118121
If one root of the quadratic equation \(a x^2+b x+\) \(c=0\) is equal to the \(n^{\text {th }}\) power of the other, then \(\left(a^n\right)^{1 / n+1}+\left(a^n c\right)^{1 / n+1}=\)
1 \(-2 \mathrm{~b}\)
2 \(-\mathrm{b}\)
3 \(b-1\)
4 \(b+1\)
Explanation:
B Given, quadratic equation- \(a x^2+b x+c=0\) Let, \(\alpha, \beta\) are the roots of the quadratic equation. Then, \(\alpha+\beta=\frac{-b}{a}\) \(\alpha \beta=\frac{c}{a}\) Given, \(\alpha=\beta^{\mathrm{x}}\) We get, \(\beta^{n+1}=\frac{c}{a}\) \(\beta=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}\) \(\alpha=\left(\frac{c}{a}\right)^{\frac{n}{n+1}}\) Adding both, we get- \(\alpha+\beta=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}=\frac{-b}{a}\) \(-b=a \cdot\left[\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}\right]\) \(-b=\left(c^n\right)^{\frac{1}{n+1}}+\left(a^n c\right)^{\frac{1}{n+1}}\)
AP EAMCET-06.07.2022
Complex Numbers and Quadratic Equation
118122
The sum of squares of roots of the equation \(\mathrm{x}^{\frac{2}{3}}+\mathrm{x}^{\frac{1}{3}}-2=0\) is
1 82
2 65
3 50
4 37
Explanation:
B Given, the equation \(\mathrm{x}^{\frac{2}{3}}+\mathrm{x}^{\frac{1}{3}}-2=0\) Let \(x^{\frac{1}{3}}=t \text {, then the give }\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) Put \(t=1\), we get \(x^{\frac{1}{3}}=1 \Rightarrow x=1\) \(t=-2\), we get \(x^{\frac{1}{3}}=-2\) \(\Rightarrow \quad \mathrm{x}=-8\) \(\therefore \quad\) Roots are \(1,-8\). \(\therefore\) Sum of the squares \(=1+64=65\)
AP EAMCET-06.07.2022
Complex Numbers and Quadratic Equation
118123
If \(\alpha\) satisfies the equation \(\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2\), then the roots of the equations \(\alpha^2 x^2+4 \alpha x+3=\) 0 are
1 1,3
2 \(-1,1\)
3 \(2,-3\)
4 3,4
Explanation:
A Given the equation, \(\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2\) Let \(\sqrt{\frac{2 \mathrm{x}+1}{\mathrm{x}}}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{t}}+\mathrm{t}=2\) Or \(\quad t^2+1=2 t\) Or \(\quad t^2+1-2 t=0 \Rightarrow t=1\) \(\therefore \quad \sqrt{\frac{2 \mathrm{x}+1}{\mathrm{x}}}=1\) \(\Rightarrow \quad 2 \mathrm{x}+1=\mathrm{x}\) \(\Rightarrow \quad \mathrm{x}=-1\) Since, \(\alpha\) satisfy the equation. \(\therefore \quad \alpha=-1\) Now, the given equation \(\alpha^2 x^2+4 \alpha x+3=0\) becomes \(x^2-4 x+3=0\). or\(\mathrm{x}^2-3 \mathrm{x}-\mathrm{x}+3=0\) Or\(\mathrm{x}(\mathrm{x}-3)-1(\mathrm{x}-3)=0\) \((\mathrm{x}-3)(\mathrm{x}-1)=0\)\(\therefore \quad \mathrm{x}=1,3\)
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118125
The solution of \(\sqrt{3 x^2-2}=2 x-1\) are
1 \((2,4)\)
2 \((1,4)\)
3 \((3,4)\)
4 \((1,3)\)
Explanation:
D We have the equation, \(\sqrt{3 x^2-2}=2 x-1\) Squaring, both the sides, we get \(3 x^2-2=(2 x-1)^2=4 x^2+1-4 x\) \(x^2-4 x+3=0\) \(x^2-3 x-x+3=0\) \((x-3)(x-1)=0\) \(x=(1,3)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
118121
If one root of the quadratic equation \(a x^2+b x+\) \(c=0\) is equal to the \(n^{\text {th }}\) power of the other, then \(\left(a^n\right)^{1 / n+1}+\left(a^n c\right)^{1 / n+1}=\)
1 \(-2 \mathrm{~b}\)
2 \(-\mathrm{b}\)
3 \(b-1\)
4 \(b+1\)
Explanation:
B Given, quadratic equation- \(a x^2+b x+c=0\) Let, \(\alpha, \beta\) are the roots of the quadratic equation. Then, \(\alpha+\beta=\frac{-b}{a}\) \(\alpha \beta=\frac{c}{a}\) Given, \(\alpha=\beta^{\mathrm{x}}\) We get, \(\beta^{n+1}=\frac{c}{a}\) \(\beta=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}\) \(\alpha=\left(\frac{c}{a}\right)^{\frac{n}{n+1}}\) Adding both, we get- \(\alpha+\beta=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}=\frac{-b}{a}\) \(-b=a \cdot\left[\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}\right]\) \(-b=\left(c^n\right)^{\frac{1}{n+1}}+\left(a^n c\right)^{\frac{1}{n+1}}\)
AP EAMCET-06.07.2022
Complex Numbers and Quadratic Equation
118122
The sum of squares of roots of the equation \(\mathrm{x}^{\frac{2}{3}}+\mathrm{x}^{\frac{1}{3}}-2=0\) is
1 82
2 65
3 50
4 37
Explanation:
B Given, the equation \(\mathrm{x}^{\frac{2}{3}}+\mathrm{x}^{\frac{1}{3}}-2=0\) Let \(x^{\frac{1}{3}}=t \text {, then the give }\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) Put \(t=1\), we get \(x^{\frac{1}{3}}=1 \Rightarrow x=1\) \(t=-2\), we get \(x^{\frac{1}{3}}=-2\) \(\Rightarrow \quad \mathrm{x}=-8\) \(\therefore \quad\) Roots are \(1,-8\). \(\therefore\) Sum of the squares \(=1+64=65\)
AP EAMCET-06.07.2022
Complex Numbers and Quadratic Equation
118123
If \(\alpha\) satisfies the equation \(\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2\), then the roots of the equations \(\alpha^2 x^2+4 \alpha x+3=\) 0 are
1 1,3
2 \(-1,1\)
3 \(2,-3\)
4 3,4
Explanation:
A Given the equation, \(\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2\) Let \(\sqrt{\frac{2 \mathrm{x}+1}{\mathrm{x}}}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{t}}+\mathrm{t}=2\) Or \(\quad t^2+1=2 t\) Or \(\quad t^2+1-2 t=0 \Rightarrow t=1\) \(\therefore \quad \sqrt{\frac{2 \mathrm{x}+1}{\mathrm{x}}}=1\) \(\Rightarrow \quad 2 \mathrm{x}+1=\mathrm{x}\) \(\Rightarrow \quad \mathrm{x}=-1\) Since, \(\alpha\) satisfy the equation. \(\therefore \quad \alpha=-1\) Now, the given equation \(\alpha^2 x^2+4 \alpha x+3=0\) becomes \(x^2-4 x+3=0\). or\(\mathrm{x}^2-3 \mathrm{x}-\mathrm{x}+3=0\) Or\(\mathrm{x}(\mathrm{x}-3)-1(\mathrm{x}-3)=0\) \((\mathrm{x}-3)(\mathrm{x}-1)=0\)\(\therefore \quad \mathrm{x}=1,3\)
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118125
The solution of \(\sqrt{3 x^2-2}=2 x-1\) are
1 \((2,4)\)
2 \((1,4)\)
3 \((3,4)\)
4 \((1,3)\)
Explanation:
D We have the equation, \(\sqrt{3 x^2-2}=2 x-1\) Squaring, both the sides, we get \(3 x^2-2=(2 x-1)^2=4 x^2+1-4 x\) \(x^2-4 x+3=0\) \(x^2-3 x-x+3=0\) \((x-3)(x-1)=0\) \(x=(1,3)\)
118121
If one root of the quadratic equation \(a x^2+b x+\) \(c=0\) is equal to the \(n^{\text {th }}\) power of the other, then \(\left(a^n\right)^{1 / n+1}+\left(a^n c\right)^{1 / n+1}=\)
1 \(-2 \mathrm{~b}\)
2 \(-\mathrm{b}\)
3 \(b-1\)
4 \(b+1\)
Explanation:
B Given, quadratic equation- \(a x^2+b x+c=0\) Let, \(\alpha, \beta\) are the roots of the quadratic equation. Then, \(\alpha+\beta=\frac{-b}{a}\) \(\alpha \beta=\frac{c}{a}\) Given, \(\alpha=\beta^{\mathrm{x}}\) We get, \(\beta^{n+1}=\frac{c}{a}\) \(\beta=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}\) \(\alpha=\left(\frac{c}{a}\right)^{\frac{n}{n+1}}\) Adding both, we get- \(\alpha+\beta=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}=\frac{-b}{a}\) \(-b=a \cdot\left[\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}\right]\) \(-b=\left(c^n\right)^{\frac{1}{n+1}}+\left(a^n c\right)^{\frac{1}{n+1}}\)
AP EAMCET-06.07.2022
Complex Numbers and Quadratic Equation
118122
The sum of squares of roots of the equation \(\mathrm{x}^{\frac{2}{3}}+\mathrm{x}^{\frac{1}{3}}-2=0\) is
1 82
2 65
3 50
4 37
Explanation:
B Given, the equation \(\mathrm{x}^{\frac{2}{3}}+\mathrm{x}^{\frac{1}{3}}-2=0\) Let \(x^{\frac{1}{3}}=t \text {, then the give }\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) Put \(t=1\), we get \(x^{\frac{1}{3}}=1 \Rightarrow x=1\) \(t=-2\), we get \(x^{\frac{1}{3}}=-2\) \(\Rightarrow \quad \mathrm{x}=-8\) \(\therefore \quad\) Roots are \(1,-8\). \(\therefore\) Sum of the squares \(=1+64=65\)
AP EAMCET-06.07.2022
Complex Numbers and Quadratic Equation
118123
If \(\alpha\) satisfies the equation \(\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2\), then the roots of the equations \(\alpha^2 x^2+4 \alpha x+3=\) 0 are
1 1,3
2 \(-1,1\)
3 \(2,-3\)
4 3,4
Explanation:
A Given the equation, \(\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2\) Let \(\sqrt{\frac{2 \mathrm{x}+1}{\mathrm{x}}}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{t}}+\mathrm{t}=2\) Or \(\quad t^2+1=2 t\) Or \(\quad t^2+1-2 t=0 \Rightarrow t=1\) \(\therefore \quad \sqrt{\frac{2 \mathrm{x}+1}{\mathrm{x}}}=1\) \(\Rightarrow \quad 2 \mathrm{x}+1=\mathrm{x}\) \(\Rightarrow \quad \mathrm{x}=-1\) Since, \(\alpha\) satisfy the equation. \(\therefore \quad \alpha=-1\) Now, the given equation \(\alpha^2 x^2+4 \alpha x+3=0\) becomes \(x^2-4 x+3=0\). or\(\mathrm{x}^2-3 \mathrm{x}-\mathrm{x}+3=0\) Or\(\mathrm{x}(\mathrm{x}-3)-1(\mathrm{x}-3)=0\) \((\mathrm{x}-3)(\mathrm{x}-1)=0\)\(\therefore \quad \mathrm{x}=1,3\)
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118125
The solution of \(\sqrt{3 x^2-2}=2 x-1\) are
1 \((2,4)\)
2 \((1,4)\)
3 \((3,4)\)
4 \((1,3)\)
Explanation:
D We have the equation, \(\sqrt{3 x^2-2}=2 x-1\) Squaring, both the sides, we get \(3 x^2-2=(2 x-1)^2=4 x^2+1-4 x\) \(x^2-4 x+3=0\) \(x^2-3 x-x+3=0\) \((x-3)(x-1)=0\) \(x=(1,3)\)
118121
If one root of the quadratic equation \(a x^2+b x+\) \(c=0\) is equal to the \(n^{\text {th }}\) power of the other, then \(\left(a^n\right)^{1 / n+1}+\left(a^n c\right)^{1 / n+1}=\)
1 \(-2 \mathrm{~b}\)
2 \(-\mathrm{b}\)
3 \(b-1\)
4 \(b+1\)
Explanation:
B Given, quadratic equation- \(a x^2+b x+c=0\) Let, \(\alpha, \beta\) are the roots of the quadratic equation. Then, \(\alpha+\beta=\frac{-b}{a}\) \(\alpha \beta=\frac{c}{a}\) Given, \(\alpha=\beta^{\mathrm{x}}\) We get, \(\beta^{n+1}=\frac{c}{a}\) \(\beta=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}\) \(\alpha=\left(\frac{c}{a}\right)^{\frac{n}{n+1}}\) Adding both, we get- \(\alpha+\beta=\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}=\frac{-b}{a}\) \(-b=a \cdot\left[\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}\right]\) \(-b=\left(c^n\right)^{\frac{1}{n+1}}+\left(a^n c\right)^{\frac{1}{n+1}}\)
AP EAMCET-06.07.2022
Complex Numbers and Quadratic Equation
118122
The sum of squares of roots of the equation \(\mathrm{x}^{\frac{2}{3}}+\mathrm{x}^{\frac{1}{3}}-2=0\) is
1 82
2 65
3 50
4 37
Explanation:
B Given, the equation \(\mathrm{x}^{\frac{2}{3}}+\mathrm{x}^{\frac{1}{3}}-2=0\) Let \(x^{\frac{1}{3}}=t \text {, then the give }\) \(t^2+t-2=0\) \(t^2+2 t-t-2=0\) \(t(t+2)-1(t+2)=0\) \((t-1)(t+2)=0\) Put \(t=1\), we get \(x^{\frac{1}{3}}=1 \Rightarrow x=1\) \(t=-2\), we get \(x^{\frac{1}{3}}=-2\) \(\Rightarrow \quad \mathrm{x}=-8\) \(\therefore \quad\) Roots are \(1,-8\). \(\therefore\) Sum of the squares \(=1+64=65\)
AP EAMCET-06.07.2022
Complex Numbers and Quadratic Equation
118123
If \(\alpha\) satisfies the equation \(\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2\), then the roots of the equations \(\alpha^2 x^2+4 \alpha x+3=\) 0 are
1 1,3
2 \(-1,1\)
3 \(2,-3\)
4 3,4
Explanation:
A Given the equation, \(\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2\) Let \(\sqrt{\frac{2 \mathrm{x}+1}{\mathrm{x}}}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{t}}+\mathrm{t}=2\) Or \(\quad t^2+1=2 t\) Or \(\quad t^2+1-2 t=0 \Rightarrow t=1\) \(\therefore \quad \sqrt{\frac{2 \mathrm{x}+1}{\mathrm{x}}}=1\) \(\Rightarrow \quad 2 \mathrm{x}+1=\mathrm{x}\) \(\Rightarrow \quad \mathrm{x}=-1\) Since, \(\alpha\) satisfy the equation. \(\therefore \quad \alpha=-1\) Now, the given equation \(\alpha^2 x^2+4 \alpha x+3=0\) becomes \(x^2-4 x+3=0\). or\(\mathrm{x}^2-3 \mathrm{x}-\mathrm{x}+3=0\) Or\(\mathrm{x}(\mathrm{x}-3)-1(\mathrm{x}-3)=0\) \((\mathrm{x}-3)(\mathrm{x}-1)=0\)\(\therefore \quad \mathrm{x}=1,3\)
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118125
The solution of \(\sqrt{3 x^2-2}=2 x-1\) are
1 \((2,4)\)
2 \((1,4)\)
3 \((3,4)\)
4 \((1,3)\)
Explanation:
D We have the equation, \(\sqrt{3 x^2-2}=2 x-1\) Squaring, both the sides, we get \(3 x^2-2=(2 x-1)^2=4 x^2+1-4 x\) \(x^2-4 x+3=0\) \(x^2-3 x-x+3=0\) \((x-3)(x-1)=0\) \(x=(1,3)\)