118215
If one root of the equation \(5 x^2+13 x+k=0\) is reciprocal of the other, then the value of \(k\) is
1 0
2 \(1 / 6\)
3 5
4 6
Explanation:
C According to given equation, Let one root be \(\alpha\) and the other root be \(\frac{1}{\alpha}\) Here, we have \(5 \mathrm{x}^2+13 \mathrm{x}+\mathrm{k}=0\) Product of roots \(=\frac{k}{5}\) \(\alpha \times \frac{1}{\alpha}=\frac{k}{5}\) \(1=\frac{k}{5}\) \(k=5\)
SRM JEEE-2010
Complex Numbers and Quadratic Equation
118216
If the roots of the equation \(x^2-p x+q=0\) differ by unity, then
1 \(\mathrm{p}^2=4 \mathrm{q}\)
2 \(\mathrm{p}^2=4 \mathrm{q}-1\)
3 \(p^2=2 q\)
4 \(p^2=4 q+1\)
Explanation:
D Given, \(x^2-p x+q=0\) let the roots be \(\alpha, \beta\) \(\therefore \alpha+\beta=\mathrm{p}, \quad \alpha \beta=\mathrm{q}\) Also \(\alpha-\beta=1\) \(\therefore(\alpha+\beta)^2=(\alpha-\beta)^2+4 \alpha \beta\) \(p^2=(1)^2+4 q\) \(p^2=1+4 q\)
SRM JEEE-2010
Complex Numbers and Quadratic Equation
118217
The condition that one root of the equation \(a x^2+b x+c=0\) may be double of the other is
1 \(\mathrm{b}^2=9 \mathrm{ac}\)
2 \(2 b^2=9 \mathrm{ac}\)
3 \(2 b^2=a c\)
4 \(b^2=\mathrm{ac}\)
Explanation:
B Let the roots of the given quadratic equation be \(\alpha, 2 \alpha\). Then, \(\alpha+2 \alpha=\frac{-b}{a} \Rightarrow \alpha=\frac{-b}{3 a}\) And, \(\alpha(2 \alpha)=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow 2 \alpha^2=\frac{\mathrm{c}}{\mathrm{a}}\) \(2\left(\frac{-\mathrm{b}}{3 \mathrm{a}}\right)^2=\frac{\mathrm{c}}{\mathrm{a}} \quad \text { (from equation (i)) }\) \(2 \frac{\mathrm{b}^2}{9 \mathrm{a}^2}=\frac{\mathrm{c}}{\mathrm{a}}\) \(2 \mathrm{~b}^2=\frac{9 \mathrm{a}^2 \mathrm{c}}{\mathrm{a}}\) \(2 \mathrm{~b}^2=9 \mathrm{ac}\)(from equation (i))
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118218
If the roots of the quadratic equation \(2 x^2-\left(a^3+1\right) x+\left(a^2-2 a\right)=0\) are real and opposite in signs, then the set of possible values of a lies in the interval
1 \((0,2)\)
2 \((0,1)\)
3 \((1,2)\)
4 \((-1,2)\)
Explanation:
A \(2 \mathrm{x}^2-\left(\mathrm{a}^3+1\right) \mathrm{x}+\left(\mathrm{a}^2-2 \mathrm{a}\right)=0\) has real and opposite signs roots. Sum of roots \(\left(x_1+x_2\right)=\frac{a^3+1}{2}\) product of roots \(x_1 \cdot x_2=\frac{a^2-2 a}{2}\) since, \(\mathrm{x}_1\) and \(\mathrm{x}_2\) are of opposite signs. \(\therefore \mathrm{x}_1 \cdot \mathrm{x}_2\lt 0\) \(\frac{\mathrm{a}^2-2 \mathrm{a}}{2}\lt 0 \Rightarrow \mathrm{a}^2-2 \mathrm{a}\lt 0\) \(\mathrm{a}(\mathrm{a}-2)\lt 0 \quad \mathrm{a} \in(0,2) .\)
118215
If one root of the equation \(5 x^2+13 x+k=0\) is reciprocal of the other, then the value of \(k\) is
1 0
2 \(1 / 6\)
3 5
4 6
Explanation:
C According to given equation, Let one root be \(\alpha\) and the other root be \(\frac{1}{\alpha}\) Here, we have \(5 \mathrm{x}^2+13 \mathrm{x}+\mathrm{k}=0\) Product of roots \(=\frac{k}{5}\) \(\alpha \times \frac{1}{\alpha}=\frac{k}{5}\) \(1=\frac{k}{5}\) \(k=5\)
SRM JEEE-2010
Complex Numbers and Quadratic Equation
118216
If the roots of the equation \(x^2-p x+q=0\) differ by unity, then
1 \(\mathrm{p}^2=4 \mathrm{q}\)
2 \(\mathrm{p}^2=4 \mathrm{q}-1\)
3 \(p^2=2 q\)
4 \(p^2=4 q+1\)
Explanation:
D Given, \(x^2-p x+q=0\) let the roots be \(\alpha, \beta\) \(\therefore \alpha+\beta=\mathrm{p}, \quad \alpha \beta=\mathrm{q}\) Also \(\alpha-\beta=1\) \(\therefore(\alpha+\beta)^2=(\alpha-\beta)^2+4 \alpha \beta\) \(p^2=(1)^2+4 q\) \(p^2=1+4 q\)
SRM JEEE-2010
Complex Numbers and Quadratic Equation
118217
The condition that one root of the equation \(a x^2+b x+c=0\) may be double of the other is
1 \(\mathrm{b}^2=9 \mathrm{ac}\)
2 \(2 b^2=9 \mathrm{ac}\)
3 \(2 b^2=a c\)
4 \(b^2=\mathrm{ac}\)
Explanation:
B Let the roots of the given quadratic equation be \(\alpha, 2 \alpha\). Then, \(\alpha+2 \alpha=\frac{-b}{a} \Rightarrow \alpha=\frac{-b}{3 a}\) And, \(\alpha(2 \alpha)=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow 2 \alpha^2=\frac{\mathrm{c}}{\mathrm{a}}\) \(2\left(\frac{-\mathrm{b}}{3 \mathrm{a}}\right)^2=\frac{\mathrm{c}}{\mathrm{a}} \quad \text { (from equation (i)) }\) \(2 \frac{\mathrm{b}^2}{9 \mathrm{a}^2}=\frac{\mathrm{c}}{\mathrm{a}}\) \(2 \mathrm{~b}^2=\frac{9 \mathrm{a}^2 \mathrm{c}}{\mathrm{a}}\) \(2 \mathrm{~b}^2=9 \mathrm{ac}\)(from equation (i))
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118218
If the roots of the quadratic equation \(2 x^2-\left(a^3+1\right) x+\left(a^2-2 a\right)=0\) are real and opposite in signs, then the set of possible values of a lies in the interval
1 \((0,2)\)
2 \((0,1)\)
3 \((1,2)\)
4 \((-1,2)\)
Explanation:
A \(2 \mathrm{x}^2-\left(\mathrm{a}^3+1\right) \mathrm{x}+\left(\mathrm{a}^2-2 \mathrm{a}\right)=0\) has real and opposite signs roots. Sum of roots \(\left(x_1+x_2\right)=\frac{a^3+1}{2}\) product of roots \(x_1 \cdot x_2=\frac{a^2-2 a}{2}\) since, \(\mathrm{x}_1\) and \(\mathrm{x}_2\) are of opposite signs. \(\therefore \mathrm{x}_1 \cdot \mathrm{x}_2\lt 0\) \(\frac{\mathrm{a}^2-2 \mathrm{a}}{2}\lt 0 \Rightarrow \mathrm{a}^2-2 \mathrm{a}\lt 0\) \(\mathrm{a}(\mathrm{a}-2)\lt 0 \quad \mathrm{a} \in(0,2) .\)
118215
If one root of the equation \(5 x^2+13 x+k=0\) is reciprocal of the other, then the value of \(k\) is
1 0
2 \(1 / 6\)
3 5
4 6
Explanation:
C According to given equation, Let one root be \(\alpha\) and the other root be \(\frac{1}{\alpha}\) Here, we have \(5 \mathrm{x}^2+13 \mathrm{x}+\mathrm{k}=0\) Product of roots \(=\frac{k}{5}\) \(\alpha \times \frac{1}{\alpha}=\frac{k}{5}\) \(1=\frac{k}{5}\) \(k=5\)
SRM JEEE-2010
Complex Numbers and Quadratic Equation
118216
If the roots of the equation \(x^2-p x+q=0\) differ by unity, then
1 \(\mathrm{p}^2=4 \mathrm{q}\)
2 \(\mathrm{p}^2=4 \mathrm{q}-1\)
3 \(p^2=2 q\)
4 \(p^2=4 q+1\)
Explanation:
D Given, \(x^2-p x+q=0\) let the roots be \(\alpha, \beta\) \(\therefore \alpha+\beta=\mathrm{p}, \quad \alpha \beta=\mathrm{q}\) Also \(\alpha-\beta=1\) \(\therefore(\alpha+\beta)^2=(\alpha-\beta)^2+4 \alpha \beta\) \(p^2=(1)^2+4 q\) \(p^2=1+4 q\)
SRM JEEE-2010
Complex Numbers and Quadratic Equation
118217
The condition that one root of the equation \(a x^2+b x+c=0\) may be double of the other is
1 \(\mathrm{b}^2=9 \mathrm{ac}\)
2 \(2 b^2=9 \mathrm{ac}\)
3 \(2 b^2=a c\)
4 \(b^2=\mathrm{ac}\)
Explanation:
B Let the roots of the given quadratic equation be \(\alpha, 2 \alpha\). Then, \(\alpha+2 \alpha=\frac{-b}{a} \Rightarrow \alpha=\frac{-b}{3 a}\) And, \(\alpha(2 \alpha)=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow 2 \alpha^2=\frac{\mathrm{c}}{\mathrm{a}}\) \(2\left(\frac{-\mathrm{b}}{3 \mathrm{a}}\right)^2=\frac{\mathrm{c}}{\mathrm{a}} \quad \text { (from equation (i)) }\) \(2 \frac{\mathrm{b}^2}{9 \mathrm{a}^2}=\frac{\mathrm{c}}{\mathrm{a}}\) \(2 \mathrm{~b}^2=\frac{9 \mathrm{a}^2 \mathrm{c}}{\mathrm{a}}\) \(2 \mathrm{~b}^2=9 \mathrm{ac}\)(from equation (i))
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118218
If the roots of the quadratic equation \(2 x^2-\left(a^3+1\right) x+\left(a^2-2 a\right)=0\) are real and opposite in signs, then the set of possible values of a lies in the interval
1 \((0,2)\)
2 \((0,1)\)
3 \((1,2)\)
4 \((-1,2)\)
Explanation:
A \(2 \mathrm{x}^2-\left(\mathrm{a}^3+1\right) \mathrm{x}+\left(\mathrm{a}^2-2 \mathrm{a}\right)=0\) has real and opposite signs roots. Sum of roots \(\left(x_1+x_2\right)=\frac{a^3+1}{2}\) product of roots \(x_1 \cdot x_2=\frac{a^2-2 a}{2}\) since, \(\mathrm{x}_1\) and \(\mathrm{x}_2\) are of opposite signs. \(\therefore \mathrm{x}_1 \cdot \mathrm{x}_2\lt 0\) \(\frac{\mathrm{a}^2-2 \mathrm{a}}{2}\lt 0 \Rightarrow \mathrm{a}^2-2 \mathrm{a}\lt 0\) \(\mathrm{a}(\mathrm{a}-2)\lt 0 \quad \mathrm{a} \in(0,2) .\)
118215
If one root of the equation \(5 x^2+13 x+k=0\) is reciprocal of the other, then the value of \(k\) is
1 0
2 \(1 / 6\)
3 5
4 6
Explanation:
C According to given equation, Let one root be \(\alpha\) and the other root be \(\frac{1}{\alpha}\) Here, we have \(5 \mathrm{x}^2+13 \mathrm{x}+\mathrm{k}=0\) Product of roots \(=\frac{k}{5}\) \(\alpha \times \frac{1}{\alpha}=\frac{k}{5}\) \(1=\frac{k}{5}\) \(k=5\)
SRM JEEE-2010
Complex Numbers and Quadratic Equation
118216
If the roots of the equation \(x^2-p x+q=0\) differ by unity, then
1 \(\mathrm{p}^2=4 \mathrm{q}\)
2 \(\mathrm{p}^2=4 \mathrm{q}-1\)
3 \(p^2=2 q\)
4 \(p^2=4 q+1\)
Explanation:
D Given, \(x^2-p x+q=0\) let the roots be \(\alpha, \beta\) \(\therefore \alpha+\beta=\mathrm{p}, \quad \alpha \beta=\mathrm{q}\) Also \(\alpha-\beta=1\) \(\therefore(\alpha+\beta)^2=(\alpha-\beta)^2+4 \alpha \beta\) \(p^2=(1)^2+4 q\) \(p^2=1+4 q\)
SRM JEEE-2010
Complex Numbers and Quadratic Equation
118217
The condition that one root of the equation \(a x^2+b x+c=0\) may be double of the other is
1 \(\mathrm{b}^2=9 \mathrm{ac}\)
2 \(2 b^2=9 \mathrm{ac}\)
3 \(2 b^2=a c\)
4 \(b^2=\mathrm{ac}\)
Explanation:
B Let the roots of the given quadratic equation be \(\alpha, 2 \alpha\). Then, \(\alpha+2 \alpha=\frac{-b}{a} \Rightarrow \alpha=\frac{-b}{3 a}\) And, \(\alpha(2 \alpha)=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow 2 \alpha^2=\frac{\mathrm{c}}{\mathrm{a}}\) \(2\left(\frac{-\mathrm{b}}{3 \mathrm{a}}\right)^2=\frac{\mathrm{c}}{\mathrm{a}} \quad \text { (from equation (i)) }\) \(2 \frac{\mathrm{b}^2}{9 \mathrm{a}^2}=\frac{\mathrm{c}}{\mathrm{a}}\) \(2 \mathrm{~b}^2=\frac{9 \mathrm{a}^2 \mathrm{c}}{\mathrm{a}}\) \(2 \mathrm{~b}^2=9 \mathrm{ac}\)(from equation (i))
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118218
If the roots of the quadratic equation \(2 x^2-\left(a^3+1\right) x+\left(a^2-2 a\right)=0\) are real and opposite in signs, then the set of possible values of a lies in the interval
1 \((0,2)\)
2 \((0,1)\)
3 \((1,2)\)
4 \((-1,2)\)
Explanation:
A \(2 \mathrm{x}^2-\left(\mathrm{a}^3+1\right) \mathrm{x}+\left(\mathrm{a}^2-2 \mathrm{a}\right)=0\) has real and opposite signs roots. Sum of roots \(\left(x_1+x_2\right)=\frac{a^3+1}{2}\) product of roots \(x_1 \cdot x_2=\frac{a^2-2 a}{2}\) since, \(\mathrm{x}_1\) and \(\mathrm{x}_2\) are of opposite signs. \(\therefore \mathrm{x}_1 \cdot \mathrm{x}_2\lt 0\) \(\frac{\mathrm{a}^2-2 \mathrm{a}}{2}\lt 0 \Rightarrow \mathrm{a}^2-2 \mathrm{a}\lt 0\) \(\mathrm{a}(\mathrm{a}-2)\lt 0 \quad \mathrm{a} \in(0,2) .\)
