118059
If \(a,-a, b\) are the roots of \(x^3-5 x^2-x+5=0\), then \(b\) is a root of
1 \(x^2-5 x+10=0\)
2 \(x^2+3 x-20=0\)
3 \(x^2+5 x-30=0\)
4 \(x^2-3 x-10=0\)
Explanation:
D Given, \(x^3-5 x^2-x+5=0\) are roots \(a,-a\), and \(b\) sum of roots, \(a-a+b=5 \Rightarrow b=5\) \(b=5\) satisfies the equation, \(f(x)=x^2-3 x-10=0\) \(f(5)=5^2-3 \times 5-10\) \(=25-15-10\) \(=0\) So, \(\mathrm{b}\) is the roots equation \(\mathrm{x}^2-3 \mathrm{x}-10=0\)
Karnataka CET-2010
Complex Numbers and Quadratic Equation
118070
If \(\alpha\) and \(\beta\) are roots of \(x^2-x+1=0\), then the equation whose roots are \(\alpha^{100}\) and \(\beta^{100}\) are
1 \(\mathrm{x}^2-\mathrm{x}+1=0\)
2 \(x^2+x-1=0\)
3 \(x^2-x-1=0\)
4 \(x^2+x+1=0\)
Explanation:
D We have \(\mathrm{x}^3+1 \equiv(\mathrm{x}+1)\left(\mathrm{x}^2-\mathrm{x}+1\right)\). Therefore, \(\alpha\) and \(\beta\) are the complex cube roots of -1 so that we may take \(\alpha=-\omega\) and \(\beta=-\omega^2\), where \(\omega \neq 1\) is a cube root of unity. Thus \(\alpha^{100}=(-\omega)^{100}=\omega\) and \(\beta^{100}=\left(-\omega^2\right)^{100}=\omega^2\) So, that the required equation is \(\mathrm{x}^2+\mathrm{x}+1=0\).
BITSAT-2013
Complex Numbers and Quadratic Equation
118072
If \(\alpha, \beta\) are the roots of the equation \(a x^2+b x+c=0\), then the roots of the equation \(\mathbf{a x ^ { 2 }}+\mathbf{b x}(\mathbf{x}+\mathbf{1})+\mathbf{c}(\mathbf{x}+\mathbf{1})^2=\mathbf{0}\) are
D The second equation can be rewritten as \(a\left(\frac{x}{x+1}\right)^2+b\left(\frac{x}{x+1}\right)+c=0\) and hence its roots correspond to \(\frac{x}{x+1}=\alpha\) and \(\frac{x}{x+1}=\beta\). Hence \(x=\frac{\alpha}{1-\alpha}\) and \(\frac{\beta}{1-\beta}\).
BITSAT-2011
Complex Numbers and Quadratic Equation
118103
Let \(\alpha, \beta\) be the roots of the quadratic equation \(\mathbf{x}^2+\sqrt{6} \mathrm{x}+3=0\). Then \(\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}\) is equal to
118059
If \(a,-a, b\) are the roots of \(x^3-5 x^2-x+5=0\), then \(b\) is a root of
1 \(x^2-5 x+10=0\)
2 \(x^2+3 x-20=0\)
3 \(x^2+5 x-30=0\)
4 \(x^2-3 x-10=0\)
Explanation:
D Given, \(x^3-5 x^2-x+5=0\) are roots \(a,-a\), and \(b\) sum of roots, \(a-a+b=5 \Rightarrow b=5\) \(b=5\) satisfies the equation, \(f(x)=x^2-3 x-10=0\) \(f(5)=5^2-3 \times 5-10\) \(=25-15-10\) \(=0\) So, \(\mathrm{b}\) is the roots equation \(\mathrm{x}^2-3 \mathrm{x}-10=0\)
Karnataka CET-2010
Complex Numbers and Quadratic Equation
118070
If \(\alpha\) and \(\beta\) are roots of \(x^2-x+1=0\), then the equation whose roots are \(\alpha^{100}\) and \(\beta^{100}\) are
1 \(\mathrm{x}^2-\mathrm{x}+1=0\)
2 \(x^2+x-1=0\)
3 \(x^2-x-1=0\)
4 \(x^2+x+1=0\)
Explanation:
D We have \(\mathrm{x}^3+1 \equiv(\mathrm{x}+1)\left(\mathrm{x}^2-\mathrm{x}+1\right)\). Therefore, \(\alpha\) and \(\beta\) are the complex cube roots of -1 so that we may take \(\alpha=-\omega\) and \(\beta=-\omega^2\), where \(\omega \neq 1\) is a cube root of unity. Thus \(\alpha^{100}=(-\omega)^{100}=\omega\) and \(\beta^{100}=\left(-\omega^2\right)^{100}=\omega^2\) So, that the required equation is \(\mathrm{x}^2+\mathrm{x}+1=0\).
BITSAT-2013
Complex Numbers and Quadratic Equation
118072
If \(\alpha, \beta\) are the roots of the equation \(a x^2+b x+c=0\), then the roots of the equation \(\mathbf{a x ^ { 2 }}+\mathbf{b x}(\mathbf{x}+\mathbf{1})+\mathbf{c}(\mathbf{x}+\mathbf{1})^2=\mathbf{0}\) are
D The second equation can be rewritten as \(a\left(\frac{x}{x+1}\right)^2+b\left(\frac{x}{x+1}\right)+c=0\) and hence its roots correspond to \(\frac{x}{x+1}=\alpha\) and \(\frac{x}{x+1}=\beta\). Hence \(x=\frac{\alpha}{1-\alpha}\) and \(\frac{\beta}{1-\beta}\).
BITSAT-2011
Complex Numbers and Quadratic Equation
118103
Let \(\alpha, \beta\) be the roots of the quadratic equation \(\mathbf{x}^2+\sqrt{6} \mathrm{x}+3=0\). Then \(\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}\) is equal to
118059
If \(a,-a, b\) are the roots of \(x^3-5 x^2-x+5=0\), then \(b\) is a root of
1 \(x^2-5 x+10=0\)
2 \(x^2+3 x-20=0\)
3 \(x^2+5 x-30=0\)
4 \(x^2-3 x-10=0\)
Explanation:
D Given, \(x^3-5 x^2-x+5=0\) are roots \(a,-a\), and \(b\) sum of roots, \(a-a+b=5 \Rightarrow b=5\) \(b=5\) satisfies the equation, \(f(x)=x^2-3 x-10=0\) \(f(5)=5^2-3 \times 5-10\) \(=25-15-10\) \(=0\) So, \(\mathrm{b}\) is the roots equation \(\mathrm{x}^2-3 \mathrm{x}-10=0\)
Karnataka CET-2010
Complex Numbers and Quadratic Equation
118070
If \(\alpha\) and \(\beta\) are roots of \(x^2-x+1=0\), then the equation whose roots are \(\alpha^{100}\) and \(\beta^{100}\) are
1 \(\mathrm{x}^2-\mathrm{x}+1=0\)
2 \(x^2+x-1=0\)
3 \(x^2-x-1=0\)
4 \(x^2+x+1=0\)
Explanation:
D We have \(\mathrm{x}^3+1 \equiv(\mathrm{x}+1)\left(\mathrm{x}^2-\mathrm{x}+1\right)\). Therefore, \(\alpha\) and \(\beta\) are the complex cube roots of -1 so that we may take \(\alpha=-\omega\) and \(\beta=-\omega^2\), where \(\omega \neq 1\) is a cube root of unity. Thus \(\alpha^{100}=(-\omega)^{100}=\omega\) and \(\beta^{100}=\left(-\omega^2\right)^{100}=\omega^2\) So, that the required equation is \(\mathrm{x}^2+\mathrm{x}+1=0\).
BITSAT-2013
Complex Numbers and Quadratic Equation
118072
If \(\alpha, \beta\) are the roots of the equation \(a x^2+b x+c=0\), then the roots of the equation \(\mathbf{a x ^ { 2 }}+\mathbf{b x}(\mathbf{x}+\mathbf{1})+\mathbf{c}(\mathbf{x}+\mathbf{1})^2=\mathbf{0}\) are
D The second equation can be rewritten as \(a\left(\frac{x}{x+1}\right)^2+b\left(\frac{x}{x+1}\right)+c=0\) and hence its roots correspond to \(\frac{x}{x+1}=\alpha\) and \(\frac{x}{x+1}=\beta\). Hence \(x=\frac{\alpha}{1-\alpha}\) and \(\frac{\beta}{1-\beta}\).
BITSAT-2011
Complex Numbers and Quadratic Equation
118103
Let \(\alpha, \beta\) be the roots of the quadratic equation \(\mathbf{x}^2+\sqrt{6} \mathrm{x}+3=0\). Then \(\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}\) is equal to
118059
If \(a,-a, b\) are the roots of \(x^3-5 x^2-x+5=0\), then \(b\) is a root of
1 \(x^2-5 x+10=0\)
2 \(x^2+3 x-20=0\)
3 \(x^2+5 x-30=0\)
4 \(x^2-3 x-10=0\)
Explanation:
D Given, \(x^3-5 x^2-x+5=0\) are roots \(a,-a\), and \(b\) sum of roots, \(a-a+b=5 \Rightarrow b=5\) \(b=5\) satisfies the equation, \(f(x)=x^2-3 x-10=0\) \(f(5)=5^2-3 \times 5-10\) \(=25-15-10\) \(=0\) So, \(\mathrm{b}\) is the roots equation \(\mathrm{x}^2-3 \mathrm{x}-10=0\)
Karnataka CET-2010
Complex Numbers and Quadratic Equation
118070
If \(\alpha\) and \(\beta\) are roots of \(x^2-x+1=0\), then the equation whose roots are \(\alpha^{100}\) and \(\beta^{100}\) are
1 \(\mathrm{x}^2-\mathrm{x}+1=0\)
2 \(x^2+x-1=0\)
3 \(x^2-x-1=0\)
4 \(x^2+x+1=0\)
Explanation:
D We have \(\mathrm{x}^3+1 \equiv(\mathrm{x}+1)\left(\mathrm{x}^2-\mathrm{x}+1\right)\). Therefore, \(\alpha\) and \(\beta\) are the complex cube roots of -1 so that we may take \(\alpha=-\omega\) and \(\beta=-\omega^2\), where \(\omega \neq 1\) is a cube root of unity. Thus \(\alpha^{100}=(-\omega)^{100}=\omega\) and \(\beta^{100}=\left(-\omega^2\right)^{100}=\omega^2\) So, that the required equation is \(\mathrm{x}^2+\mathrm{x}+1=0\).
BITSAT-2013
Complex Numbers and Quadratic Equation
118072
If \(\alpha, \beta\) are the roots of the equation \(a x^2+b x+c=0\), then the roots of the equation \(\mathbf{a x ^ { 2 }}+\mathbf{b x}(\mathbf{x}+\mathbf{1})+\mathbf{c}(\mathbf{x}+\mathbf{1})^2=\mathbf{0}\) are
D The second equation can be rewritten as \(a\left(\frac{x}{x+1}\right)^2+b\left(\frac{x}{x+1}\right)+c=0\) and hence its roots correspond to \(\frac{x}{x+1}=\alpha\) and \(\frac{x}{x+1}=\beta\). Hence \(x=\frac{\alpha}{1-\alpha}\) and \(\frac{\beta}{1-\beta}\).
BITSAT-2011
Complex Numbers and Quadratic Equation
118103
Let \(\alpha, \beta\) be the roots of the quadratic equation \(\mathbf{x}^2+\sqrt{6} \mathrm{x}+3=0\). Then \(\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}\) is equal to
