118058
The real root of the equation \(x^3-6 x+9=0\) is
1 6
2 -3
3 -6
4 -9
Explanation:
B Given, \(x^3-6 x+9=0\) Now, put \(\mathrm{x}=-3\) \((-3)^3-6(-3)+9\) \(=-27+18+9=0\) \(\therefore \quad \mathrm{x}=-3\) is root of \(\mathrm{eq}^{\mathrm{n}} \mathrm{x}^3-6 \mathrm{x}+9=0\)
Karnataka CET-2008
Complex Numbers and Quadratic Equation
118060
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3-8 x+8=0\), then \(\quad \Sigma \alpha^2\) and \(\quad \Sigma \frac{1}{\alpha \beta}\) are respectively
118061
If \(2 x=-1+\sqrt{3} i\), then the value of \(\left(1-x^2+x\right)^6-\left(1-x+x^2\right)^6=\)
1 32
2 -64
3 64
4 0
Explanation:
D Given, \(2 \mathrm{x}=-1+\sqrt{3} \mathrm{i}\) Then, \(x=-\frac{1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) We know that, \(\omega=-\frac{1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) Where \(\omega\) is cube root of unity \(\therefore \mathrm{x}=\omega=\frac{-1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) Now, \(\left(1-\mathrm{x}^2+\mathrm{x}\right)^{\frac{2}{6}}-\left(1-\mathrm{x}+\mathrm{x}^2\right)^6\) \(=\left(1-\omega^2+\omega\right)^6-\left(1-\omega+\omega^2\right)^6\) \(=\left(-\omega^2-\omega^2\right)^6-(-\omega-\omega)^6\left\{\because 1+\omega+\omega^2=0\right\}\) \(=\left(-2 \omega^2\right)^6-(-2 \omega)^6\) \(=2^6-2^6=0 \quad\left\{\because \omega^3=1\right\}\)
Karnataka CET-2006
Complex Numbers and Quadratic Equation
118062
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-3 x^2+2 x-1=0\) then the value of \([(\mathbf{1}-\boldsymbol{\alpha})(\mathbf{1}-\boldsymbol{\beta})(\mathbf{1}-\gamma)]\) is
118058
The real root of the equation \(x^3-6 x+9=0\) is
1 6
2 -3
3 -6
4 -9
Explanation:
B Given, \(x^3-6 x+9=0\) Now, put \(\mathrm{x}=-3\) \((-3)^3-6(-3)+9\) \(=-27+18+9=0\) \(\therefore \quad \mathrm{x}=-3\) is root of \(\mathrm{eq}^{\mathrm{n}} \mathrm{x}^3-6 \mathrm{x}+9=0\)
Karnataka CET-2008
Complex Numbers and Quadratic Equation
118060
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3-8 x+8=0\), then \(\quad \Sigma \alpha^2\) and \(\quad \Sigma \frac{1}{\alpha \beta}\) are respectively
118061
If \(2 x=-1+\sqrt{3} i\), then the value of \(\left(1-x^2+x\right)^6-\left(1-x+x^2\right)^6=\)
1 32
2 -64
3 64
4 0
Explanation:
D Given, \(2 \mathrm{x}=-1+\sqrt{3} \mathrm{i}\) Then, \(x=-\frac{1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) We know that, \(\omega=-\frac{1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) Where \(\omega\) is cube root of unity \(\therefore \mathrm{x}=\omega=\frac{-1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) Now, \(\left(1-\mathrm{x}^2+\mathrm{x}\right)^{\frac{2}{6}}-\left(1-\mathrm{x}+\mathrm{x}^2\right)^6\) \(=\left(1-\omega^2+\omega\right)^6-\left(1-\omega+\omega^2\right)^6\) \(=\left(-\omega^2-\omega^2\right)^6-(-\omega-\omega)^6\left\{\because 1+\omega+\omega^2=0\right\}\) \(=\left(-2 \omega^2\right)^6-(-2 \omega)^6\) \(=2^6-2^6=0 \quad\left\{\because \omega^3=1\right\}\)
Karnataka CET-2006
Complex Numbers and Quadratic Equation
118062
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-3 x^2+2 x-1=0\) then the value of \([(\mathbf{1}-\boldsymbol{\alpha})(\mathbf{1}-\boldsymbol{\beta})(\mathbf{1}-\gamma)]\) is
118058
The real root of the equation \(x^3-6 x+9=0\) is
1 6
2 -3
3 -6
4 -9
Explanation:
B Given, \(x^3-6 x+9=0\) Now, put \(\mathrm{x}=-3\) \((-3)^3-6(-3)+9\) \(=-27+18+9=0\) \(\therefore \quad \mathrm{x}=-3\) is root of \(\mathrm{eq}^{\mathrm{n}} \mathrm{x}^3-6 \mathrm{x}+9=0\)
Karnataka CET-2008
Complex Numbers and Quadratic Equation
118060
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3-8 x+8=0\), then \(\quad \Sigma \alpha^2\) and \(\quad \Sigma \frac{1}{\alpha \beta}\) are respectively
118061
If \(2 x=-1+\sqrt{3} i\), then the value of \(\left(1-x^2+x\right)^6-\left(1-x+x^2\right)^6=\)
1 32
2 -64
3 64
4 0
Explanation:
D Given, \(2 \mathrm{x}=-1+\sqrt{3} \mathrm{i}\) Then, \(x=-\frac{1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) We know that, \(\omega=-\frac{1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) Where \(\omega\) is cube root of unity \(\therefore \mathrm{x}=\omega=\frac{-1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) Now, \(\left(1-\mathrm{x}^2+\mathrm{x}\right)^{\frac{2}{6}}-\left(1-\mathrm{x}+\mathrm{x}^2\right)^6\) \(=\left(1-\omega^2+\omega\right)^6-\left(1-\omega+\omega^2\right)^6\) \(=\left(-\omega^2-\omega^2\right)^6-(-\omega-\omega)^6\left\{\because 1+\omega+\omega^2=0\right\}\) \(=\left(-2 \omega^2\right)^6-(-2 \omega)^6\) \(=2^6-2^6=0 \quad\left\{\because \omega^3=1\right\}\)
Karnataka CET-2006
Complex Numbers and Quadratic Equation
118062
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-3 x^2+2 x-1=0\) then the value of \([(\mathbf{1}-\boldsymbol{\alpha})(\mathbf{1}-\boldsymbol{\beta})(\mathbf{1}-\gamma)]\) is
118058
The real root of the equation \(x^3-6 x+9=0\) is
1 6
2 -3
3 -6
4 -9
Explanation:
B Given, \(x^3-6 x+9=0\) Now, put \(\mathrm{x}=-3\) \((-3)^3-6(-3)+9\) \(=-27+18+9=0\) \(\therefore \quad \mathrm{x}=-3\) is root of \(\mathrm{eq}^{\mathrm{n}} \mathrm{x}^3-6 \mathrm{x}+9=0\)
Karnataka CET-2008
Complex Numbers and Quadratic Equation
118060
If \(\alpha, \beta\) and \(\gamma\) are the roots of the equation \(x^3-8 x+8=0\), then \(\quad \Sigma \alpha^2\) and \(\quad \Sigma \frac{1}{\alpha \beta}\) are respectively
118061
If \(2 x=-1+\sqrt{3} i\), then the value of \(\left(1-x^2+x\right)^6-\left(1-x+x^2\right)^6=\)
1 32
2 -64
3 64
4 0
Explanation:
D Given, \(2 \mathrm{x}=-1+\sqrt{3} \mathrm{i}\) Then, \(x=-\frac{1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) We know that, \(\omega=-\frac{1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) Where \(\omega\) is cube root of unity \(\therefore \mathrm{x}=\omega=\frac{-1}{2}+\frac{\sqrt{3} \mathrm{i}}{2}\) Now, \(\left(1-\mathrm{x}^2+\mathrm{x}\right)^{\frac{2}{6}}-\left(1-\mathrm{x}+\mathrm{x}^2\right)^6\) \(=\left(1-\omega^2+\omega\right)^6-\left(1-\omega+\omega^2\right)^6\) \(=\left(-\omega^2-\omega^2\right)^6-(-\omega-\omega)^6\left\{\because 1+\omega+\omega^2=0\right\}\) \(=\left(-2 \omega^2\right)^6-(-2 \omega)^6\) \(=2^6-2^6=0 \quad\left\{\because \omega^3=1\right\}\)
Karnataka CET-2006
Complex Numbers and Quadratic Equation
118062
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^3-3 x^2+2 x-1=0\) then the value of \([(\mathbf{1}-\boldsymbol{\alpha})(\mathbf{1}-\boldsymbol{\beta})(\mathbf{1}-\gamma)]\) is
