118183
The square-root of \(\frac{(0.75)^3}{1-(0.75)}+\left(0.75+(0.75)^2+1\right) \text { is }\)
1 1
2 2
3 3
4 4
Explanation:
B Given, \(\frac{(0.75)^3}{1-0.75}+\left((0.75)+(0.75)^2+1\right)\) \(=\frac{\left(\frac{3}{4}\right)^3}{1-\frac{3}{4}}+\frac{3}{4}+\left(\frac{3}{4}\right)^2+1\) \(=\frac{3^3}{4^2}+\frac{12}{4^2}+\frac{9}{4^2}+\frac{16}{4^2}\) \(=\frac{1}{4^2}[27+12+4+16]=\frac{64}{16}=4\)\(\therefore\) The square root of the given expression is 2 .
C Given, \(i x^2-3 x-2 i=0\) Multiply by \(i\) to both the sides we get- \(i^2 x^2-3 x i-2 i^2=0\) \(-x^2-3 x i+2=0\) \(x^2+3 x i-2=0\) \(x=\frac{-3 i \pm \sqrt{(3 i)^2}+8}{2}\) \(=\frac{-3 i \pm \sqrt{-9+8}}{2}\) \(=\frac{-3 i \pm i}{2} \Rightarrow \frac{-2 i}{2}=-i,\) \(=\frac{-4 i}{2}=-2 i\)
AP EAMCET-23.09.2020
Complex Numbers and Quadratic Equation
118185
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-2 x^2+3 x-4=0\), then find \(\sum a \beta(\alpha+\beta)\)
1 -2
2 -6
3 6
4 2
Explanation:
B Given, \(\alpha, \beta, \gamma\) are the roots of \(\mathrm{x}^3-2 \mathrm{x}^2+3 \mathrm{x}-4=0\), then \(\alpha+\beta+\gamma=2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=4\) Now, \(\alpha \beta(\alpha+\beta)\) \(= \alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\alpha \gamma(\alpha+\gamma)\) \(= \alpha+\beta=(2-\gamma)\) \(= \beta+\gamma=(2-\alpha)\) \(= \alpha+\gamma=(2-\beta)\) \(\therefore \quad\) The above is \(\alpha \beta(2-\gamma)+\beta \gamma(2-\alpha)+\alpha \gamma(2-\beta)\) \(=2[\alpha \beta+\beta \gamma+\alpha \gamma]-3 \alpha \beta \gamma\) \(=(2 \times 3)-(3 \times 4)=6-12=-6\)
AP EAMCET-23.09.2020
Complex Numbers and Quadratic Equation
118187
The product of real of the equation \(|x|^{6 / 5}-\) 26 \(|\mathbf{x}|^{3 / 5}-27=0\)
1 \(-3^{10}\)
2 \(-3^{12}\)
3 \(-3^{12 / 5}\)
4 \(-3^{21 / 5}\)
Explanation:
A Given, \(|x|^{6 / 5}-26|x|^{3 / 5}-27=0\) Let, \(\quad|\mathrm{x}|^{3 / 5}=\mathrm{t}\), then we have \(\mathrm{t}^2-26 \mathrm{t}-27=0\) \(\mathrm{t}^2-27 \mathrm{t}+\mathrm{t}-27=0\) \(\mathrm{t}(\mathrm{t}-27)+1(\mathrm{t}-27)=0\) \((\mathrm{t}-27)(\mathrm{t}+1)=0\) \(\mathrm{t}=27,-1\) \(|x|^{3 / 5}=27=3^3\) \(\therefore \quad|\mathrm{x}|^{1 / 5}=3 \Rightarrow|\mathrm{x}|=3^5\). \(\therefore \quad \mathrm{x}= \pm 3^5\) \(|\mathrm{x}|^{3 / 5}=-1\) yields no real root. \(\therefore\) Product of the real root \(=3^5 \times\left(-3^5\right)\) \(=-3^{10}\)
118183
The square-root of \(\frac{(0.75)^3}{1-(0.75)}+\left(0.75+(0.75)^2+1\right) \text { is }\)
1 1
2 2
3 3
4 4
Explanation:
B Given, \(\frac{(0.75)^3}{1-0.75}+\left((0.75)+(0.75)^2+1\right)\) \(=\frac{\left(\frac{3}{4}\right)^3}{1-\frac{3}{4}}+\frac{3}{4}+\left(\frac{3}{4}\right)^2+1\) \(=\frac{3^3}{4^2}+\frac{12}{4^2}+\frac{9}{4^2}+\frac{16}{4^2}\) \(=\frac{1}{4^2}[27+12+4+16]=\frac{64}{16}=4\)\(\therefore\) The square root of the given expression is 2 .
C Given, \(i x^2-3 x-2 i=0\) Multiply by \(i\) to both the sides we get- \(i^2 x^2-3 x i-2 i^2=0\) \(-x^2-3 x i+2=0\) \(x^2+3 x i-2=0\) \(x=\frac{-3 i \pm \sqrt{(3 i)^2}+8}{2}\) \(=\frac{-3 i \pm \sqrt{-9+8}}{2}\) \(=\frac{-3 i \pm i}{2} \Rightarrow \frac{-2 i}{2}=-i,\) \(=\frac{-4 i}{2}=-2 i\)
AP EAMCET-23.09.2020
Complex Numbers and Quadratic Equation
118185
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-2 x^2+3 x-4=0\), then find \(\sum a \beta(\alpha+\beta)\)
1 -2
2 -6
3 6
4 2
Explanation:
B Given, \(\alpha, \beta, \gamma\) are the roots of \(\mathrm{x}^3-2 \mathrm{x}^2+3 \mathrm{x}-4=0\), then \(\alpha+\beta+\gamma=2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=4\) Now, \(\alpha \beta(\alpha+\beta)\) \(= \alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\alpha \gamma(\alpha+\gamma)\) \(= \alpha+\beta=(2-\gamma)\) \(= \beta+\gamma=(2-\alpha)\) \(= \alpha+\gamma=(2-\beta)\) \(\therefore \quad\) The above is \(\alpha \beta(2-\gamma)+\beta \gamma(2-\alpha)+\alpha \gamma(2-\beta)\) \(=2[\alpha \beta+\beta \gamma+\alpha \gamma]-3 \alpha \beta \gamma\) \(=(2 \times 3)-(3 \times 4)=6-12=-6\)
AP EAMCET-23.09.2020
Complex Numbers and Quadratic Equation
118187
The product of real of the equation \(|x|^{6 / 5}-\) 26 \(|\mathbf{x}|^{3 / 5}-27=0\)
1 \(-3^{10}\)
2 \(-3^{12}\)
3 \(-3^{12 / 5}\)
4 \(-3^{21 / 5}\)
Explanation:
A Given, \(|x|^{6 / 5}-26|x|^{3 / 5}-27=0\) Let, \(\quad|\mathrm{x}|^{3 / 5}=\mathrm{t}\), then we have \(\mathrm{t}^2-26 \mathrm{t}-27=0\) \(\mathrm{t}^2-27 \mathrm{t}+\mathrm{t}-27=0\) \(\mathrm{t}(\mathrm{t}-27)+1(\mathrm{t}-27)=0\) \((\mathrm{t}-27)(\mathrm{t}+1)=0\) \(\mathrm{t}=27,-1\) \(|x|^{3 / 5}=27=3^3\) \(\therefore \quad|\mathrm{x}|^{1 / 5}=3 \Rightarrow|\mathrm{x}|=3^5\). \(\therefore \quad \mathrm{x}= \pm 3^5\) \(|\mathrm{x}|^{3 / 5}=-1\) yields no real root. \(\therefore\) Product of the real root \(=3^5 \times\left(-3^5\right)\) \(=-3^{10}\)
118183
The square-root of \(\frac{(0.75)^3}{1-(0.75)}+\left(0.75+(0.75)^2+1\right) \text { is }\)
1 1
2 2
3 3
4 4
Explanation:
B Given, \(\frac{(0.75)^3}{1-0.75}+\left((0.75)+(0.75)^2+1\right)\) \(=\frac{\left(\frac{3}{4}\right)^3}{1-\frac{3}{4}}+\frac{3}{4}+\left(\frac{3}{4}\right)^2+1\) \(=\frac{3^3}{4^2}+\frac{12}{4^2}+\frac{9}{4^2}+\frac{16}{4^2}\) \(=\frac{1}{4^2}[27+12+4+16]=\frac{64}{16}=4\)\(\therefore\) The square root of the given expression is 2 .
C Given, \(i x^2-3 x-2 i=0\) Multiply by \(i\) to both the sides we get- \(i^2 x^2-3 x i-2 i^2=0\) \(-x^2-3 x i+2=0\) \(x^2+3 x i-2=0\) \(x=\frac{-3 i \pm \sqrt{(3 i)^2}+8}{2}\) \(=\frac{-3 i \pm \sqrt{-9+8}}{2}\) \(=\frac{-3 i \pm i}{2} \Rightarrow \frac{-2 i}{2}=-i,\) \(=\frac{-4 i}{2}=-2 i\)
AP EAMCET-23.09.2020
Complex Numbers and Quadratic Equation
118185
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-2 x^2+3 x-4=0\), then find \(\sum a \beta(\alpha+\beta)\)
1 -2
2 -6
3 6
4 2
Explanation:
B Given, \(\alpha, \beta, \gamma\) are the roots of \(\mathrm{x}^3-2 \mathrm{x}^2+3 \mathrm{x}-4=0\), then \(\alpha+\beta+\gamma=2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=4\) Now, \(\alpha \beta(\alpha+\beta)\) \(= \alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\alpha \gamma(\alpha+\gamma)\) \(= \alpha+\beta=(2-\gamma)\) \(= \beta+\gamma=(2-\alpha)\) \(= \alpha+\gamma=(2-\beta)\) \(\therefore \quad\) The above is \(\alpha \beta(2-\gamma)+\beta \gamma(2-\alpha)+\alpha \gamma(2-\beta)\) \(=2[\alpha \beta+\beta \gamma+\alpha \gamma]-3 \alpha \beta \gamma\) \(=(2 \times 3)-(3 \times 4)=6-12=-6\)
AP EAMCET-23.09.2020
Complex Numbers and Quadratic Equation
118187
The product of real of the equation \(|x|^{6 / 5}-\) 26 \(|\mathbf{x}|^{3 / 5}-27=0\)
1 \(-3^{10}\)
2 \(-3^{12}\)
3 \(-3^{12 / 5}\)
4 \(-3^{21 / 5}\)
Explanation:
A Given, \(|x|^{6 / 5}-26|x|^{3 / 5}-27=0\) Let, \(\quad|\mathrm{x}|^{3 / 5}=\mathrm{t}\), then we have \(\mathrm{t}^2-26 \mathrm{t}-27=0\) \(\mathrm{t}^2-27 \mathrm{t}+\mathrm{t}-27=0\) \(\mathrm{t}(\mathrm{t}-27)+1(\mathrm{t}-27)=0\) \((\mathrm{t}-27)(\mathrm{t}+1)=0\) \(\mathrm{t}=27,-1\) \(|x|^{3 / 5}=27=3^3\) \(\therefore \quad|\mathrm{x}|^{1 / 5}=3 \Rightarrow|\mathrm{x}|=3^5\). \(\therefore \quad \mathrm{x}= \pm 3^5\) \(|\mathrm{x}|^{3 / 5}=-1\) yields no real root. \(\therefore\) Product of the real root \(=3^5 \times\left(-3^5\right)\) \(=-3^{10}\)
118183
The square-root of \(\frac{(0.75)^3}{1-(0.75)}+\left(0.75+(0.75)^2+1\right) \text { is }\)
1 1
2 2
3 3
4 4
Explanation:
B Given, \(\frac{(0.75)^3}{1-0.75}+\left((0.75)+(0.75)^2+1\right)\) \(=\frac{\left(\frac{3}{4}\right)^3}{1-\frac{3}{4}}+\frac{3}{4}+\left(\frac{3}{4}\right)^2+1\) \(=\frac{3^3}{4^2}+\frac{12}{4^2}+\frac{9}{4^2}+\frac{16}{4^2}\) \(=\frac{1}{4^2}[27+12+4+16]=\frac{64}{16}=4\)\(\therefore\) The square root of the given expression is 2 .
C Given, \(i x^2-3 x-2 i=0\) Multiply by \(i\) to both the sides we get- \(i^2 x^2-3 x i-2 i^2=0\) \(-x^2-3 x i+2=0\) \(x^2+3 x i-2=0\) \(x=\frac{-3 i \pm \sqrt{(3 i)^2}+8}{2}\) \(=\frac{-3 i \pm \sqrt{-9+8}}{2}\) \(=\frac{-3 i \pm i}{2} \Rightarrow \frac{-2 i}{2}=-i,\) \(=\frac{-4 i}{2}=-2 i\)
AP EAMCET-23.09.2020
Complex Numbers and Quadratic Equation
118185
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-2 x^2+3 x-4=0\), then find \(\sum a \beta(\alpha+\beta)\)
1 -2
2 -6
3 6
4 2
Explanation:
B Given, \(\alpha, \beta, \gamma\) are the roots of \(\mathrm{x}^3-2 \mathrm{x}^2+3 \mathrm{x}-4=0\), then \(\alpha+\beta+\gamma=2\) \(\alpha \beta+\beta \gamma+\alpha \gamma=3\) \(\alpha \beta \gamma=4\) Now, \(\alpha \beta(\alpha+\beta)\) \(= \alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\alpha \gamma(\alpha+\gamma)\) \(= \alpha+\beta=(2-\gamma)\) \(= \beta+\gamma=(2-\alpha)\) \(= \alpha+\gamma=(2-\beta)\) \(\therefore \quad\) The above is \(\alpha \beta(2-\gamma)+\beta \gamma(2-\alpha)+\alpha \gamma(2-\beta)\) \(=2[\alpha \beta+\beta \gamma+\alpha \gamma]-3 \alpha \beta \gamma\) \(=(2 \times 3)-(3 \times 4)=6-12=-6\)
AP EAMCET-23.09.2020
Complex Numbers and Quadratic Equation
118187
The product of real of the equation \(|x|^{6 / 5}-\) 26 \(|\mathbf{x}|^{3 / 5}-27=0\)
1 \(-3^{10}\)
2 \(-3^{12}\)
3 \(-3^{12 / 5}\)
4 \(-3^{21 / 5}\)
Explanation:
A Given, \(|x|^{6 / 5}-26|x|^{3 / 5}-27=0\) Let, \(\quad|\mathrm{x}|^{3 / 5}=\mathrm{t}\), then we have \(\mathrm{t}^2-26 \mathrm{t}-27=0\) \(\mathrm{t}^2-27 \mathrm{t}+\mathrm{t}-27=0\) \(\mathrm{t}(\mathrm{t}-27)+1(\mathrm{t}-27)=0\) \((\mathrm{t}-27)(\mathrm{t}+1)=0\) \(\mathrm{t}=27,-1\) \(|x|^{3 / 5}=27=3^3\) \(\therefore \quad|\mathrm{x}|^{1 / 5}=3 \Rightarrow|\mathrm{x}|=3^5\). \(\therefore \quad \mathrm{x}= \pm 3^5\) \(|\mathrm{x}|^{3 / 5}=-1\) yields no real root. \(\therefore\) Product of the real root \(=3^5 \times\left(-3^5\right)\) \(=-3^{10}\)
