The sum of roots is (\frac{2(m-1)}{m+1}). Setting it equal to (2) gives (m-1=m+1), which is impossible.
Step 2
Why this answer is correct
The correct answer is D. नहीं, कोई मान नहीं / No, no value. The sum of roots is (\frac{2(m-1)}{m+1}). Setting it equal to (2) gives (m-1=m+1), which is impossible.
Step 3
Exam Tip
जड़ों का योग (\frac{2(m-1)}{m+1}) है। इसे (2) रखने पर (m-1=m+1) आता है, जो असंभव है।
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). With \(\alpha+\beta=-4\) and \(\alpha\beta=1\), the value is (14).
Step 2
Why this answer is correct
The correct answer is C. (14). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). With \(\alpha+\beta=-4\) and \(\alpha\beta=1\), the value is (14).
Step 3
Exam Tip
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\) होता है। \(\alpha+\beta=-4\) और \(\alpha\beta=1\) से मान (14) आता है।
For real roots, \(D=36-36k\ge0\) is required. Thus \(k\le1\), and \(k\ne0\) is also needed for a quadratic equation.
Step 2
Why this answer is correct
The correct answer is A. \(k\le 1,\ k\ne0\). For real roots, \(D=36-36k\ge0\) is required. Thus \(k\le1\), and \(k\ne0\) is also needed for a quadratic equation.
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D=36-36k\ge0\) होना चाहिए। इसलिए \(k\le1\) और द्विघात के लिए \(k\ne0\) भी जरूरी है।
Let the roots be (r) and (4r). Then \(4r^2=4\), so \(r=\pm1\); using \(5r=-\frac{p}{3}\), we get \(p=\pm15\).
Step 2
Why this answer is correct
The correct answer is A. (15) या (-15) / (15) or (-15). Let the roots be (r) and (4r). Then \(4r^2=4\), so \(r=\pm1\); using \(5r=-\frac{p}{3}\), we get \(p=\pm15\).
Step 3
Exam Tip
जड़ें (r) और (4r) मानने पर \(4r^2=4\), इसलिए \(r=\pm1\)। योग \(5r=-\frac{p}{3}\) से \(p=\pm15\) मिलता है।
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।
Here \(\alpha+\beta=6\) and \(\alpha\beta=m\). Using (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)), we get (m=8).
Step 2
Why this answer is correct
The correct answer is C. (8). Here \(\alpha+\beta=6\) and \(\alpha\beta=m\). Using (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)), we get (m=8).
Step 3
Exam Tip
\(\alpha+\beta=6\) और \(\alpha\beta=m\) है। (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)) से (m=8) 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।
A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\)/\(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
Step 1
Concept
Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\). Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 3
Exam Tip
जड़ें (r) और (2r) मानने पर (3r=3-k) और \(2r^2=k\) मिलता है। हल करने पर \(2k^2-21k+18=0\), इसलिए दिए गए दोनों मान मिलते हैं।
Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-10x+3=0\). Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{10}{3}\) और \(\alpha\beta=1\) है। व्युत्क्रम जड़ों का योग \(\frac{10}{3}\) और गुणनफल (1) ही रहता है।
Here \(\alpha+\beta=\frac{5}{2}\) and \(\alpha\beta=\frac{m}{2}\). Using (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta), we get (m=3).
Step 2
Why this answer is correct
The correct answer is B. (3). Here \(\alpha+\beta=\frac{5}{2}\) and \(\alpha\beta=\frac{m}{2}\). Using (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta), we get (m=3).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{5}{2}\) और \(\alpha\beta=\frac{m}{2}\) है। (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) से (m=3) मिलता है।
