A Since \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) \(\therefore \quad \sin \alpha+\cos \alpha=-\frac{q}{p}\) On squaring both sides- \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(1+\sin 2 \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) Also, \(\quad \sin \alpha \cos \alpha=\frac{r}{\mathrm{p}}\) \(2 \sin \alpha \cos \alpha =\frac{2 r}{p}\) \(\sin 2 \alpha =\frac{2 r}{p}\) On putting the value \(\sin 2 \alpha\) in equation (i)- \(1+\frac{2 r}{p}=\frac{q^2}{p^2}\) \(\Rightarrow p(p+2 r)=q^2\) \(\Rightarrow p^2-q^2+2 p r=0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118455
If the quadratic equation \(z^2+(a+i b) z+c+\) id \(=0\), where \(a, b, c, d\) are non-zero real numbers, has a real root, then
1 \(a b d=b^2 c+d^2\)
2 \(a b c=b c^2+d^2\)
3 \(a b d=b c^2+a d^2\)
4 None of these
Explanation:
A Let \(\alpha\) be a real root of the equation Then, \(\mathrm{z}=\alpha\) \(z^2+(a+i b) z+c+i d=0\) \(\Rightarrow \alpha^2+(a+i b) \alpha+c+i d=0\) \(\Rightarrow\left(\alpha^2+a \alpha+c\right)+i(b \alpha+d)=0\) Equating the real and imaginary parts, we have- \(\alpha=\frac{-\mathrm{d}}{\mathrm{b}}\) Put, \(\alpha=\frac{-d}{b}\) in \(\left(\alpha^2+\mathrm{a} \alpha+\mathrm{c}\right)\) \(\Rightarrow\left(-\frac{d}{b}\right)^2+a\left(\frac{-d}{b}\right)+c=0\) \(\Rightarrow \frac{\mathrm{d}^2}{\mathrm{~b}^2}-\frac{\mathrm{ad}}{\mathrm{b}}+\mathrm{c}=0\) \(\Rightarrow d^2-a b d+b^2 c=0\) \(\Rightarrow d^2+b^2 c=a b d\)
COMEDK-2016
Complex Numbers and Quadratic Equation
118456
If \(\alpha, \beta\) are the roots of the equations \(x^2-2 x-1=0\), then what is the value of \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\)
1 -2
2 0
3 30
4 34
Explanation:
D Given, If \(\alpha, \beta\) are the roots of the equations \(\mathrm{x}^2-2 \mathrm{x}-1=0\) Sum of the roots \((\alpha+\beta)=-\left(\frac{-2}{1}\right)=2\) Products of the roots \((\alpha \beta)=\frac{-1}{1}=-1\) \(\alpha^2 \beta^2=1\) Now, \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\) \(=\frac{\alpha^2}{\beta^2}+\frac{\beta^2}{\alpha^2}\) \(=\frac{\alpha^4+\beta^4}{\alpha^2 \beta^2}\) \(=\frac{\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(\alpha+\beta)^2-2 \alpha \beta\right]^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(2)^2-2 \times(-1)\right]^2-2 \times(1)}{1}\) \(=34\)
BITSAT-2016
Complex Numbers and Quadratic Equation
118466
If the roots of the equation \(x^3-6 x^2+11 x-6\) \(=0\) are \(\alpha, \beta\) and \(\gamma\). Then the equation whose roots are \(\alpha^2, \beta^2, \gamma^2\) among the following is \(\qquad\)
1 \(x^3+14 x^2+49 x-36=0\)
2 \(x^3-14 x^2+49 x-36=0\)
3 \(x^3-14 x^2-49 x+36=0\)
4 \(x^3-14 x^2-49 x-36=0\)
Explanation:
B Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(x^3-6 x^2+11 x-6=0\) Now, to find the equation whose roots are \(\alpha^2, \beta^2\) and \(\gamma^2\) Put \(\alpha^2=\mathrm{x} \Rightarrow \alpha=\sqrt{\mathrm{x}}\) Since, \(\alpha\) is the root of the given equation, so \(\mathrm{x}^{3 / 2}-6 \mathrm{x}+11 \mathrm{x}^{1 / 2}-6=0\) \(x^{1 / 2}(x+11)=6(x+1)\) On squaring both sides, we get- \(x(x+11)^2=36(x+1)^2\) \(x\left[x^2+22 x+121\right]=36\left[x^2+2 x+1\right]\) \(x^3+22 x^2+121 x=36 x^2+72 x+36\) \(x^3-14 x^2+49 x-36=0\)
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118458
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\), such that \(|\alpha-\beta|=\sqrt{10}\), then \(p\) belongs to the set :
1 \(\{2,-5\}\)
2 \(\{-3,2\}\)
3 \(\{-2,5\}\)
4 \(\{3,-5\}\)
Explanation:
A Given, If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\) \(\therefore \quad \alpha+\beta=\frac{-p}{1}=p\) \(\alpha \cdot \beta=\frac{\frac{3 p}{4}}{1}=\frac{3 p}{4}\) \(|\alpha-\beta|=\sqrt{10}\) \((\alpha-\beta)^2=10\) We know that, \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \((\alpha-\beta)^2=\alpha^2+\beta^2-2 \alpha \beta\) \(\therefore \quad(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta\) \(10=\mathrm{p}^2-4 \times \frac{3 \mathrm{p}}{4}\) \(10=\mathrm{p}^2-3 \mathrm{p}\) \(\mathrm{p}^2-3 \mathrm{p}-10=0\) \(\mathrm{p}^2-5 \mathrm{p}+2 \mathrm{p}-10=0\) \(\mathrm{p}(\mathrm{p}-5)+2(\mathrm{p}-5)=0\) \((p-5)(p+2)=0\) \(\mathrm{p}=\{-2,5\}\)
A Since \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) \(\therefore \quad \sin \alpha+\cos \alpha=-\frac{q}{p}\) On squaring both sides- \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(1+\sin 2 \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) Also, \(\quad \sin \alpha \cos \alpha=\frac{r}{\mathrm{p}}\) \(2 \sin \alpha \cos \alpha =\frac{2 r}{p}\) \(\sin 2 \alpha =\frac{2 r}{p}\) On putting the value \(\sin 2 \alpha\) in equation (i)- \(1+\frac{2 r}{p}=\frac{q^2}{p^2}\) \(\Rightarrow p(p+2 r)=q^2\) \(\Rightarrow p^2-q^2+2 p r=0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118455
If the quadratic equation \(z^2+(a+i b) z+c+\) id \(=0\), where \(a, b, c, d\) are non-zero real numbers, has a real root, then
1 \(a b d=b^2 c+d^2\)
2 \(a b c=b c^2+d^2\)
3 \(a b d=b c^2+a d^2\)
4 None of these
Explanation:
A Let \(\alpha\) be a real root of the equation Then, \(\mathrm{z}=\alpha\) \(z^2+(a+i b) z+c+i d=0\) \(\Rightarrow \alpha^2+(a+i b) \alpha+c+i d=0\) \(\Rightarrow\left(\alpha^2+a \alpha+c\right)+i(b \alpha+d)=0\) Equating the real and imaginary parts, we have- \(\alpha=\frac{-\mathrm{d}}{\mathrm{b}}\) Put, \(\alpha=\frac{-d}{b}\) in \(\left(\alpha^2+\mathrm{a} \alpha+\mathrm{c}\right)\) \(\Rightarrow\left(-\frac{d}{b}\right)^2+a\left(\frac{-d}{b}\right)+c=0\) \(\Rightarrow \frac{\mathrm{d}^2}{\mathrm{~b}^2}-\frac{\mathrm{ad}}{\mathrm{b}}+\mathrm{c}=0\) \(\Rightarrow d^2-a b d+b^2 c=0\) \(\Rightarrow d^2+b^2 c=a b d\)
COMEDK-2016
Complex Numbers and Quadratic Equation
118456
If \(\alpha, \beta\) are the roots of the equations \(x^2-2 x-1=0\), then what is the value of \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\)
1 -2
2 0
3 30
4 34
Explanation:
D Given, If \(\alpha, \beta\) are the roots of the equations \(\mathrm{x}^2-2 \mathrm{x}-1=0\) Sum of the roots \((\alpha+\beta)=-\left(\frac{-2}{1}\right)=2\) Products of the roots \((\alpha \beta)=\frac{-1}{1}=-1\) \(\alpha^2 \beta^2=1\) Now, \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\) \(=\frac{\alpha^2}{\beta^2}+\frac{\beta^2}{\alpha^2}\) \(=\frac{\alpha^4+\beta^4}{\alpha^2 \beta^2}\) \(=\frac{\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(\alpha+\beta)^2-2 \alpha \beta\right]^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(2)^2-2 \times(-1)\right]^2-2 \times(1)}{1}\) \(=34\)
BITSAT-2016
Complex Numbers and Quadratic Equation
118466
If the roots of the equation \(x^3-6 x^2+11 x-6\) \(=0\) are \(\alpha, \beta\) and \(\gamma\). Then the equation whose roots are \(\alpha^2, \beta^2, \gamma^2\) among the following is \(\qquad\)
1 \(x^3+14 x^2+49 x-36=0\)
2 \(x^3-14 x^2+49 x-36=0\)
3 \(x^3-14 x^2-49 x+36=0\)
4 \(x^3-14 x^2-49 x-36=0\)
Explanation:
B Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(x^3-6 x^2+11 x-6=0\) Now, to find the equation whose roots are \(\alpha^2, \beta^2\) and \(\gamma^2\) Put \(\alpha^2=\mathrm{x} \Rightarrow \alpha=\sqrt{\mathrm{x}}\) Since, \(\alpha\) is the root of the given equation, so \(\mathrm{x}^{3 / 2}-6 \mathrm{x}+11 \mathrm{x}^{1 / 2}-6=0\) \(x^{1 / 2}(x+11)=6(x+1)\) On squaring both sides, we get- \(x(x+11)^2=36(x+1)^2\) \(x\left[x^2+22 x+121\right]=36\left[x^2+2 x+1\right]\) \(x^3+22 x^2+121 x=36 x^2+72 x+36\) \(x^3-14 x^2+49 x-36=0\)
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118458
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\), such that \(|\alpha-\beta|=\sqrt{10}\), then \(p\) belongs to the set :
1 \(\{2,-5\}\)
2 \(\{-3,2\}\)
3 \(\{-2,5\}\)
4 \(\{3,-5\}\)
Explanation:
A Given, If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\) \(\therefore \quad \alpha+\beta=\frac{-p}{1}=p\) \(\alpha \cdot \beta=\frac{\frac{3 p}{4}}{1}=\frac{3 p}{4}\) \(|\alpha-\beta|=\sqrt{10}\) \((\alpha-\beta)^2=10\) We know that, \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \((\alpha-\beta)^2=\alpha^2+\beta^2-2 \alpha \beta\) \(\therefore \quad(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta\) \(10=\mathrm{p}^2-4 \times \frac{3 \mathrm{p}}{4}\) \(10=\mathrm{p}^2-3 \mathrm{p}\) \(\mathrm{p}^2-3 \mathrm{p}-10=0\) \(\mathrm{p}^2-5 \mathrm{p}+2 \mathrm{p}-10=0\) \(\mathrm{p}(\mathrm{p}-5)+2(\mathrm{p}-5)=0\) \((p-5)(p+2)=0\) \(\mathrm{p}=\{-2,5\}\)
A Since \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) \(\therefore \quad \sin \alpha+\cos \alpha=-\frac{q}{p}\) On squaring both sides- \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(1+\sin 2 \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) Also, \(\quad \sin \alpha \cos \alpha=\frac{r}{\mathrm{p}}\) \(2 \sin \alpha \cos \alpha =\frac{2 r}{p}\) \(\sin 2 \alpha =\frac{2 r}{p}\) On putting the value \(\sin 2 \alpha\) in equation (i)- \(1+\frac{2 r}{p}=\frac{q^2}{p^2}\) \(\Rightarrow p(p+2 r)=q^2\) \(\Rightarrow p^2-q^2+2 p r=0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118455
If the quadratic equation \(z^2+(a+i b) z+c+\) id \(=0\), where \(a, b, c, d\) are non-zero real numbers, has a real root, then
1 \(a b d=b^2 c+d^2\)
2 \(a b c=b c^2+d^2\)
3 \(a b d=b c^2+a d^2\)
4 None of these
Explanation:
A Let \(\alpha\) be a real root of the equation Then, \(\mathrm{z}=\alpha\) \(z^2+(a+i b) z+c+i d=0\) \(\Rightarrow \alpha^2+(a+i b) \alpha+c+i d=0\) \(\Rightarrow\left(\alpha^2+a \alpha+c\right)+i(b \alpha+d)=0\) Equating the real and imaginary parts, we have- \(\alpha=\frac{-\mathrm{d}}{\mathrm{b}}\) Put, \(\alpha=\frac{-d}{b}\) in \(\left(\alpha^2+\mathrm{a} \alpha+\mathrm{c}\right)\) \(\Rightarrow\left(-\frac{d}{b}\right)^2+a\left(\frac{-d}{b}\right)+c=0\) \(\Rightarrow \frac{\mathrm{d}^2}{\mathrm{~b}^2}-\frac{\mathrm{ad}}{\mathrm{b}}+\mathrm{c}=0\) \(\Rightarrow d^2-a b d+b^2 c=0\) \(\Rightarrow d^2+b^2 c=a b d\)
COMEDK-2016
Complex Numbers and Quadratic Equation
118456
If \(\alpha, \beta\) are the roots of the equations \(x^2-2 x-1=0\), then what is the value of \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\)
1 -2
2 0
3 30
4 34
Explanation:
D Given, If \(\alpha, \beta\) are the roots of the equations \(\mathrm{x}^2-2 \mathrm{x}-1=0\) Sum of the roots \((\alpha+\beta)=-\left(\frac{-2}{1}\right)=2\) Products of the roots \((\alpha \beta)=\frac{-1}{1}=-1\) \(\alpha^2 \beta^2=1\) Now, \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\) \(=\frac{\alpha^2}{\beta^2}+\frac{\beta^2}{\alpha^2}\) \(=\frac{\alpha^4+\beta^4}{\alpha^2 \beta^2}\) \(=\frac{\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(\alpha+\beta)^2-2 \alpha \beta\right]^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(2)^2-2 \times(-1)\right]^2-2 \times(1)}{1}\) \(=34\)
BITSAT-2016
Complex Numbers and Quadratic Equation
118466
If the roots of the equation \(x^3-6 x^2+11 x-6\) \(=0\) are \(\alpha, \beta\) and \(\gamma\). Then the equation whose roots are \(\alpha^2, \beta^2, \gamma^2\) among the following is \(\qquad\)
1 \(x^3+14 x^2+49 x-36=0\)
2 \(x^3-14 x^2+49 x-36=0\)
3 \(x^3-14 x^2-49 x+36=0\)
4 \(x^3-14 x^2-49 x-36=0\)
Explanation:
B Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(x^3-6 x^2+11 x-6=0\) Now, to find the equation whose roots are \(\alpha^2, \beta^2\) and \(\gamma^2\) Put \(\alpha^2=\mathrm{x} \Rightarrow \alpha=\sqrt{\mathrm{x}}\) Since, \(\alpha\) is the root of the given equation, so \(\mathrm{x}^{3 / 2}-6 \mathrm{x}+11 \mathrm{x}^{1 / 2}-6=0\) \(x^{1 / 2}(x+11)=6(x+1)\) On squaring both sides, we get- \(x(x+11)^2=36(x+1)^2\) \(x\left[x^2+22 x+121\right]=36\left[x^2+2 x+1\right]\) \(x^3+22 x^2+121 x=36 x^2+72 x+36\) \(x^3-14 x^2+49 x-36=0\)
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118458
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\), such that \(|\alpha-\beta|=\sqrt{10}\), then \(p\) belongs to the set :
1 \(\{2,-5\}\)
2 \(\{-3,2\}\)
3 \(\{-2,5\}\)
4 \(\{3,-5\}\)
Explanation:
A Given, If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\) \(\therefore \quad \alpha+\beta=\frac{-p}{1}=p\) \(\alpha \cdot \beta=\frac{\frac{3 p}{4}}{1}=\frac{3 p}{4}\) \(|\alpha-\beta|=\sqrt{10}\) \((\alpha-\beta)^2=10\) We know that, \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \((\alpha-\beta)^2=\alpha^2+\beta^2-2 \alpha \beta\) \(\therefore \quad(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta\) \(10=\mathrm{p}^2-4 \times \frac{3 \mathrm{p}}{4}\) \(10=\mathrm{p}^2-3 \mathrm{p}\) \(\mathrm{p}^2-3 \mathrm{p}-10=0\) \(\mathrm{p}^2-5 \mathrm{p}+2 \mathrm{p}-10=0\) \(\mathrm{p}(\mathrm{p}-5)+2(\mathrm{p}-5)=0\) \((p-5)(p+2)=0\) \(\mathrm{p}=\{-2,5\}\)
A Since \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) \(\therefore \quad \sin \alpha+\cos \alpha=-\frac{q}{p}\) On squaring both sides- \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(1+\sin 2 \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) Also, \(\quad \sin \alpha \cos \alpha=\frac{r}{\mathrm{p}}\) \(2 \sin \alpha \cos \alpha =\frac{2 r}{p}\) \(\sin 2 \alpha =\frac{2 r}{p}\) On putting the value \(\sin 2 \alpha\) in equation (i)- \(1+\frac{2 r}{p}=\frac{q^2}{p^2}\) \(\Rightarrow p(p+2 r)=q^2\) \(\Rightarrow p^2-q^2+2 p r=0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118455
If the quadratic equation \(z^2+(a+i b) z+c+\) id \(=0\), where \(a, b, c, d\) are non-zero real numbers, has a real root, then
1 \(a b d=b^2 c+d^2\)
2 \(a b c=b c^2+d^2\)
3 \(a b d=b c^2+a d^2\)
4 None of these
Explanation:
A Let \(\alpha\) be a real root of the equation Then, \(\mathrm{z}=\alpha\) \(z^2+(a+i b) z+c+i d=0\) \(\Rightarrow \alpha^2+(a+i b) \alpha+c+i d=0\) \(\Rightarrow\left(\alpha^2+a \alpha+c\right)+i(b \alpha+d)=0\) Equating the real and imaginary parts, we have- \(\alpha=\frac{-\mathrm{d}}{\mathrm{b}}\) Put, \(\alpha=\frac{-d}{b}\) in \(\left(\alpha^2+\mathrm{a} \alpha+\mathrm{c}\right)\) \(\Rightarrow\left(-\frac{d}{b}\right)^2+a\left(\frac{-d}{b}\right)+c=0\) \(\Rightarrow \frac{\mathrm{d}^2}{\mathrm{~b}^2}-\frac{\mathrm{ad}}{\mathrm{b}}+\mathrm{c}=0\) \(\Rightarrow d^2-a b d+b^2 c=0\) \(\Rightarrow d^2+b^2 c=a b d\)
COMEDK-2016
Complex Numbers and Quadratic Equation
118456
If \(\alpha, \beta\) are the roots of the equations \(x^2-2 x-1=0\), then what is the value of \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\)
1 -2
2 0
3 30
4 34
Explanation:
D Given, If \(\alpha, \beta\) are the roots of the equations \(\mathrm{x}^2-2 \mathrm{x}-1=0\) Sum of the roots \((\alpha+\beta)=-\left(\frac{-2}{1}\right)=2\) Products of the roots \((\alpha \beta)=\frac{-1}{1}=-1\) \(\alpha^2 \beta^2=1\) Now, \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\) \(=\frac{\alpha^2}{\beta^2}+\frac{\beta^2}{\alpha^2}\) \(=\frac{\alpha^4+\beta^4}{\alpha^2 \beta^2}\) \(=\frac{\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(\alpha+\beta)^2-2 \alpha \beta\right]^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(2)^2-2 \times(-1)\right]^2-2 \times(1)}{1}\) \(=34\)
BITSAT-2016
Complex Numbers and Quadratic Equation
118466
If the roots of the equation \(x^3-6 x^2+11 x-6\) \(=0\) are \(\alpha, \beta\) and \(\gamma\). Then the equation whose roots are \(\alpha^2, \beta^2, \gamma^2\) among the following is \(\qquad\)
1 \(x^3+14 x^2+49 x-36=0\)
2 \(x^3-14 x^2+49 x-36=0\)
3 \(x^3-14 x^2-49 x+36=0\)
4 \(x^3-14 x^2-49 x-36=0\)
Explanation:
B Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(x^3-6 x^2+11 x-6=0\) Now, to find the equation whose roots are \(\alpha^2, \beta^2\) and \(\gamma^2\) Put \(\alpha^2=\mathrm{x} \Rightarrow \alpha=\sqrt{\mathrm{x}}\) Since, \(\alpha\) is the root of the given equation, so \(\mathrm{x}^{3 / 2}-6 \mathrm{x}+11 \mathrm{x}^{1 / 2}-6=0\) \(x^{1 / 2}(x+11)=6(x+1)\) On squaring both sides, we get- \(x(x+11)^2=36(x+1)^2\) \(x\left[x^2+22 x+121\right]=36\left[x^2+2 x+1\right]\) \(x^3+22 x^2+121 x=36 x^2+72 x+36\) \(x^3-14 x^2+49 x-36=0\)
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118458
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\), such that \(|\alpha-\beta|=\sqrt{10}\), then \(p\) belongs to the set :
1 \(\{2,-5\}\)
2 \(\{-3,2\}\)
3 \(\{-2,5\}\)
4 \(\{3,-5\}\)
Explanation:
A Given, If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\) \(\therefore \quad \alpha+\beta=\frac{-p}{1}=p\) \(\alpha \cdot \beta=\frac{\frac{3 p}{4}}{1}=\frac{3 p}{4}\) \(|\alpha-\beta|=\sqrt{10}\) \((\alpha-\beta)^2=10\) We know that, \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \((\alpha-\beta)^2=\alpha^2+\beta^2-2 \alpha \beta\) \(\therefore \quad(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta\) \(10=\mathrm{p}^2-4 \times \frac{3 \mathrm{p}}{4}\) \(10=\mathrm{p}^2-3 \mathrm{p}\) \(\mathrm{p}^2-3 \mathrm{p}-10=0\) \(\mathrm{p}^2-5 \mathrm{p}+2 \mathrm{p}-10=0\) \(\mathrm{p}(\mathrm{p}-5)+2(\mathrm{p}-5)=0\) \((p-5)(p+2)=0\) \(\mathrm{p}=\{-2,5\}\)
A Since \(\sin \alpha\) and \(\cos \alpha\) are the roots of the equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) \(\therefore \quad \sin \alpha+\cos \alpha=-\frac{q}{p}\) On squaring both sides- \(\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) \(1+\sin 2 \alpha=\frac{\mathrm{q}^2}{\mathrm{p}^2}\) Also, \(\quad \sin \alpha \cos \alpha=\frac{r}{\mathrm{p}}\) \(2 \sin \alpha \cos \alpha =\frac{2 r}{p}\) \(\sin 2 \alpha =\frac{2 r}{p}\) On putting the value \(\sin 2 \alpha\) in equation (i)- \(1+\frac{2 r}{p}=\frac{q^2}{p^2}\) \(\Rightarrow p(p+2 r)=q^2\) \(\Rightarrow p^2-q^2+2 p r=0\)
SRM JEEE-2009
Complex Numbers and Quadratic Equation
118455
If the quadratic equation \(z^2+(a+i b) z+c+\) id \(=0\), where \(a, b, c, d\) are non-zero real numbers, has a real root, then
1 \(a b d=b^2 c+d^2\)
2 \(a b c=b c^2+d^2\)
3 \(a b d=b c^2+a d^2\)
4 None of these
Explanation:
A Let \(\alpha\) be a real root of the equation Then, \(\mathrm{z}=\alpha\) \(z^2+(a+i b) z+c+i d=0\) \(\Rightarrow \alpha^2+(a+i b) \alpha+c+i d=0\) \(\Rightarrow\left(\alpha^2+a \alpha+c\right)+i(b \alpha+d)=0\) Equating the real and imaginary parts, we have- \(\alpha=\frac{-\mathrm{d}}{\mathrm{b}}\) Put, \(\alpha=\frac{-d}{b}\) in \(\left(\alpha^2+\mathrm{a} \alpha+\mathrm{c}\right)\) \(\Rightarrow\left(-\frac{d}{b}\right)^2+a\left(\frac{-d}{b}\right)+c=0\) \(\Rightarrow \frac{\mathrm{d}^2}{\mathrm{~b}^2}-\frac{\mathrm{ad}}{\mathrm{b}}+\mathrm{c}=0\) \(\Rightarrow d^2-a b d+b^2 c=0\) \(\Rightarrow d^2+b^2 c=a b d\)
COMEDK-2016
Complex Numbers and Quadratic Equation
118456
If \(\alpha, \beta\) are the roots of the equations \(x^2-2 x-1=0\), then what is the value of \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\)
1 -2
2 0
3 30
4 34
Explanation:
D Given, If \(\alpha, \beta\) are the roots of the equations \(\mathrm{x}^2-2 \mathrm{x}-1=0\) Sum of the roots \((\alpha+\beta)=-\left(\frac{-2}{1}\right)=2\) Products of the roots \((\alpha \beta)=\frac{-1}{1}=-1\) \(\alpha^2 \beta^2=1\) Now, \(\alpha^2 \beta^{-2}+\alpha^{-2} \beta^2\) \(=\frac{\alpha^2}{\beta^2}+\frac{\beta^2}{\alpha^2}\) \(=\frac{\alpha^4+\beta^4}{\alpha^2 \beta^2}\) \(=\frac{\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(\alpha+\beta)^2-2 \alpha \beta\right]^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}\) \(=\frac{\left[(2)^2-2 \times(-1)\right]^2-2 \times(1)}{1}\) \(=34\)
BITSAT-2016
Complex Numbers and Quadratic Equation
118466
If the roots of the equation \(x^3-6 x^2+11 x-6\) \(=0\) are \(\alpha, \beta\) and \(\gamma\). Then the equation whose roots are \(\alpha^2, \beta^2, \gamma^2\) among the following is \(\qquad\)
1 \(x^3+14 x^2+49 x-36=0\)
2 \(x^3-14 x^2+49 x-36=0\)
3 \(x^3-14 x^2-49 x+36=0\)
4 \(x^3-14 x^2-49 x-36=0\)
Explanation:
B Given, \(\alpha, \beta\) and \(\gamma\) roots of the equation \(x^3-6 x^2+11 x-6=0\) Now, to find the equation whose roots are \(\alpha^2, \beta^2\) and \(\gamma^2\) Put \(\alpha^2=\mathrm{x} \Rightarrow \alpha=\sqrt{\mathrm{x}}\) Since, \(\alpha\) is the root of the given equation, so \(\mathrm{x}^{3 / 2}-6 \mathrm{x}+11 \mathrm{x}^{1 / 2}-6=0\) \(x^{1 / 2}(x+11)=6(x+1)\) On squaring both sides, we get- \(x(x+11)^2=36(x+1)^2\) \(x\left[x^2+22 x+121\right]=36\left[x^2+2 x+1\right]\) \(x^3+22 x^2+121 x=36 x^2+72 x+36\) \(x^3-14 x^2+49 x-36=0\)
AP EAMCET-22.09.2020
Complex Numbers and Quadratic Equation
118458
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\), such that \(|\alpha-\beta|=\sqrt{10}\), then \(p\) belongs to the set :
1 \(\{2,-5\}\)
2 \(\{-3,2\}\)
3 \(\{-2,5\}\)
4 \(\{3,-5\}\)
Explanation:
A Given, If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+p x+\frac{3 p}{4}=0\) \(\therefore \quad \alpha+\beta=\frac{-p}{1}=p\) \(\alpha \cdot \beta=\frac{\frac{3 p}{4}}{1}=\frac{3 p}{4}\) \(|\alpha-\beta|=\sqrt{10}\) \((\alpha-\beta)^2=10\) We know that, \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\) \((\alpha-\beta)^2=\alpha^2+\beta^2-2 \alpha \beta\) \(\therefore \quad(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta\) \(10=\mathrm{p}^2-4 \times \frac{3 \mathrm{p}}{4}\) \(10=\mathrm{p}^2-3 \mathrm{p}\) \(\mathrm{p}^2-3 \mathrm{p}-10=0\) \(\mathrm{p}^2-5 \mathrm{p}+2 \mathrm{p}-10=0\) \(\mathrm{p}(\mathrm{p}-5)+2(\mathrm{p}-5)=0\) \((p-5)(p+2)=0\) \(\mathrm{p}=\{-2,5\}\)