118468
If \(\alpha, \beta\) are the roots of the quadratic equation \(x^2+a x+b=0,(b \neq 0)\), then the quadratic equation whose roots are \(\alpha-\frac{1}{\beta}, \beta-\frac{1}{\alpha}\), is
1 \(a x^2+a(b-1) x+(a-1)^2=0\)
2 \(b x^2+a(b-1) x+(b-1)^2=0\)
3 \(x^2+a x+b v=0\)
4 \(a b x^2+b x+a=0\)
Explanation:
B Given, \(x^2+a x+b=0 \quad b \neq 0\) \(\alpha+\beta=-a\) \(\alpha \cdot \beta=\mathrm{b}\) Now, \(\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \cdot \beta}\right)\) \(=(-a)-\left(\frac{-a}{b}\right)\) \(=-a+\frac{a}{b}\) \(=-\frac{a}{b}(1-b)\) And, \(\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}\) \(=\mathrm{b}+\frac{1}{\mathrm{~b}}-2\) \(=\frac{1}{\mathrm{~b}}\left(\mathrm{~b}^2-2 \mathrm{~b}+1\right)=\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2\) Required of equation whose roots are \(\left(\alpha-\frac{1}{\beta}\right) \text { and }\left(\beta-\frac{1}{\alpha}\right) \text { is, }\) \(\mathrm{x}^2-\left[\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right] \mathrm{x}+\left[\left(\alpha-\frac{1}{\beta}\right) \cdot\left(\beta-\frac{1}{\alpha}\right)\right]=0\) \(\mathrm{x}^2-\frac{\mathrm{a}}{\mathrm{b}}(1-\mathrm{b}) \mathrm{x}+\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2=0\) \(\mathrm{bx}^2+\mathrm{a}(\mathrm{b}-1) \mathrm{x}+(\mathrm{b}-1)^2=0\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118469
The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is
1 \(x^2+11 x-30=0\)
2 \(x^2+11 x+30=0\)
3 \(x^2-11 x-30=0\)
4 \(x^2-11 x+30=0\)
Explanation:
D Sum of the roots \((\alpha+\beta)=11\) Square of the \(\left(\alpha^2+\beta^2\right)=61\) We know that, \((\alpha+\beta)^2=\left(\alpha^2+\beta^2\right)+2 \alpha \beta\) \((11)^2=61+2 \alpha \beta\) \(\alpha \beta=\frac{121-61}{2}\) \(\alpha \beta=30\) The quadratic equation, \(x^2-(\alpha+\beta) x+(\alpha \beta)=0\) \(x^2-11 x+30=0\)
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118470
Let \(p, q \in R\). If \(2-\sqrt{3}\) is a root of the quadratic equation, \(x^2+\mathbf{p}+\mathbf{q}=\mathbf{0}\), then
1 \(q^2-4 p-16=0\)
2 \(p^2-4 q-12=0\)
3 \(\mathrm{p}^2-4 \mathrm{p}+12=0\)
4 \(q^2+4 p+14=0\)
Explanation:
B Given, Let \(\mathrm{p}, \mathrm{q} \in \mathrm{R}\). If we take root as any real number \(\beta\), then quadratic equation will be \(x^2-(\beta+2-\sqrt{3}) x+\beta(2-\sqrt{3})=0\) Now, we can have none or any of the options can be correct depending upon ' \(\beta\) ' Instead of \(p, q \in R\) it should be \(p, q \in R\) it should be \(p\), \(\mathrm{q} \in \mathrm{Q}\) then other root will be \(2+\sqrt{3}\) \(\Rightarrow \quad \mathrm{p}=-(\mathrm{q}+\sqrt{3}-2-\sqrt{3})=-4\) And, \(\quad \mathrm{q}=(2+\sqrt{3})(2-\sqrt{3})=1\) \(\Rightarrow \quad p^2-4 q-12=(-4)^2-4-12=0\)Hence the correct (b)
JEE Main 09.04.2019
Complex Numbers and Quadratic Equation
118471
The quardratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), will be
1 \(7 x^2-6 x+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(x^2-6 x+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}} \text { and } \frac{1}{3-\sqrt{2}}\) So, the required quadratic equation is \(\mathrm{x}^2\) (sum of the roots) \(\mathrm{x}+\) product of the roots \(=0\) \(x^2-\left(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\right) x+\left(\frac{1}{3+\sqrt{2}} \times \frac{1}{3-\sqrt{2}}\right)=0\) \(x^2-\left(\frac{3-\sqrt{2}+3+\sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}\right) x+\frac{1}{3^2-(\sqrt{2})^2}=0\) \(x^2-\frac{6}{9-2} x+\frac{1}{9-2}=0\) \(x^2-\frac{6}{7} x+\frac{1}{7}=0\) \(7 x^2-6 x+1=0\)
Manipal UGET-2020
Complex Numbers and Quadratic Equation
118472
If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\), then the quadratic equation whose roots are \(\sqrt{5} \alpha, \sqrt{5} \beta\) is
A Given, If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\) \(\alpha+\beta=-\frac{b}{a}\) \(\alpha \cdot \beta=\frac{c}{a}\) If the roots of quadratic equation is \(\sqrt{5} \alpha\) and \(\sqrt{5} \beta\) then the equation. \(x^2-(\text { Sum of roots }) x+(\text { product of roots })=0\) \(\Rightarrow \alpha=\frac{-b}{a} \beta=\frac{c}{a}\) \(\Rightarrow x^2-(\sqrt{5} \alpha+\sqrt{5} \beta) x+(\sqrt{5} \alpha \times \sqrt{5} \beta)=0\) \(\Rightarrow x^2-\sqrt{5}(\alpha+\beta) x+5 \alpha \beta=0\) \(\Rightarrow x^2-\sqrt{5}\left(\frac{-b}{a}\right) x+\frac{5 c}{a}=0\) \(\Rightarrow \mathrm{ax}^2+\sqrt{5} b x+5 c=0\)
118468
If \(\alpha, \beta\) are the roots of the quadratic equation \(x^2+a x+b=0,(b \neq 0)\), then the quadratic equation whose roots are \(\alpha-\frac{1}{\beta}, \beta-\frac{1}{\alpha}\), is
1 \(a x^2+a(b-1) x+(a-1)^2=0\)
2 \(b x^2+a(b-1) x+(b-1)^2=0\)
3 \(x^2+a x+b v=0\)
4 \(a b x^2+b x+a=0\)
Explanation:
B Given, \(x^2+a x+b=0 \quad b \neq 0\) \(\alpha+\beta=-a\) \(\alpha \cdot \beta=\mathrm{b}\) Now, \(\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \cdot \beta}\right)\) \(=(-a)-\left(\frac{-a}{b}\right)\) \(=-a+\frac{a}{b}\) \(=-\frac{a}{b}(1-b)\) And, \(\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}\) \(=\mathrm{b}+\frac{1}{\mathrm{~b}}-2\) \(=\frac{1}{\mathrm{~b}}\left(\mathrm{~b}^2-2 \mathrm{~b}+1\right)=\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2\) Required of equation whose roots are \(\left(\alpha-\frac{1}{\beta}\right) \text { and }\left(\beta-\frac{1}{\alpha}\right) \text { is, }\) \(\mathrm{x}^2-\left[\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right] \mathrm{x}+\left[\left(\alpha-\frac{1}{\beta}\right) \cdot\left(\beta-\frac{1}{\alpha}\right)\right]=0\) \(\mathrm{x}^2-\frac{\mathrm{a}}{\mathrm{b}}(1-\mathrm{b}) \mathrm{x}+\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2=0\) \(\mathrm{bx}^2+\mathrm{a}(\mathrm{b}-1) \mathrm{x}+(\mathrm{b}-1)^2=0\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118469
The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is
1 \(x^2+11 x-30=0\)
2 \(x^2+11 x+30=0\)
3 \(x^2-11 x-30=0\)
4 \(x^2-11 x+30=0\)
Explanation:
D Sum of the roots \((\alpha+\beta)=11\) Square of the \(\left(\alpha^2+\beta^2\right)=61\) We know that, \((\alpha+\beta)^2=\left(\alpha^2+\beta^2\right)+2 \alpha \beta\) \((11)^2=61+2 \alpha \beta\) \(\alpha \beta=\frac{121-61}{2}\) \(\alpha \beta=30\) The quadratic equation, \(x^2-(\alpha+\beta) x+(\alpha \beta)=0\) \(x^2-11 x+30=0\)
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118470
Let \(p, q \in R\). If \(2-\sqrt{3}\) is a root of the quadratic equation, \(x^2+\mathbf{p}+\mathbf{q}=\mathbf{0}\), then
1 \(q^2-4 p-16=0\)
2 \(p^2-4 q-12=0\)
3 \(\mathrm{p}^2-4 \mathrm{p}+12=0\)
4 \(q^2+4 p+14=0\)
Explanation:
B Given, Let \(\mathrm{p}, \mathrm{q} \in \mathrm{R}\). If we take root as any real number \(\beta\), then quadratic equation will be \(x^2-(\beta+2-\sqrt{3}) x+\beta(2-\sqrt{3})=0\) Now, we can have none or any of the options can be correct depending upon ' \(\beta\) ' Instead of \(p, q \in R\) it should be \(p, q \in R\) it should be \(p\), \(\mathrm{q} \in \mathrm{Q}\) then other root will be \(2+\sqrt{3}\) \(\Rightarrow \quad \mathrm{p}=-(\mathrm{q}+\sqrt{3}-2-\sqrt{3})=-4\) And, \(\quad \mathrm{q}=(2+\sqrt{3})(2-\sqrt{3})=1\) \(\Rightarrow \quad p^2-4 q-12=(-4)^2-4-12=0\)Hence the correct (b)
JEE Main 09.04.2019
Complex Numbers and Quadratic Equation
118471
The quardratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), will be
1 \(7 x^2-6 x+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(x^2-6 x+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}} \text { and } \frac{1}{3-\sqrt{2}}\) So, the required quadratic equation is \(\mathrm{x}^2\) (sum of the roots) \(\mathrm{x}+\) product of the roots \(=0\) \(x^2-\left(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\right) x+\left(\frac{1}{3+\sqrt{2}} \times \frac{1}{3-\sqrt{2}}\right)=0\) \(x^2-\left(\frac{3-\sqrt{2}+3+\sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}\right) x+\frac{1}{3^2-(\sqrt{2})^2}=0\) \(x^2-\frac{6}{9-2} x+\frac{1}{9-2}=0\) \(x^2-\frac{6}{7} x+\frac{1}{7}=0\) \(7 x^2-6 x+1=0\)
Manipal UGET-2020
Complex Numbers and Quadratic Equation
118472
If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\), then the quadratic equation whose roots are \(\sqrt{5} \alpha, \sqrt{5} \beta\) is
A Given, If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\) \(\alpha+\beta=-\frac{b}{a}\) \(\alpha \cdot \beta=\frac{c}{a}\) If the roots of quadratic equation is \(\sqrt{5} \alpha\) and \(\sqrt{5} \beta\) then the equation. \(x^2-(\text { Sum of roots }) x+(\text { product of roots })=0\) \(\Rightarrow \alpha=\frac{-b}{a} \beta=\frac{c}{a}\) \(\Rightarrow x^2-(\sqrt{5} \alpha+\sqrt{5} \beta) x+(\sqrt{5} \alpha \times \sqrt{5} \beta)=0\) \(\Rightarrow x^2-\sqrt{5}(\alpha+\beta) x+5 \alpha \beta=0\) \(\Rightarrow x^2-\sqrt{5}\left(\frac{-b}{a}\right) x+\frac{5 c}{a}=0\) \(\Rightarrow \mathrm{ax}^2+\sqrt{5} b x+5 c=0\)
118468
If \(\alpha, \beta\) are the roots of the quadratic equation \(x^2+a x+b=0,(b \neq 0)\), then the quadratic equation whose roots are \(\alpha-\frac{1}{\beta}, \beta-\frac{1}{\alpha}\), is
1 \(a x^2+a(b-1) x+(a-1)^2=0\)
2 \(b x^2+a(b-1) x+(b-1)^2=0\)
3 \(x^2+a x+b v=0\)
4 \(a b x^2+b x+a=0\)
Explanation:
B Given, \(x^2+a x+b=0 \quad b \neq 0\) \(\alpha+\beta=-a\) \(\alpha \cdot \beta=\mathrm{b}\) Now, \(\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \cdot \beta}\right)\) \(=(-a)-\left(\frac{-a}{b}\right)\) \(=-a+\frac{a}{b}\) \(=-\frac{a}{b}(1-b)\) And, \(\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}\) \(=\mathrm{b}+\frac{1}{\mathrm{~b}}-2\) \(=\frac{1}{\mathrm{~b}}\left(\mathrm{~b}^2-2 \mathrm{~b}+1\right)=\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2\) Required of equation whose roots are \(\left(\alpha-\frac{1}{\beta}\right) \text { and }\left(\beta-\frac{1}{\alpha}\right) \text { is, }\) \(\mathrm{x}^2-\left[\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right] \mathrm{x}+\left[\left(\alpha-\frac{1}{\beta}\right) \cdot\left(\beta-\frac{1}{\alpha}\right)\right]=0\) \(\mathrm{x}^2-\frac{\mathrm{a}}{\mathrm{b}}(1-\mathrm{b}) \mathrm{x}+\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2=0\) \(\mathrm{bx}^2+\mathrm{a}(\mathrm{b}-1) \mathrm{x}+(\mathrm{b}-1)^2=0\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118469
The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is
1 \(x^2+11 x-30=0\)
2 \(x^2+11 x+30=0\)
3 \(x^2-11 x-30=0\)
4 \(x^2-11 x+30=0\)
Explanation:
D Sum of the roots \((\alpha+\beta)=11\) Square of the \(\left(\alpha^2+\beta^2\right)=61\) We know that, \((\alpha+\beta)^2=\left(\alpha^2+\beta^2\right)+2 \alpha \beta\) \((11)^2=61+2 \alpha \beta\) \(\alpha \beta=\frac{121-61}{2}\) \(\alpha \beta=30\) The quadratic equation, \(x^2-(\alpha+\beta) x+(\alpha \beta)=0\) \(x^2-11 x+30=0\)
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118470
Let \(p, q \in R\). If \(2-\sqrt{3}\) is a root of the quadratic equation, \(x^2+\mathbf{p}+\mathbf{q}=\mathbf{0}\), then
1 \(q^2-4 p-16=0\)
2 \(p^2-4 q-12=0\)
3 \(\mathrm{p}^2-4 \mathrm{p}+12=0\)
4 \(q^2+4 p+14=0\)
Explanation:
B Given, Let \(\mathrm{p}, \mathrm{q} \in \mathrm{R}\). If we take root as any real number \(\beta\), then quadratic equation will be \(x^2-(\beta+2-\sqrt{3}) x+\beta(2-\sqrt{3})=0\) Now, we can have none or any of the options can be correct depending upon ' \(\beta\) ' Instead of \(p, q \in R\) it should be \(p, q \in R\) it should be \(p\), \(\mathrm{q} \in \mathrm{Q}\) then other root will be \(2+\sqrt{3}\) \(\Rightarrow \quad \mathrm{p}=-(\mathrm{q}+\sqrt{3}-2-\sqrt{3})=-4\) And, \(\quad \mathrm{q}=(2+\sqrt{3})(2-\sqrt{3})=1\) \(\Rightarrow \quad p^2-4 q-12=(-4)^2-4-12=0\)Hence the correct (b)
JEE Main 09.04.2019
Complex Numbers and Quadratic Equation
118471
The quardratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), will be
1 \(7 x^2-6 x+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(x^2-6 x+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}} \text { and } \frac{1}{3-\sqrt{2}}\) So, the required quadratic equation is \(\mathrm{x}^2\) (sum of the roots) \(\mathrm{x}+\) product of the roots \(=0\) \(x^2-\left(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\right) x+\left(\frac{1}{3+\sqrt{2}} \times \frac{1}{3-\sqrt{2}}\right)=0\) \(x^2-\left(\frac{3-\sqrt{2}+3+\sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}\right) x+\frac{1}{3^2-(\sqrt{2})^2}=0\) \(x^2-\frac{6}{9-2} x+\frac{1}{9-2}=0\) \(x^2-\frac{6}{7} x+\frac{1}{7}=0\) \(7 x^2-6 x+1=0\)
Manipal UGET-2020
Complex Numbers and Quadratic Equation
118472
If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\), then the quadratic equation whose roots are \(\sqrt{5} \alpha, \sqrt{5} \beta\) is
A Given, If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\) \(\alpha+\beta=-\frac{b}{a}\) \(\alpha \cdot \beta=\frac{c}{a}\) If the roots of quadratic equation is \(\sqrt{5} \alpha\) and \(\sqrt{5} \beta\) then the equation. \(x^2-(\text { Sum of roots }) x+(\text { product of roots })=0\) \(\Rightarrow \alpha=\frac{-b}{a} \beta=\frac{c}{a}\) \(\Rightarrow x^2-(\sqrt{5} \alpha+\sqrt{5} \beta) x+(\sqrt{5} \alpha \times \sqrt{5} \beta)=0\) \(\Rightarrow x^2-\sqrt{5}(\alpha+\beta) x+5 \alpha \beta=0\) \(\Rightarrow x^2-\sqrt{5}\left(\frac{-b}{a}\right) x+\frac{5 c}{a}=0\) \(\Rightarrow \mathrm{ax}^2+\sqrt{5} b x+5 c=0\)
118468
If \(\alpha, \beta\) are the roots of the quadratic equation \(x^2+a x+b=0,(b \neq 0)\), then the quadratic equation whose roots are \(\alpha-\frac{1}{\beta}, \beta-\frac{1}{\alpha}\), is
1 \(a x^2+a(b-1) x+(a-1)^2=0\)
2 \(b x^2+a(b-1) x+(b-1)^2=0\)
3 \(x^2+a x+b v=0\)
4 \(a b x^2+b x+a=0\)
Explanation:
B Given, \(x^2+a x+b=0 \quad b \neq 0\) \(\alpha+\beta=-a\) \(\alpha \cdot \beta=\mathrm{b}\) Now, \(\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \cdot \beta}\right)\) \(=(-a)-\left(\frac{-a}{b}\right)\) \(=-a+\frac{a}{b}\) \(=-\frac{a}{b}(1-b)\) And, \(\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}\) \(=\mathrm{b}+\frac{1}{\mathrm{~b}}-2\) \(=\frac{1}{\mathrm{~b}}\left(\mathrm{~b}^2-2 \mathrm{~b}+1\right)=\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2\) Required of equation whose roots are \(\left(\alpha-\frac{1}{\beta}\right) \text { and }\left(\beta-\frac{1}{\alpha}\right) \text { is, }\) \(\mathrm{x}^2-\left[\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right] \mathrm{x}+\left[\left(\alpha-\frac{1}{\beta}\right) \cdot\left(\beta-\frac{1}{\alpha}\right)\right]=0\) \(\mathrm{x}^2-\frac{\mathrm{a}}{\mathrm{b}}(1-\mathrm{b}) \mathrm{x}+\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2=0\) \(\mathrm{bx}^2+\mathrm{a}(\mathrm{b}-1) \mathrm{x}+(\mathrm{b}-1)^2=0\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118469
The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is
1 \(x^2+11 x-30=0\)
2 \(x^2+11 x+30=0\)
3 \(x^2-11 x-30=0\)
4 \(x^2-11 x+30=0\)
Explanation:
D Sum of the roots \((\alpha+\beta)=11\) Square of the \(\left(\alpha^2+\beta^2\right)=61\) We know that, \((\alpha+\beta)^2=\left(\alpha^2+\beta^2\right)+2 \alpha \beta\) \((11)^2=61+2 \alpha \beta\) \(\alpha \beta=\frac{121-61}{2}\) \(\alpha \beta=30\) The quadratic equation, \(x^2-(\alpha+\beta) x+(\alpha \beta)=0\) \(x^2-11 x+30=0\)
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118470
Let \(p, q \in R\). If \(2-\sqrt{3}\) is a root of the quadratic equation, \(x^2+\mathbf{p}+\mathbf{q}=\mathbf{0}\), then
1 \(q^2-4 p-16=0\)
2 \(p^2-4 q-12=0\)
3 \(\mathrm{p}^2-4 \mathrm{p}+12=0\)
4 \(q^2+4 p+14=0\)
Explanation:
B Given, Let \(\mathrm{p}, \mathrm{q} \in \mathrm{R}\). If we take root as any real number \(\beta\), then quadratic equation will be \(x^2-(\beta+2-\sqrt{3}) x+\beta(2-\sqrt{3})=0\) Now, we can have none or any of the options can be correct depending upon ' \(\beta\) ' Instead of \(p, q \in R\) it should be \(p, q \in R\) it should be \(p\), \(\mathrm{q} \in \mathrm{Q}\) then other root will be \(2+\sqrt{3}\) \(\Rightarrow \quad \mathrm{p}=-(\mathrm{q}+\sqrt{3}-2-\sqrt{3})=-4\) And, \(\quad \mathrm{q}=(2+\sqrt{3})(2-\sqrt{3})=1\) \(\Rightarrow \quad p^2-4 q-12=(-4)^2-4-12=0\)Hence the correct (b)
JEE Main 09.04.2019
Complex Numbers and Quadratic Equation
118471
The quardratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), will be
1 \(7 x^2-6 x+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(x^2-6 x+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}} \text { and } \frac{1}{3-\sqrt{2}}\) So, the required quadratic equation is \(\mathrm{x}^2\) (sum of the roots) \(\mathrm{x}+\) product of the roots \(=0\) \(x^2-\left(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\right) x+\left(\frac{1}{3+\sqrt{2}} \times \frac{1}{3-\sqrt{2}}\right)=0\) \(x^2-\left(\frac{3-\sqrt{2}+3+\sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}\right) x+\frac{1}{3^2-(\sqrt{2})^2}=0\) \(x^2-\frac{6}{9-2} x+\frac{1}{9-2}=0\) \(x^2-\frac{6}{7} x+\frac{1}{7}=0\) \(7 x^2-6 x+1=0\)
Manipal UGET-2020
Complex Numbers and Quadratic Equation
118472
If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\), then the quadratic equation whose roots are \(\sqrt{5} \alpha, \sqrt{5} \beta\) is
A Given, If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\) \(\alpha+\beta=-\frac{b}{a}\) \(\alpha \cdot \beta=\frac{c}{a}\) If the roots of quadratic equation is \(\sqrt{5} \alpha\) and \(\sqrt{5} \beta\) then the equation. \(x^2-(\text { Sum of roots }) x+(\text { product of roots })=0\) \(\Rightarrow \alpha=\frac{-b}{a} \beta=\frac{c}{a}\) \(\Rightarrow x^2-(\sqrt{5} \alpha+\sqrt{5} \beta) x+(\sqrt{5} \alpha \times \sqrt{5} \beta)=0\) \(\Rightarrow x^2-\sqrt{5}(\alpha+\beta) x+5 \alpha \beta=0\) \(\Rightarrow x^2-\sqrt{5}\left(\frac{-b}{a}\right) x+\frac{5 c}{a}=0\) \(\Rightarrow \mathrm{ax}^2+\sqrt{5} b x+5 c=0\)
118468
If \(\alpha, \beta\) are the roots of the quadratic equation \(x^2+a x+b=0,(b \neq 0)\), then the quadratic equation whose roots are \(\alpha-\frac{1}{\beta}, \beta-\frac{1}{\alpha}\), is
1 \(a x^2+a(b-1) x+(a-1)^2=0\)
2 \(b x^2+a(b-1) x+(b-1)^2=0\)
3 \(x^2+a x+b v=0\)
4 \(a b x^2+b x+a=0\)
Explanation:
B Given, \(x^2+a x+b=0 \quad b \neq 0\) \(\alpha+\beta=-a\) \(\alpha \cdot \beta=\mathrm{b}\) Now, \(\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \cdot \beta}\right)\) \(=(-a)-\left(\frac{-a}{b}\right)\) \(=-a+\frac{a}{b}\) \(=-\frac{a}{b}(1-b)\) And, \(\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}\) \(=\mathrm{b}+\frac{1}{\mathrm{~b}}-2\) \(=\frac{1}{\mathrm{~b}}\left(\mathrm{~b}^2-2 \mathrm{~b}+1\right)=\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2\) Required of equation whose roots are \(\left(\alpha-\frac{1}{\beta}\right) \text { and }\left(\beta-\frac{1}{\alpha}\right) \text { is, }\) \(\mathrm{x}^2-\left[\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right] \mathrm{x}+\left[\left(\alpha-\frac{1}{\beta}\right) \cdot\left(\beta-\frac{1}{\alpha}\right)\right]=0\) \(\mathrm{x}^2-\frac{\mathrm{a}}{\mathrm{b}}(1-\mathrm{b}) \mathrm{x}+\frac{1}{\mathrm{~b}}(\mathrm{~b}-1)^2=0\) \(\mathrm{bx}^2+\mathrm{a}(\mathrm{b}-1) \mathrm{x}+(\mathrm{b}-1)^2=0\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118469
The quadratic equation whose sum of the roots is 11 and sum of squares of the roots is 61 is
1 \(x^2+11 x-30=0\)
2 \(x^2+11 x+30=0\)
3 \(x^2-11 x-30=0\)
4 \(x^2-11 x+30=0\)
Explanation:
D Sum of the roots \((\alpha+\beta)=11\) Square of the \(\left(\alpha^2+\beta^2\right)=61\) We know that, \((\alpha+\beta)^2=\left(\alpha^2+\beta^2\right)+2 \alpha \beta\) \((11)^2=61+2 \alpha \beta\) \(\alpha \beta=\frac{121-61}{2}\) \(\alpha \beta=30\) The quadratic equation, \(x^2-(\alpha+\beta) x+(\alpha \beta)=0\) \(x^2-11 x+30=0\)
AP EAMCET-08.07.2022
Complex Numbers and Quadratic Equation
118470
Let \(p, q \in R\). If \(2-\sqrt{3}\) is a root of the quadratic equation, \(x^2+\mathbf{p}+\mathbf{q}=\mathbf{0}\), then
1 \(q^2-4 p-16=0\)
2 \(p^2-4 q-12=0\)
3 \(\mathrm{p}^2-4 \mathrm{p}+12=0\)
4 \(q^2+4 p+14=0\)
Explanation:
B Given, Let \(\mathrm{p}, \mathrm{q} \in \mathrm{R}\). If we take root as any real number \(\beta\), then quadratic equation will be \(x^2-(\beta+2-\sqrt{3}) x+\beta(2-\sqrt{3})=0\) Now, we can have none or any of the options can be correct depending upon ' \(\beta\) ' Instead of \(p, q \in R\) it should be \(p, q \in R\) it should be \(p\), \(\mathrm{q} \in \mathrm{Q}\) then other root will be \(2+\sqrt{3}\) \(\Rightarrow \quad \mathrm{p}=-(\mathrm{q}+\sqrt{3}-2-\sqrt{3})=-4\) And, \(\quad \mathrm{q}=(2+\sqrt{3})(2-\sqrt{3})=1\) \(\Rightarrow \quad p^2-4 q-12=(-4)^2-4-12=0\)Hence the correct (b)
JEE Main 09.04.2019
Complex Numbers and Quadratic Equation
118471
The quardratic equation whose roots are \(\frac{1}{3+\sqrt{2}}\) and \(\frac{1}{3-\sqrt{2}}\), will be
1 \(7 x^2-6 x+1=0\)
2 \(6 x^2-7 x+1=0\)
3 \(x^2-6 x+7=0\)
4 \(x^2-7 x+6=0\)
Explanation:
A Given, The quadratic equation whose roots are \(\frac{1}{3+\sqrt{2}} \text { and } \frac{1}{3-\sqrt{2}}\) So, the required quadratic equation is \(\mathrm{x}^2\) (sum of the roots) \(\mathrm{x}+\) product of the roots \(=0\) \(x^2-\left(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}\right) x+\left(\frac{1}{3+\sqrt{2}} \times \frac{1}{3-\sqrt{2}}\right)=0\) \(x^2-\left(\frac{3-\sqrt{2}+3+\sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}\right) x+\frac{1}{3^2-(\sqrt{2})^2}=0\) \(x^2-\frac{6}{9-2} x+\frac{1}{9-2}=0\) \(x^2-\frac{6}{7} x+\frac{1}{7}=0\) \(7 x^2-6 x+1=0\)
Manipal UGET-2020
Complex Numbers and Quadratic Equation
118472
If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\), then the quadratic equation whose roots are \(\sqrt{5} \alpha, \sqrt{5} \beta\) is
A Given, If \(\alpha, \beta\) are the roots of \(a x^2+b x+c=0\) \(\alpha+\beta=-\frac{b}{a}\) \(\alpha \cdot \beta=\frac{c}{a}\) If the roots of quadratic equation is \(\sqrt{5} \alpha\) and \(\sqrt{5} \beta\) then the equation. \(x^2-(\text { Sum of roots }) x+(\text { product of roots })=0\) \(\Rightarrow \alpha=\frac{-b}{a} \beta=\frac{c}{a}\) \(\Rightarrow x^2-(\sqrt{5} \alpha+\sqrt{5} \beta) x+(\sqrt{5} \alpha \times \sqrt{5} \beta)=0\) \(\Rightarrow x^2-\sqrt{5}(\alpha+\beta) x+5 \alpha \beta=0\) \(\Rightarrow x^2-\sqrt{5}\left(\frac{-b}{a}\right) x+\frac{5 c}{a}=0\) \(\Rightarrow \mathrm{ax}^2+\sqrt{5} b x+5 c=0\)
