118463
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-6 x^2+11 x-6=\) 0 , then the equation having the \(\alpha^2, \beta^2, \gamma^2\), and \(\gamma^2\) \(+\boldsymbol{\alpha}^2\) is
A Given, \(x^3-6 x^2+11 x-6=0\) \((x-1)(x-2)(x-3)=0\) \(x=1,2,3\) \(\because \alpha, \beta, \mathrm{r}\) are the roots of the equation (i) So, \(\alpha=1, \beta=2, r=3\) Therefore, \(\alpha^2+\beta^2=(1)^2+(2)^2=5=\alpha^{\prime}\) (say) \(\beta^2+\mathrm{r}^2=(2)^2+(3)^2\) \(\beta^2+\mathrm{r}^2=13=\beta^{\prime} \text { (say) }\) And, \(\gamma^2+\alpha^2=(3)^2+1=10=\mathrm{r}^{\prime}\) (say) Equation of the having the roots \(\alpha^{\prime}, \beta^{\prime}\) and \(r^{\prime}\), \(x^3-\left(\alpha^{\prime}+\beta^{\prime}+r^{\prime}\right) x^2+\left(\alpha^{\prime} \beta^{\prime}+\beta^{\prime} r^{\prime}+r^{\prime} \alpha^{\prime}\right) x-\alpha^{\prime} \beta^{\prime} r=0\) \(x^3-(5+13+10) x^2+(5 \times 13+13 \times 10+10 \times 5) x-5\) \(\times 13 \times 10=0\) \(x^3-28 x^2+245 x-650=0\)
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118464
Which among the following equations has roots which are negatives of the roots of the equation \(x^3-x^2+x-4=0\) ?
1 \(x^3-x^2+x-4=0\)
2 \(x^3+x^2+x+4=0\)
3 \(x^3-x^2+x-4=0\)
4 \(x^3-x^2-x-4=0\)
Explanation:
B Let \(\alpha\) is the roots of equation, \(x^3-x^2+x-4=0\) Now, put \((-\alpha)=x\), \((-x)^3-(-x)^2+(-x)-4=0\) \(-x^3-x^2-x-4=0\) \(x^3+x^2+x+4=0\)
AP EAMCET-18.09.2020
Complex Numbers and Quadratic Equation
118465
If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13 , then find that equation.
1 \(x^2+x-6=0\)
2 \(x^2-x+6=0\)
3 \(x^2-x-6=0\)
4 \(x^2+x+6=0\)
Explanation:
C Let \(\alpha\) and \(\beta\) are roots of the equation, Given, Sum of \(\operatorname{root}(\alpha+\beta)=1\) And sum of square of the roots \(\left(\alpha^2+\beta^2\right)=13\) \(\therefore \quad \alpha \beta=\frac{1}{2}\left[(\alpha+\beta)^2-\left(\alpha^2+\beta^2\right)\right]\) \(\quad=\frac{1}{2}[(1-13)]=-6\) Equation of required quadratic is, \(x^2-(\alpha+\beta) x+\alpha \cdot \beta=0\) \(x^2-x-6=0\)
AP EAMCET-18.09.2020
Complex Numbers and Quadratic Equation
118467
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^4+3 x^3\) \(-6 x^2+2 x-4=0\), then find the equation having \(\operatorname{roots} \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)
A Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(\mathrm{x}^4+3 \mathrm{x}^3-6 \mathrm{x}^2+2 \mathrm{x}-4=0\) Required equation having roots \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\) can be obtained by replacing \(\mathrm{x}\) by \(\frac{1}{\mathrm{x}}\) in given equation, \(\left(\frac{1}{x}\right)^4+3\left(\frac{1}{x}\right)^3-6\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x}\right)-4=0\) \(4 x^4-2 x^3+6 x^2-3 x-1=0\)
118463
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-6 x^2+11 x-6=\) 0 , then the equation having the \(\alpha^2, \beta^2, \gamma^2\), and \(\gamma^2\) \(+\boldsymbol{\alpha}^2\) is
A Given, \(x^3-6 x^2+11 x-6=0\) \((x-1)(x-2)(x-3)=0\) \(x=1,2,3\) \(\because \alpha, \beta, \mathrm{r}\) are the roots of the equation (i) So, \(\alpha=1, \beta=2, r=3\) Therefore, \(\alpha^2+\beta^2=(1)^2+(2)^2=5=\alpha^{\prime}\) (say) \(\beta^2+\mathrm{r}^2=(2)^2+(3)^2\) \(\beta^2+\mathrm{r}^2=13=\beta^{\prime} \text { (say) }\) And, \(\gamma^2+\alpha^2=(3)^2+1=10=\mathrm{r}^{\prime}\) (say) Equation of the having the roots \(\alpha^{\prime}, \beta^{\prime}\) and \(r^{\prime}\), \(x^3-\left(\alpha^{\prime}+\beta^{\prime}+r^{\prime}\right) x^2+\left(\alpha^{\prime} \beta^{\prime}+\beta^{\prime} r^{\prime}+r^{\prime} \alpha^{\prime}\right) x-\alpha^{\prime} \beta^{\prime} r=0\) \(x^3-(5+13+10) x^2+(5 \times 13+13 \times 10+10 \times 5) x-5\) \(\times 13 \times 10=0\) \(x^3-28 x^2+245 x-650=0\)
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118464
Which among the following equations has roots which are negatives of the roots of the equation \(x^3-x^2+x-4=0\) ?
1 \(x^3-x^2+x-4=0\)
2 \(x^3+x^2+x+4=0\)
3 \(x^3-x^2+x-4=0\)
4 \(x^3-x^2-x-4=0\)
Explanation:
B Let \(\alpha\) is the roots of equation, \(x^3-x^2+x-4=0\) Now, put \((-\alpha)=x\), \((-x)^3-(-x)^2+(-x)-4=0\) \(-x^3-x^2-x-4=0\) \(x^3+x^2+x+4=0\)
AP EAMCET-18.09.2020
Complex Numbers and Quadratic Equation
118465
If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13 , then find that equation.
1 \(x^2+x-6=0\)
2 \(x^2-x+6=0\)
3 \(x^2-x-6=0\)
4 \(x^2+x+6=0\)
Explanation:
C Let \(\alpha\) and \(\beta\) are roots of the equation, Given, Sum of \(\operatorname{root}(\alpha+\beta)=1\) And sum of square of the roots \(\left(\alpha^2+\beta^2\right)=13\) \(\therefore \quad \alpha \beta=\frac{1}{2}\left[(\alpha+\beta)^2-\left(\alpha^2+\beta^2\right)\right]\) \(\quad=\frac{1}{2}[(1-13)]=-6\) Equation of required quadratic is, \(x^2-(\alpha+\beta) x+\alpha \cdot \beta=0\) \(x^2-x-6=0\)
AP EAMCET-18.09.2020
Complex Numbers and Quadratic Equation
118467
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^4+3 x^3\) \(-6 x^2+2 x-4=0\), then find the equation having \(\operatorname{roots} \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)
A Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(\mathrm{x}^4+3 \mathrm{x}^3-6 \mathrm{x}^2+2 \mathrm{x}-4=0\) Required equation having roots \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\) can be obtained by replacing \(\mathrm{x}\) by \(\frac{1}{\mathrm{x}}\) in given equation, \(\left(\frac{1}{x}\right)^4+3\left(\frac{1}{x}\right)^3-6\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x}\right)-4=0\) \(4 x^4-2 x^3+6 x^2-3 x-1=0\)
118463
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-6 x^2+11 x-6=\) 0 , then the equation having the \(\alpha^2, \beta^2, \gamma^2\), and \(\gamma^2\) \(+\boldsymbol{\alpha}^2\) is
A Given, \(x^3-6 x^2+11 x-6=0\) \((x-1)(x-2)(x-3)=0\) \(x=1,2,3\) \(\because \alpha, \beta, \mathrm{r}\) are the roots of the equation (i) So, \(\alpha=1, \beta=2, r=3\) Therefore, \(\alpha^2+\beta^2=(1)^2+(2)^2=5=\alpha^{\prime}\) (say) \(\beta^2+\mathrm{r}^2=(2)^2+(3)^2\) \(\beta^2+\mathrm{r}^2=13=\beta^{\prime} \text { (say) }\) And, \(\gamma^2+\alpha^2=(3)^2+1=10=\mathrm{r}^{\prime}\) (say) Equation of the having the roots \(\alpha^{\prime}, \beta^{\prime}\) and \(r^{\prime}\), \(x^3-\left(\alpha^{\prime}+\beta^{\prime}+r^{\prime}\right) x^2+\left(\alpha^{\prime} \beta^{\prime}+\beta^{\prime} r^{\prime}+r^{\prime} \alpha^{\prime}\right) x-\alpha^{\prime} \beta^{\prime} r=0\) \(x^3-(5+13+10) x^2+(5 \times 13+13 \times 10+10 \times 5) x-5\) \(\times 13 \times 10=0\) \(x^3-28 x^2+245 x-650=0\)
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118464
Which among the following equations has roots which are negatives of the roots of the equation \(x^3-x^2+x-4=0\) ?
1 \(x^3-x^2+x-4=0\)
2 \(x^3+x^2+x+4=0\)
3 \(x^3-x^2+x-4=0\)
4 \(x^3-x^2-x-4=0\)
Explanation:
B Let \(\alpha\) is the roots of equation, \(x^3-x^2+x-4=0\) Now, put \((-\alpha)=x\), \((-x)^3-(-x)^2+(-x)-4=0\) \(-x^3-x^2-x-4=0\) \(x^3+x^2+x+4=0\)
AP EAMCET-18.09.2020
Complex Numbers and Quadratic Equation
118465
If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13 , then find that equation.
1 \(x^2+x-6=0\)
2 \(x^2-x+6=0\)
3 \(x^2-x-6=0\)
4 \(x^2+x+6=0\)
Explanation:
C Let \(\alpha\) and \(\beta\) are roots of the equation, Given, Sum of \(\operatorname{root}(\alpha+\beta)=1\) And sum of square of the roots \(\left(\alpha^2+\beta^2\right)=13\) \(\therefore \quad \alpha \beta=\frac{1}{2}\left[(\alpha+\beta)^2-\left(\alpha^2+\beta^2\right)\right]\) \(\quad=\frac{1}{2}[(1-13)]=-6\) Equation of required quadratic is, \(x^2-(\alpha+\beta) x+\alpha \cdot \beta=0\) \(x^2-x-6=0\)
AP EAMCET-18.09.2020
Complex Numbers and Quadratic Equation
118467
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^4+3 x^3\) \(-6 x^2+2 x-4=0\), then find the equation having \(\operatorname{roots} \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)
A Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(\mathrm{x}^4+3 \mathrm{x}^3-6 \mathrm{x}^2+2 \mathrm{x}-4=0\) Required equation having roots \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\) can be obtained by replacing \(\mathrm{x}\) by \(\frac{1}{\mathrm{x}}\) in given equation, \(\left(\frac{1}{x}\right)^4+3\left(\frac{1}{x}\right)^3-6\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x}\right)-4=0\) \(4 x^4-2 x^3+6 x^2-3 x-1=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118463
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-6 x^2+11 x-6=\) 0 , then the equation having the \(\alpha^2, \beta^2, \gamma^2\), and \(\gamma^2\) \(+\boldsymbol{\alpha}^2\) is
A Given, \(x^3-6 x^2+11 x-6=0\) \((x-1)(x-2)(x-3)=0\) \(x=1,2,3\) \(\because \alpha, \beta, \mathrm{r}\) are the roots of the equation (i) So, \(\alpha=1, \beta=2, r=3\) Therefore, \(\alpha^2+\beta^2=(1)^2+(2)^2=5=\alpha^{\prime}\) (say) \(\beta^2+\mathrm{r}^2=(2)^2+(3)^2\) \(\beta^2+\mathrm{r}^2=13=\beta^{\prime} \text { (say) }\) And, \(\gamma^2+\alpha^2=(3)^2+1=10=\mathrm{r}^{\prime}\) (say) Equation of the having the roots \(\alpha^{\prime}, \beta^{\prime}\) and \(r^{\prime}\), \(x^3-\left(\alpha^{\prime}+\beta^{\prime}+r^{\prime}\right) x^2+\left(\alpha^{\prime} \beta^{\prime}+\beta^{\prime} r^{\prime}+r^{\prime} \alpha^{\prime}\right) x-\alpha^{\prime} \beta^{\prime} r=0\) \(x^3-(5+13+10) x^2+(5 \times 13+13 \times 10+10 \times 5) x-5\) \(\times 13 \times 10=0\) \(x^3-28 x^2+245 x-650=0\)
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118464
Which among the following equations has roots which are negatives of the roots of the equation \(x^3-x^2+x-4=0\) ?
1 \(x^3-x^2+x-4=0\)
2 \(x^3+x^2+x+4=0\)
3 \(x^3-x^2+x-4=0\)
4 \(x^3-x^2-x-4=0\)
Explanation:
B Let \(\alpha\) is the roots of equation, \(x^3-x^2+x-4=0\) Now, put \((-\alpha)=x\), \((-x)^3-(-x)^2+(-x)-4=0\) \(-x^3-x^2-x-4=0\) \(x^3+x^2+x+4=0\)
AP EAMCET-18.09.2020
Complex Numbers and Quadratic Equation
118465
If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13 , then find that equation.
1 \(x^2+x-6=0\)
2 \(x^2-x+6=0\)
3 \(x^2-x-6=0\)
4 \(x^2+x+6=0\)
Explanation:
C Let \(\alpha\) and \(\beta\) are roots of the equation, Given, Sum of \(\operatorname{root}(\alpha+\beta)=1\) And sum of square of the roots \(\left(\alpha^2+\beta^2\right)=13\) \(\therefore \quad \alpha \beta=\frac{1}{2}\left[(\alpha+\beta)^2-\left(\alpha^2+\beta^2\right)\right]\) \(\quad=\frac{1}{2}[(1-13)]=-6\) Equation of required quadratic is, \(x^2-(\alpha+\beta) x+\alpha \cdot \beta=0\) \(x^2-x-6=0\)
AP EAMCET-18.09.2020
Complex Numbers and Quadratic Equation
118467
If \(\alpha, \beta, \gamma\) are the roots of the equation \(x^4+3 x^3\) \(-6 x^2+2 x-4=0\), then find the equation having \(\operatorname{roots} \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)
A Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(\mathrm{x}^4+3 \mathrm{x}^3-6 \mathrm{x}^2+2 \mathrm{x}-4=0\) Required equation having roots \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\) can be obtained by replacing \(\mathrm{x}\) by \(\frac{1}{\mathrm{x}}\) in given equation, \(\left(\frac{1}{x}\right)^4+3\left(\frac{1}{x}\right)^3-6\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x}\right)-4=0\) \(4 x^4-2 x^3+6 x^2-3 x-1=0\)
