right\)) and (\left\(-\frac{7}{6},0\right\)). From \(36x^2-49=0\), \(x=\pm\frac{7}{6}\). Tip: treat \(36x^2\) as ((6x)2).
Step 3
Exam Tip
\(36x^2-49=0\) से \(x=\pm\frac{7}{6}\) मिलता है। टिप: \(36x^2\) को ((6x)2) समझें।
It is ((x-c)2-25), so \(x-c=\pm5\) and the zeroes are (c-5), (c+5). Tip: use difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (c-5) और (c+5) / (c-5) and (c+5). It is ((x-c)2-25), so \(x-c=\pm5\) and the zeroes are (c-5), (c+5). Tip: use difference of squares.
Step 3
Exam Tip
यह ((x-c)2-25) है, इसलिए \(x-c=\pm5\) और शून्यक (c-5), (c+5) हैं। टिप: वर्गों के अंतर का उपयोग करें।
A. (\left\(\frac{6}{5},0\right\)) और (\left\(-\frac{6}{5},0\right\))/(\left\(\frac{6}{5},0\right\)) and (\left\(-\frac{6}{5},0\right\))
Step 1
Concept
From \(25x^2-36=0\), \(x=\pm\frac{6}{5}\). Tip: treat \(25x^2\) as ((5x)2).
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{6}{5},0\right\)) और (\left\(-\frac{6}{5},0\right\)) / (\left\(\frac{6}{5},0\right\)) and (\left\(-\frac{6}{5},0\right\)). From \(25x^2-36=0\), \(x=\pm\frac{6}{5}\). Tip: treat \(25x^2\) as ((5x)2).
Step 3
Exam Tip
\(25x^2-36=0\) से \(x=\pm\frac{6}{5}\) मिलता है। टिप: \(25x^2\) को ((5x)2) समझें।
It is ((x-b)2-16), so \(x-b=\pm4\) and the zeroes are (b-4), (b+4). Tip: use difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (b-4) और (b+4) / (b-4) and (b+4). It is ((x-b)2-16), so \(x-b=\pm4\) and the zeroes are (b-4), (b+4). Tip: use difference of squares.
Step 3
Exam Tip
यह ((x-b)2-16) है इसलिए \(x-b=\pm4\) और शून्यक (b-4), (b+4) हैं। टिप: वर्गों के अंतर का उपयोग करें।
A. (\left\(\frac{3}{4},0\right\)) और (\left\(-\frac{3}{4},0\right\))/(\left\(\frac{3}{4},0\right\)) and (\left\(-\frac{3}{4},0\right\))
Step 1
Concept
From \(16x^2-9=0\), \(x=\pm\frac{3}{4}\). Tip: treat \(16x^2\) as ((4x)2).
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{3}{4},0\right\)) और (\left\(-\frac{3}{4},0\right\)) / (\left\(\frac{3}{4},0\right\)) and (\left\(-\frac{3}{4},0\right\)). From \(16x^2-9=0\), \(x=\pm\frac{3}{4}\). Tip: treat \(16x^2\) as ((4x)2).
Step 3
Exam Tip
\(16x^2-9=0\) से \(x=\pm\frac{3}{4}\) मिलता है। टिप: \(16x^2\) को ((4x)2) समझें।
It is ((x-a)2-9), so \(x-a=\pm3\) and the zeroes are (a-3), (a+3). Tip: use difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (a-3) और (a+3) / (a-3) and (a+3). It is ((x-a)2-9), so \(x-a=\pm3\) and the zeroes are (a-3), (a+3). Tip: use difference of squares.
Step 3
Exam Tip
यह ((x-a)2-9) है, इसलिए \(x-a=\pm3\) और शून्यक (a-3), (a+3) हैं। टिप: वर्गों के अंतर का उपयोग करें।
A. (\left\(\frac{4}{3},0\right\)) और (\left\(-\frac{4}{3},0\right\))/(\left\(\frac{4}{3},0\right\)) and (\left\(-\frac{4}{3},0\right\))
Step 1
Concept
From \(9x^2-16=0\), \(x=\pm\frac{4}{3}\). Tip: treat \(9x^2\) as ((3x)2).
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{4}{3},0\right\)) और (\left\(-\frac{4}{3},0\right\)) / (\left\(\frac{4}{3},0\right\)) and (\left\(-\frac{4}{3},0\right\)). From \(9x^2-16=0\), \(x=\pm\frac{4}{3}\). Tip: treat \(9x^2\) as ((3x)2).
Step 3
Exam Tip
\(9x^2-16=0\) से \(x=\pm\frac{4}{3}\) मिलता है। टिप: \(9x^2\) को ((3x)2) समझें।
A. (\left\(\frac{5}{2},0\right\)) और (\left\(-\frac{5}{2},0\right\))/(\left\(\frac{5}{2},0\right\)) and (\left\(-\frac{5}{2},0\right\))
Step 1
Concept
From \(4x^2-25=0\), \(x=\pm\frac{5}{2}\). Tip: treat \(4x^2\) as ((2x)2).
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{5}{2},0\right\)) और (\left\(-\frac{5}{2},0\right\)) / (\left\(\frac{5}{2},0\right\)) and (\left\(-\frac{5}{2},0\right\)). From \(4x^2-25=0\), \(x=\pm\frac{5}{2}\). Tip: treat \(4x^2\) as ((2x)2).
Step 3
Exam Tip
\(4x^2-25=0\) से \(x=\pm\frac{5}{2}\) मिलता है। टिप: \(4x^2\) को ((2x)2) समझें।
In such forms, identify the difference of squares before expanding. चरण 1: यह संयुग्मी गुणन है। चरण 2: (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=13-12=1)। चरण 3: ऐसे रूपों में विस्तार करने से पहले अंतर के वर्ग को पहचानें।
Multiplying conjugate surds often removes the irrational part. चरण 1: यह ((u+v)(u-v)) के रूप में है। चरण 2: मान (11-3=8) आता है, जो परिमेय है। चरण 3: संयुग्मी पदों का गुणन अक्सर अपरिमेय भाग हटा देता है।
View (ab) as (\(\sqrt{3}+\sqrt{2}\)\(\sqrt{3}-\sqrt{2}\)).
Step 2
Why this answer is correct
This equals (\(\sqrt{3}\)2-\(\sqrt{2}\)2=3-2=1).
Step 3
Exam Tip
Since addition order does not change the sum, recognize the conjugate form. चरण 1: (ab=\(\sqrt{3}+\sqrt{2}\)\(\sqrt{3}-\sqrt{2}\)) के रूप में देखा जा सकता है। चरण 2: यह (\(\sqrt{3}\)2-\(\sqrt{2}\)2=3-2=1) है। चरण 3: क्रम बदलने से योग नहीं बदलता, इसलिए संयुग्मी रूप पहचानें।
In conjugate multiplication, the middle irrational terms cancel. चरण 1: यह संयुग्मी संख्याओं का गुणन है। चरण 2: (\(1+\sqrt{2}\)\(1-\sqrt{2}\)=1-\(\sqrt{2}\)2=1-2=-1)। चरण 3: संयुग्मी गुणन में बीच के अपरिमेय पद कट जाते हैं।
