Here (D=(-13)2-4(1)(7)=141), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{141}\). In exams, the difference of roots can be found directly from (D).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{141}\). Here (D=(-13)2-4(1)(7)=141), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{141}\). In exams, the difference of roots can be found directly from (D).
Step 3
Exam Tip
यहां (D=(-13)2-4(1)(7)=141), इसलिए मूलों का अंतर \(\frac{\sqrt{D}}{|a|}=\sqrt{141}\) है। परीक्षा में मूलों का अंतर सीधे (D) से मिल सकता है।
Here (D=(-11)2-4(1)(6)=97), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{97}\). In exams, the difference of roots can be found directly from (D).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{97}\). Here (D=(-11)2-4(1)(6)=97), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{97}\). In exams, the difference of roots can be found directly from (D).
Step 3
Exam Tip
यहां (D=(-11)2-4(1)(6)=97), इसलिए मूलों का अंतर \(\frac{\sqrt{D}}{|a|}=\sqrt{97}\) है। परीक्षा में मूलों का अंतर सीधे (D) से मिल सकता है।
Here (D=(-9)2-4(1)(5)=61), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{61}\). In exams, the difference of roots can be found directly from (D).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{61}\). Here (D=(-9)2-4(1)(5)=61), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{61}\). In exams, the difference of roots can be found directly from (D).
Step 3
Exam Tip
यहां (D=(-9)2-4(1)(5)=61), इसलिए मूलों का अंतर \(\frac{\sqrt{D}}{|a|}=\sqrt{61}\) है। परीक्षा में मूलों का अंतर सीधे (D) से मिल सकता है।
Here (D=(-7)2-4(1)(3)=37), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{37}\). In exams, the difference of roots can be found directly from (D).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{37}\). Here (D=(-7)2-4(1)(3)=37), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{37}\). In exams, the difference of roots can be found directly from (D).
Step 3
Exam Tip
यहां (D=(-7)2-4(1)(3)=37), इसलिए मूलों का अंतर \(\frac{\sqrt{D}}{|a|}=\sqrt{37}\) है। परीक्षा में मूलों का अंतर सीधे (D) से मिल सकता है।
Here (D=(-5)2-4(1)(2)=17), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{17}\). In exams, the difference of roots can be found directly from (D).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{17}\). Here (D=(-5)2-4(1)(2)=17), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{17}\). In exams, the difference of roots can be found directly from (D).
Step 3
Exam Tip
यहां (D=(-5)2-4(1)(2)=17), इसलिए मूलों का अंतर \(\frac{\sqrt{D}}{|a|}=\sqrt{17}\) है। परीक्षा में मूलों का अंतर सीधे (D) से मिल सकता है।
The square of the difference is \(D/a^2=5\), so the difference is \(\sqrt{5}\). In exams, the difference of roots is \(\frac{\sqrt{D}}{|a|}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}\). The square of the difference is \(D/a^2=5\), so the difference is \(\sqrt{5}\). In exams, the difference of roots is \(\frac{\sqrt{D}}{|a|}\).
Step 3
Exam Tip
अंतर का वर्ग \(D/a^2=5\) है, इसलिए अंतर \(\sqrt{5}\) होगा। परीक्षा में मूलों का अंतर \(\frac{\sqrt{D}}{|a|}\) से मिलता है।
Here \(\alpha+\beta=2a+1\) and \(\alpha\beta=a^2+a-6\). Since (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=25), the positive difference is (5).
Step 2
Why this answer is correct
The correct answer is A. (5). Here \(\alpha+\beta=2a+1\) and \(\alpha\beta=a^2+a-6\). Since (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=25), the positive difference is (5).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=2a+1\) और \(\alpha\beta=a^2+a-6\) है। (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=25), इसलिए धनात्मक अंतर (5) है।
Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Since (\(\alpha-\beta\)2=25-9=16), the positive difference is (4).
Step 2
Why this answer is correct
The correct answer is A. (4). Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Since (\(\alpha-\beta\)2=25-9=16), the positive difference is (4).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=\frac{9}{4}\) है। (\(\alpha-\beta\)2=25-9=16), इसलिए धनात्मक अंतर (4) है।
Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Thus (\(\alpha-\beta\)2=25-9=16), so the positive difference is (4); option (A) should be correct.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{7}{2}\). Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Thus (\(\alpha-\beta\)2=25-9=16), so the positive difference is (4); option (A) should be correct.
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=\frac{9}{4}\) है। (\(\alpha-\beta\)2=25-9=16), इसलिए धनात्मक अंतर (4) है, अतः विकल्प (A) सही होना चाहिए।
Here \(\alpha+\beta=\frac{16}{5}\) and \(\alpha\beta=\frac{p}{5}\). Using (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta), we get (p=11).
Step 2
Why this answer is correct
The correct answer is C. (11). Here \(\alpha+\beta=\frac{16}{5}\) and \(\alpha\beta=\frac{p}{5}\). Using (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta), we get (p=11).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{16}{5}\) और \(\alpha\beta=\frac{p}{5}\) है। (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) से (p=11) मिलता है।
Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). With \(\alpha+\beta=\frac{13}{3}\) and \(\alpha\beta=\frac{4}{3}\), the positive difference is \(\frac{11}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{11}{3}\). Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). With \(\alpha+\beta=\frac{13}{3}\) and \(\alpha\beta=\frac{4}{3}\), the positive difference is \(\frac{11}{3}\).
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) लगाएँ। \(\alpha+\beta=\frac{13}{3}\) और \(\alpha\beta=\frac{4}{3}\), इसलिए धनात्मक अंतर \(\frac{11}{3}\) है।
In the given equation, the sum of roots is (2r+5) and the product is (r-2+5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). In the given equation, the sum of roots is (2r+5) and the product is (r-2+5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).
Step 3
Exam Tip
दिए गए समीकरण में जड़ों का योग (2r+5) और गुणनफल (r-2+5r+6=(r+2)(r+3)) है। इसलिए जड़ें (r+2) और (r+3) हैं, अतः धनात्मक अंतर (1) है।
Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). With \(\alpha+\beta=\frac{7}{2}\) and \(\alpha\beta=\frac{3}{2}\), the positive difference is \(\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{2}\). Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). With \(\alpha+\beta=\frac{7}{2}\) and \(\alpha\beta=\frac{3}{2}\), the positive difference is \(\frac{5}{2}\).
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) लगाएँ। \(\alpha+\beta=\frac{7}{2}\) और \(\alpha\beta=\frac{3}{2}\), इसलिए धनात्मक अंतर \(\frac{5}{2}\) है।
Here \(\alpha+\beta=\frac{11}{3}\) and (\(\alpha-\beta\)2=\frac{25}{9}). Using (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta), we get \(\alpha\beta=\frac{8}{3}\), so (p=8).
Step 2
Why this answer is correct
The correct answer is A. (8). Here \(\alpha+\beta=\frac{11}{3}\) and (\(\alpha-\beta\)2=\frac{25}{9}). Using (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta), we get \(\alpha\beta=\frac{8}{3}\), so (p=8).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{11}{3}\) और (\(\alpha-\beta\)2=\frac{25}{9}) है। सूत्र (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) से \(\alpha\beta=\frac{8}{3}\), इसलिए (p=8)।
Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). Since (16=\(\alpha+\beta\)2-20), we get (\(\alpha+\beta\)2=36) and \(m^2=36\).
Step 2
Why this answer is correct
The correct answer is C. (36). Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). Since (16=\(\alpha+\beta\)2-20), we get (\(\alpha+\beta\)2=36) and \(m^2=36\).
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) लगाएँ। (16=\(\alpha+\beta\)2-20), इसलिए (\(\alpha+\beta\)2=36) और \(m^2=36\)।
A. \(\sqrt{57}\) और \(-\sqrt{57}\)/\(\sqrt{57}\) and \(-\sqrt{57}\)
Step 1
Concept
Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\). Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 3
Exam Tip
जड़ें (r) और (r+3) मानने पर (r(r+3)=12) मिलता है। इससे जड़ों का योग \(\pm\sqrt{57}\) होता है, इसलिए \(a=\mp\sqrt{57}\)।
A. \(4+\sqrt{21}\) या \(4-\sqrt{21}\)/\(4+\sqrt{21}\) or \(4-\sqrt{21}\)
Step 1
Concept
Put (\(\alpha-\beta\)2=9) in (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). This gives \(a^2-8a-5=0\), so \(a=4\pm\sqrt{21}\).
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{21}\) या \(4-\sqrt{21}\) / \(4+\sqrt{21}\) or \(4-\sqrt{21}\). Put (\(\alpha-\beta\)2=9) in (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). This gives \(a^2-8a-5=0\), so \(a=4\pm\sqrt{21}\).
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) में (\(\alpha-\beta\)2=9) रखें। इससे \(a^2-8a-5=0\) और \(a=4\pm\sqrt{21}\) मिलता है।
Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). Here \(\alpha+\beta=\frac{7}{2}\) and \(\alpha\beta=\frac{3}{2}\), so the value is \(\frac{25}{4}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{25}{4}\). Use (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta). Here \(\alpha+\beta=\frac{7}{2}\) and \(\alpha\beta=\frac{3}{2}\), so the value is \(\frac{25}{4}\).
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta) का प्रयोग करें। यहाँ \(\alpha+\beta=\frac{7}{2}\) और \(\alpha\beta=\frac{3}{2}\), इसलिए मान \(\frac{25}{4}\) है।
