118080
If \(5 p^2-7 p-3=0\) and \(5 q^2-7 q-3=0, p \neq q\), then the equation whose roots are \(5 p-4 q\) and \(5 q-4 p\) is
1 \(5 \mathrm{x}^2+7 \mathrm{x}-439=0\)
2 \(5 x^2-7 x-439=0\)
3 \(5 \mathrm{x}^2+7 \mathrm{x}+439=0\)
4 \(5 \mathrm{x}^2+\mathrm{x}-439=0\)
Explanation:
B Given equations are- \(5 p^2-7 p-3=0 \text { and } 5 q^2-7 q-3=0\) \(\text { From above equations, we can say that, }\) \(\text { p \& q satisfy the equation } 5 x^2-7 x-3=0\) So, \(\mathrm{p}+\mathrm{q}=\frac{7}{5} \& \mathrm{p} . \mathrm{q}=\frac{-3}{5}\) Now, the equation whose roots are \(5 p-4 q\) and \(5 q-4 p\) Sum of roots \(=(5 p-4 q)+(5 q-4 p)\) \(=(\mathrm{p}+\mathrm{q})=\frac{7}{5}\) and, product of roots \((5 p-4 q)(5 q-4 p)\) \(=25 \mathrm{pq}-20 \mathrm{p}^2-20 \mathrm{q}^2+16 \mathrm{pq}\) \(=41 \mathrm{pq}-20\left(\mathrm{p}^2+\mathrm{q}^2\right)\) \(=41 \mathrm{pq}-20\left\{(\mathrm{p}+\mathrm{q})^2-2 \mathrm{pq}\right\}\) \(=41 \mathrm{pq}-20(\mathrm{p}+\mathrm{q})^2+40 \mathrm{pq}\) \(=81 \mathrm{pq}-20\left(\frac{7}{5}\right)^2\) \(=81 \times\left(\frac{-3}{5}\right)-20 \times \frac{49}{25}\) \(=-\frac{243}{5}-\frac{196}{5}=\frac{-439}{5}\) Therefore, equation is \(\mathrm{x}^2-\) (Sum of roots) \(\mathrm{x}+\) product of roots \(=0\) \(=x^2-\frac{7}{5} x+\left(\frac{-439}{5}\right)=0\) \(5 x^2-7 x-439=0\)
UPSEE-2017
Complex Numbers and Quadratic Equation
118081
Let \(\alpha, \beta\) be the roots of \(x^2+3 x+5=0\), then the equation whose roots are \(-\frac{1}{\alpha}\) and \(-\frac{1}{\beta}\) is
1 \(5 x^2+3 x-4=0\)
2 \(5 x^2-3 x+4=0\)
3 \(5 x^2+3 x-1=0\)
4 \(5 x^2-3 x+1=0\)
Explanation:
D \(\alpha, \beta\) are the roots of given equation \(x^2+3 x+5=0\) then, \(\quad \alpha+\beta=\frac{-3}{1}, \alpha, \beta=5\) Now, equation whose roots are \(\frac{-1}{\alpha}\) and \(\frac{-1}{\beta}\) So, Sum of roots \(=\frac{-1}{\alpha}+\frac{-1}{\beta}=\frac{-(\alpha+\beta)}{\alpha \cdot \beta}\) \(=\frac{3}{5}\) Product of roots \(=\left(\frac{-1}{\alpha}\right) \cdot\left(\frac{-1}{\beta}\right)=\frac{1}{\alpha \cdot \beta}=\frac{1}{5}\) Therefore, equation is- \(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) (Product of roots) \(=0\) \(\mathrm{x}^2-\frac{3}{5}(\mathrm{x})+\frac{1}{5}=0\) \(5 \mathrm{x}^2-3 \mathrm{x}+1=0\)
118080
If \(5 p^2-7 p-3=0\) and \(5 q^2-7 q-3=0, p \neq q\), then the equation whose roots are \(5 p-4 q\) and \(5 q-4 p\) is
1 \(5 \mathrm{x}^2+7 \mathrm{x}-439=0\)
2 \(5 x^2-7 x-439=0\)
3 \(5 \mathrm{x}^2+7 \mathrm{x}+439=0\)
4 \(5 \mathrm{x}^2+\mathrm{x}-439=0\)
Explanation:
B Given equations are- \(5 p^2-7 p-3=0 \text { and } 5 q^2-7 q-3=0\) \(\text { From above equations, we can say that, }\) \(\text { p \& q satisfy the equation } 5 x^2-7 x-3=0\) So, \(\mathrm{p}+\mathrm{q}=\frac{7}{5} \& \mathrm{p} . \mathrm{q}=\frac{-3}{5}\) Now, the equation whose roots are \(5 p-4 q\) and \(5 q-4 p\) Sum of roots \(=(5 p-4 q)+(5 q-4 p)\) \(=(\mathrm{p}+\mathrm{q})=\frac{7}{5}\) and, product of roots \((5 p-4 q)(5 q-4 p)\) \(=25 \mathrm{pq}-20 \mathrm{p}^2-20 \mathrm{q}^2+16 \mathrm{pq}\) \(=41 \mathrm{pq}-20\left(\mathrm{p}^2+\mathrm{q}^2\right)\) \(=41 \mathrm{pq}-20\left\{(\mathrm{p}+\mathrm{q})^2-2 \mathrm{pq}\right\}\) \(=41 \mathrm{pq}-20(\mathrm{p}+\mathrm{q})^2+40 \mathrm{pq}\) \(=81 \mathrm{pq}-20\left(\frac{7}{5}\right)^2\) \(=81 \times\left(\frac{-3}{5}\right)-20 \times \frac{49}{25}\) \(=-\frac{243}{5}-\frac{196}{5}=\frac{-439}{5}\) Therefore, equation is \(\mathrm{x}^2-\) (Sum of roots) \(\mathrm{x}+\) product of roots \(=0\) \(=x^2-\frac{7}{5} x+\left(\frac{-439}{5}\right)=0\) \(5 x^2-7 x-439=0\)
UPSEE-2017
Complex Numbers and Quadratic Equation
118081
Let \(\alpha, \beta\) be the roots of \(x^2+3 x+5=0\), then the equation whose roots are \(-\frac{1}{\alpha}\) and \(-\frac{1}{\beta}\) is
1 \(5 x^2+3 x-4=0\)
2 \(5 x^2-3 x+4=0\)
3 \(5 x^2+3 x-1=0\)
4 \(5 x^2-3 x+1=0\)
Explanation:
D \(\alpha, \beta\) are the roots of given equation \(x^2+3 x+5=0\) then, \(\quad \alpha+\beta=\frac{-3}{1}, \alpha, \beta=5\) Now, equation whose roots are \(\frac{-1}{\alpha}\) and \(\frac{-1}{\beta}\) So, Sum of roots \(=\frac{-1}{\alpha}+\frac{-1}{\beta}=\frac{-(\alpha+\beta)}{\alpha \cdot \beta}\) \(=\frac{3}{5}\) Product of roots \(=\left(\frac{-1}{\alpha}\right) \cdot\left(\frac{-1}{\beta}\right)=\frac{1}{\alpha \cdot \beta}=\frac{1}{5}\) Therefore, equation is- \(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) (Product of roots) \(=0\) \(\mathrm{x}^2-\frac{3}{5}(\mathrm{x})+\frac{1}{5}=0\) \(5 \mathrm{x}^2-3 \mathrm{x}+1=0\)
118080
If \(5 p^2-7 p-3=0\) and \(5 q^2-7 q-3=0, p \neq q\), then the equation whose roots are \(5 p-4 q\) and \(5 q-4 p\) is
1 \(5 \mathrm{x}^2+7 \mathrm{x}-439=0\)
2 \(5 x^2-7 x-439=0\)
3 \(5 \mathrm{x}^2+7 \mathrm{x}+439=0\)
4 \(5 \mathrm{x}^2+\mathrm{x}-439=0\)
Explanation:
B Given equations are- \(5 p^2-7 p-3=0 \text { and } 5 q^2-7 q-3=0\) \(\text { From above equations, we can say that, }\) \(\text { p \& q satisfy the equation } 5 x^2-7 x-3=0\) So, \(\mathrm{p}+\mathrm{q}=\frac{7}{5} \& \mathrm{p} . \mathrm{q}=\frac{-3}{5}\) Now, the equation whose roots are \(5 p-4 q\) and \(5 q-4 p\) Sum of roots \(=(5 p-4 q)+(5 q-4 p)\) \(=(\mathrm{p}+\mathrm{q})=\frac{7}{5}\) and, product of roots \((5 p-4 q)(5 q-4 p)\) \(=25 \mathrm{pq}-20 \mathrm{p}^2-20 \mathrm{q}^2+16 \mathrm{pq}\) \(=41 \mathrm{pq}-20\left(\mathrm{p}^2+\mathrm{q}^2\right)\) \(=41 \mathrm{pq}-20\left\{(\mathrm{p}+\mathrm{q})^2-2 \mathrm{pq}\right\}\) \(=41 \mathrm{pq}-20(\mathrm{p}+\mathrm{q})^2+40 \mathrm{pq}\) \(=81 \mathrm{pq}-20\left(\frac{7}{5}\right)^2\) \(=81 \times\left(\frac{-3}{5}\right)-20 \times \frac{49}{25}\) \(=-\frac{243}{5}-\frac{196}{5}=\frac{-439}{5}\) Therefore, equation is \(\mathrm{x}^2-\) (Sum of roots) \(\mathrm{x}+\) product of roots \(=0\) \(=x^2-\frac{7}{5} x+\left(\frac{-439}{5}\right)=0\) \(5 x^2-7 x-439=0\)
UPSEE-2017
Complex Numbers and Quadratic Equation
118081
Let \(\alpha, \beta\) be the roots of \(x^2+3 x+5=0\), then the equation whose roots are \(-\frac{1}{\alpha}\) and \(-\frac{1}{\beta}\) is
1 \(5 x^2+3 x-4=0\)
2 \(5 x^2-3 x+4=0\)
3 \(5 x^2+3 x-1=0\)
4 \(5 x^2-3 x+1=0\)
Explanation:
D \(\alpha, \beta\) are the roots of given equation \(x^2+3 x+5=0\) then, \(\quad \alpha+\beta=\frac{-3}{1}, \alpha, \beta=5\) Now, equation whose roots are \(\frac{-1}{\alpha}\) and \(\frac{-1}{\beta}\) So, Sum of roots \(=\frac{-1}{\alpha}+\frac{-1}{\beta}=\frac{-(\alpha+\beta)}{\alpha \cdot \beta}\) \(=\frac{3}{5}\) Product of roots \(=\left(\frac{-1}{\alpha}\right) \cdot\left(\frac{-1}{\beta}\right)=\frac{1}{\alpha \cdot \beta}=\frac{1}{5}\) Therefore, equation is- \(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) (Product of roots) \(=0\) \(\mathrm{x}^2-\frac{3}{5}(\mathrm{x})+\frac{1}{5}=0\) \(5 \mathrm{x}^2-3 \mathrm{x}+1=0\)
118080
If \(5 p^2-7 p-3=0\) and \(5 q^2-7 q-3=0, p \neq q\), then the equation whose roots are \(5 p-4 q\) and \(5 q-4 p\) is
1 \(5 \mathrm{x}^2+7 \mathrm{x}-439=0\)
2 \(5 x^2-7 x-439=0\)
3 \(5 \mathrm{x}^2+7 \mathrm{x}+439=0\)
4 \(5 \mathrm{x}^2+\mathrm{x}-439=0\)
Explanation:
B Given equations are- \(5 p^2-7 p-3=0 \text { and } 5 q^2-7 q-3=0\) \(\text { From above equations, we can say that, }\) \(\text { p \& q satisfy the equation } 5 x^2-7 x-3=0\) So, \(\mathrm{p}+\mathrm{q}=\frac{7}{5} \& \mathrm{p} . \mathrm{q}=\frac{-3}{5}\) Now, the equation whose roots are \(5 p-4 q\) and \(5 q-4 p\) Sum of roots \(=(5 p-4 q)+(5 q-4 p)\) \(=(\mathrm{p}+\mathrm{q})=\frac{7}{5}\) and, product of roots \((5 p-4 q)(5 q-4 p)\) \(=25 \mathrm{pq}-20 \mathrm{p}^2-20 \mathrm{q}^2+16 \mathrm{pq}\) \(=41 \mathrm{pq}-20\left(\mathrm{p}^2+\mathrm{q}^2\right)\) \(=41 \mathrm{pq}-20\left\{(\mathrm{p}+\mathrm{q})^2-2 \mathrm{pq}\right\}\) \(=41 \mathrm{pq}-20(\mathrm{p}+\mathrm{q})^2+40 \mathrm{pq}\) \(=81 \mathrm{pq}-20\left(\frac{7}{5}\right)^2\) \(=81 \times\left(\frac{-3}{5}\right)-20 \times \frac{49}{25}\) \(=-\frac{243}{5}-\frac{196}{5}=\frac{-439}{5}\) Therefore, equation is \(\mathrm{x}^2-\) (Sum of roots) \(\mathrm{x}+\) product of roots \(=0\) \(=x^2-\frac{7}{5} x+\left(\frac{-439}{5}\right)=0\) \(5 x^2-7 x-439=0\)
UPSEE-2017
Complex Numbers and Quadratic Equation
118081
Let \(\alpha, \beta\) be the roots of \(x^2+3 x+5=0\), then the equation whose roots are \(-\frac{1}{\alpha}\) and \(-\frac{1}{\beta}\) is
1 \(5 x^2+3 x-4=0\)
2 \(5 x^2-3 x+4=0\)
3 \(5 x^2+3 x-1=0\)
4 \(5 x^2-3 x+1=0\)
Explanation:
D \(\alpha, \beta\) are the roots of given equation \(x^2+3 x+5=0\) then, \(\quad \alpha+\beta=\frac{-3}{1}, \alpha, \beta=5\) Now, equation whose roots are \(\frac{-1}{\alpha}\) and \(\frac{-1}{\beta}\) So, Sum of roots \(=\frac{-1}{\alpha}+\frac{-1}{\beta}=\frac{-(\alpha+\beta)}{\alpha \cdot \beta}\) \(=\frac{3}{5}\) Product of roots \(=\left(\frac{-1}{\alpha}\right) \cdot\left(\frac{-1}{\beta}\right)=\frac{1}{\alpha \cdot \beta}=\frac{1}{5}\) Therefore, equation is- \(\mathrm{x}^2-(\) sum of roots) \(\mathrm{x}+\) (Product of roots) \(=0\) \(\mathrm{x}^2-\frac{3}{5}(\mathrm{x})+\frac{1}{5}=0\) \(5 \mathrm{x}^2-3 \mathrm{x}+1=0\)
