118279
If \(\alpha, \beta\) are the real roots of \(x^2+p x+q=0\) and \(\alpha^4, \beta^4\) are the roots of \(x^2-r x+s=0\), then the equation \(x^2-4 q x+2 q^2-r=0\) has always
1 two positive roots
2 two negative roots
3 one positive root and one negative root
4 two imaginary roots
Explanation:
C Given, \(\alpha, \beta\) are the real roots of \(x^2+p x+q=0=0\) \(\alpha+\beta=-p\) \(\alpha \cdot \beta=q\) We know that, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(\quad=p^2-2 q\) \(\alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2\) \(\quad=\left(p^2-2 q\right)^2-2 q^2\) \(\alpha^4, \beta^4\) are the roots of \(x^2-r x+s=0\) \(\alpha^4+\beta^4=r\) \(\alpha^4 \times \beta^4=s\) \(\alpha^4+\beta^4=\left(p^2-2 q\right)^2-2 q^2\) \(r=\left(p^2-2 q\right)^2-2 q^2\) \(r+2 q^2=\left(p^2-2 q\right)^2\) The roots of the equation \(x^2-4 q x+2 q^2-r=0\) \(x=\frac{4 q \pm \sqrt{(4 q)^2-4\left(2 q^2-r\right)}}{2}\) \(x=2 q \pm \sqrt{2 q^2+r}\) From equation (i), \(x=2 q \pm \sqrt{\left(p^2-2 q\right)^2}\) \(x=2 q \pm\left(p^2-2 q\right)\) \(\text { One of the root }=2 q+p^2-2 q\) \(\quad=p^2, \text { which is positive }\) \(\text { Second root is }=2 q-p^2+2 q\) \(\quad=4 q-p^2\) \(\therefore p^2-4 q>0, \text { since the first quadratic has two real }\) \(\text { roots }\) \(4 q-p^2 \text { is negative. }\) Second root is \(=2 q-p^2+2 q\) \(\therefore p^2-4 q>0\), since the first quadratic has two real roots \(4 q-p^2\) is negative.
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118280
If \(\alpha, \beta\) are the roots of the equation \(x^2+2 x+4\) \(=0\), then \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\) equal to
1 \(\frac{-1}{4}\)
2 \(\frac{1}{4}\)
3 32
4 \(\frac{1}{32}\)
Explanation:
A Given, \(\alpha, \beta\) are the roots of the equations \(x^2+2 x+4=0\) \(\alpha+\beta=-2\) \(\alpha \cdot \beta=4\) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \cdot \beta}{(\alpha \cdot \beta)^2}\) \(=\frac{(-2)^2-2(4)}{(4)^2}=\frac{4-8}{16}=\frac{-1}{4}\)
AP EAMCET-05.10.2021
Complex Numbers and Quadratic Equation
118282
If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(a x^2+b x+c=0\) and \(3 b^2=16 a c\), then
1 \(\alpha=4 \beta\) or \(\beta=4 \alpha\)
2 \(\alpha=-4 \beta\) or \(\beta=4 \alpha\)
3 \(\alpha=3 \beta\) or \(\beta=3 \alpha\)
4 \(\alpha=-3 \beta\) or \(\beta=-3 \alpha\)
Explanation:
C Given, \(a x^2+b x+c=0\) Sum of roots, \(\alpha+\beta=\frac{-b}{a}\) Product of roots \(\alpha \cdot \beta=\frac{\mathrm{c}}{\mathrm{a}}\) Given, \(3 b^2=16 a c\) \(3 a^2(\alpha+\beta)^2=16 a \cdot a(\alpha \beta)\) \(3 \alpha^2+3 \beta^2+6 \alpha \beta=16 \alpha \beta\) \(3 \alpha^2-10 \alpha \beta+3 \beta^2=0\) \((\alpha-3 \beta)(3 \alpha-\beta)=0\) \(\alpha=3 \beta, \beta=3 \alpha\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118283
If one root of \(x^2-x-k=0\) is square of the other, then \(k\) is equal to
1 \(2 \pm \sqrt{3}\)
2 \(3 \pm \sqrt{2}\)
3 \(2 \pm \sqrt{5}\)
4 \(5 \pm \sqrt{2}\)
Explanation:
C Given that one root of \(\mathrm{x}^2-\mathrm{x}-\mathrm{k}=0\) is square of the other then we have to find \(\mathrm{k}\) is = ? Let us consider the roots be \(\alpha, \alpha^2\) The quadratic equation is \((x-\alpha)\left(x-\alpha^2\right)=0\) \(x^2-\left(\alpha+\alpha^2\right) x+\alpha^3=0\) Comparing this with the given quadratic equation \(x^2-x-k=x^2-\left(\alpha+\alpha^2\right) x+\alpha^3\) Comparing the coefficient we get \(\alpha+\alpha^2=1 \text { and } \mathrm{k}=-\alpha^3\) \(\alpha+\alpha^2=1\) \(\left(\alpha+\alpha^2\right)^3=1\) \(\alpha^3+\alpha^6+3 \alpha^3\left(\alpha+\alpha^2\right)=1\) \(\mathrm{k}=-\alpha^3\) Substituting this in equation (i) \(-\mathrm{k}+\mathrm{k}^2+3(-\mathrm{k})(1)=1\) \(\mathrm{k}^2-4 \mathrm{k}-1=0\) \(\mathrm{k}=\frac{4 \pm \sqrt{16+4}}{2}\) \(\mathrm{k}=2 \pm \sqrt{5}\)
118279
If \(\alpha, \beta\) are the real roots of \(x^2+p x+q=0\) and \(\alpha^4, \beta^4\) are the roots of \(x^2-r x+s=0\), then the equation \(x^2-4 q x+2 q^2-r=0\) has always
1 two positive roots
2 two negative roots
3 one positive root and one negative root
4 two imaginary roots
Explanation:
C Given, \(\alpha, \beta\) are the real roots of \(x^2+p x+q=0=0\) \(\alpha+\beta=-p\) \(\alpha \cdot \beta=q\) We know that, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(\quad=p^2-2 q\) \(\alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2\) \(\quad=\left(p^2-2 q\right)^2-2 q^2\) \(\alpha^4, \beta^4\) are the roots of \(x^2-r x+s=0\) \(\alpha^4+\beta^4=r\) \(\alpha^4 \times \beta^4=s\) \(\alpha^4+\beta^4=\left(p^2-2 q\right)^2-2 q^2\) \(r=\left(p^2-2 q\right)^2-2 q^2\) \(r+2 q^2=\left(p^2-2 q\right)^2\) The roots of the equation \(x^2-4 q x+2 q^2-r=0\) \(x=\frac{4 q \pm \sqrt{(4 q)^2-4\left(2 q^2-r\right)}}{2}\) \(x=2 q \pm \sqrt{2 q^2+r}\) From equation (i), \(x=2 q \pm \sqrt{\left(p^2-2 q\right)^2}\) \(x=2 q \pm\left(p^2-2 q\right)\) \(\text { One of the root }=2 q+p^2-2 q\) \(\quad=p^2, \text { which is positive }\) \(\text { Second root is }=2 q-p^2+2 q\) \(\quad=4 q-p^2\) \(\therefore p^2-4 q>0, \text { since the first quadratic has two real }\) \(\text { roots }\) \(4 q-p^2 \text { is negative. }\) Second root is \(=2 q-p^2+2 q\) \(\therefore p^2-4 q>0\), since the first quadratic has two real roots \(4 q-p^2\) is negative.
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118280
If \(\alpha, \beta\) are the roots of the equation \(x^2+2 x+4\) \(=0\), then \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\) equal to
1 \(\frac{-1}{4}\)
2 \(\frac{1}{4}\)
3 32
4 \(\frac{1}{32}\)
Explanation:
A Given, \(\alpha, \beta\) are the roots of the equations \(x^2+2 x+4=0\) \(\alpha+\beta=-2\) \(\alpha \cdot \beta=4\) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \cdot \beta}{(\alpha \cdot \beta)^2}\) \(=\frac{(-2)^2-2(4)}{(4)^2}=\frac{4-8}{16}=\frac{-1}{4}\)
AP EAMCET-05.10.2021
Complex Numbers and Quadratic Equation
118282
If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(a x^2+b x+c=0\) and \(3 b^2=16 a c\), then
1 \(\alpha=4 \beta\) or \(\beta=4 \alpha\)
2 \(\alpha=-4 \beta\) or \(\beta=4 \alpha\)
3 \(\alpha=3 \beta\) or \(\beta=3 \alpha\)
4 \(\alpha=-3 \beta\) or \(\beta=-3 \alpha\)
Explanation:
C Given, \(a x^2+b x+c=0\) Sum of roots, \(\alpha+\beta=\frac{-b}{a}\) Product of roots \(\alpha \cdot \beta=\frac{\mathrm{c}}{\mathrm{a}}\) Given, \(3 b^2=16 a c\) \(3 a^2(\alpha+\beta)^2=16 a \cdot a(\alpha \beta)\) \(3 \alpha^2+3 \beta^2+6 \alpha \beta=16 \alpha \beta\) \(3 \alpha^2-10 \alpha \beta+3 \beta^2=0\) \((\alpha-3 \beta)(3 \alpha-\beta)=0\) \(\alpha=3 \beta, \beta=3 \alpha\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118283
If one root of \(x^2-x-k=0\) is square of the other, then \(k\) is equal to
1 \(2 \pm \sqrt{3}\)
2 \(3 \pm \sqrt{2}\)
3 \(2 \pm \sqrt{5}\)
4 \(5 \pm \sqrt{2}\)
Explanation:
C Given that one root of \(\mathrm{x}^2-\mathrm{x}-\mathrm{k}=0\) is square of the other then we have to find \(\mathrm{k}\) is = ? Let us consider the roots be \(\alpha, \alpha^2\) The quadratic equation is \((x-\alpha)\left(x-\alpha^2\right)=0\) \(x^2-\left(\alpha+\alpha^2\right) x+\alpha^3=0\) Comparing this with the given quadratic equation \(x^2-x-k=x^2-\left(\alpha+\alpha^2\right) x+\alpha^3\) Comparing the coefficient we get \(\alpha+\alpha^2=1 \text { and } \mathrm{k}=-\alpha^3\) \(\alpha+\alpha^2=1\) \(\left(\alpha+\alpha^2\right)^3=1\) \(\alpha^3+\alpha^6+3 \alpha^3\left(\alpha+\alpha^2\right)=1\) \(\mathrm{k}=-\alpha^3\) Substituting this in equation (i) \(-\mathrm{k}+\mathrm{k}^2+3(-\mathrm{k})(1)=1\) \(\mathrm{k}^2-4 \mathrm{k}-1=0\) \(\mathrm{k}=\frac{4 \pm \sqrt{16+4}}{2}\) \(\mathrm{k}=2 \pm \sqrt{5}\)
118279
If \(\alpha, \beta\) are the real roots of \(x^2+p x+q=0\) and \(\alpha^4, \beta^4\) are the roots of \(x^2-r x+s=0\), then the equation \(x^2-4 q x+2 q^2-r=0\) has always
1 two positive roots
2 two negative roots
3 one positive root and one negative root
4 two imaginary roots
Explanation:
C Given, \(\alpha, \beta\) are the real roots of \(x^2+p x+q=0=0\) \(\alpha+\beta=-p\) \(\alpha \cdot \beta=q\) We know that, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(\quad=p^2-2 q\) \(\alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2\) \(\quad=\left(p^2-2 q\right)^2-2 q^2\) \(\alpha^4, \beta^4\) are the roots of \(x^2-r x+s=0\) \(\alpha^4+\beta^4=r\) \(\alpha^4 \times \beta^4=s\) \(\alpha^4+\beta^4=\left(p^2-2 q\right)^2-2 q^2\) \(r=\left(p^2-2 q\right)^2-2 q^2\) \(r+2 q^2=\left(p^2-2 q\right)^2\) The roots of the equation \(x^2-4 q x+2 q^2-r=0\) \(x=\frac{4 q \pm \sqrt{(4 q)^2-4\left(2 q^2-r\right)}}{2}\) \(x=2 q \pm \sqrt{2 q^2+r}\) From equation (i), \(x=2 q \pm \sqrt{\left(p^2-2 q\right)^2}\) \(x=2 q \pm\left(p^2-2 q\right)\) \(\text { One of the root }=2 q+p^2-2 q\) \(\quad=p^2, \text { which is positive }\) \(\text { Second root is }=2 q-p^2+2 q\) \(\quad=4 q-p^2\) \(\therefore p^2-4 q>0, \text { since the first quadratic has two real }\) \(\text { roots }\) \(4 q-p^2 \text { is negative. }\) Second root is \(=2 q-p^2+2 q\) \(\therefore p^2-4 q>0\), since the first quadratic has two real roots \(4 q-p^2\) is negative.
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118280
If \(\alpha, \beta\) are the roots of the equation \(x^2+2 x+4\) \(=0\), then \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\) equal to
1 \(\frac{-1}{4}\)
2 \(\frac{1}{4}\)
3 32
4 \(\frac{1}{32}\)
Explanation:
A Given, \(\alpha, \beta\) are the roots of the equations \(x^2+2 x+4=0\) \(\alpha+\beta=-2\) \(\alpha \cdot \beta=4\) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \cdot \beta}{(\alpha \cdot \beta)^2}\) \(=\frac{(-2)^2-2(4)}{(4)^2}=\frac{4-8}{16}=\frac{-1}{4}\)
AP EAMCET-05.10.2021
Complex Numbers and Quadratic Equation
118282
If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(a x^2+b x+c=0\) and \(3 b^2=16 a c\), then
1 \(\alpha=4 \beta\) or \(\beta=4 \alpha\)
2 \(\alpha=-4 \beta\) or \(\beta=4 \alpha\)
3 \(\alpha=3 \beta\) or \(\beta=3 \alpha\)
4 \(\alpha=-3 \beta\) or \(\beta=-3 \alpha\)
Explanation:
C Given, \(a x^2+b x+c=0\) Sum of roots, \(\alpha+\beta=\frac{-b}{a}\) Product of roots \(\alpha \cdot \beta=\frac{\mathrm{c}}{\mathrm{a}}\) Given, \(3 b^2=16 a c\) \(3 a^2(\alpha+\beta)^2=16 a \cdot a(\alpha \beta)\) \(3 \alpha^2+3 \beta^2+6 \alpha \beta=16 \alpha \beta\) \(3 \alpha^2-10 \alpha \beta+3 \beta^2=0\) \((\alpha-3 \beta)(3 \alpha-\beta)=0\) \(\alpha=3 \beta, \beta=3 \alpha\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118283
If one root of \(x^2-x-k=0\) is square of the other, then \(k\) is equal to
1 \(2 \pm \sqrt{3}\)
2 \(3 \pm \sqrt{2}\)
3 \(2 \pm \sqrt{5}\)
4 \(5 \pm \sqrt{2}\)
Explanation:
C Given that one root of \(\mathrm{x}^2-\mathrm{x}-\mathrm{k}=0\) is square of the other then we have to find \(\mathrm{k}\) is = ? Let us consider the roots be \(\alpha, \alpha^2\) The quadratic equation is \((x-\alpha)\left(x-\alpha^2\right)=0\) \(x^2-\left(\alpha+\alpha^2\right) x+\alpha^3=0\) Comparing this with the given quadratic equation \(x^2-x-k=x^2-\left(\alpha+\alpha^2\right) x+\alpha^3\) Comparing the coefficient we get \(\alpha+\alpha^2=1 \text { and } \mathrm{k}=-\alpha^3\) \(\alpha+\alpha^2=1\) \(\left(\alpha+\alpha^2\right)^3=1\) \(\alpha^3+\alpha^6+3 \alpha^3\left(\alpha+\alpha^2\right)=1\) \(\mathrm{k}=-\alpha^3\) Substituting this in equation (i) \(-\mathrm{k}+\mathrm{k}^2+3(-\mathrm{k})(1)=1\) \(\mathrm{k}^2-4 \mathrm{k}-1=0\) \(\mathrm{k}=\frac{4 \pm \sqrt{16+4}}{2}\) \(\mathrm{k}=2 \pm \sqrt{5}\)
118279
If \(\alpha, \beta\) are the real roots of \(x^2+p x+q=0\) and \(\alpha^4, \beta^4\) are the roots of \(x^2-r x+s=0\), then the equation \(x^2-4 q x+2 q^2-r=0\) has always
1 two positive roots
2 two negative roots
3 one positive root and one negative root
4 two imaginary roots
Explanation:
C Given, \(\alpha, \beta\) are the real roots of \(x^2+p x+q=0=0\) \(\alpha+\beta=-p\) \(\alpha \cdot \beta=q\) We know that, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\) \(\quad=p^2-2 q\) \(\alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2\) \(\quad=\left(p^2-2 q\right)^2-2 q^2\) \(\alpha^4, \beta^4\) are the roots of \(x^2-r x+s=0\) \(\alpha^4+\beta^4=r\) \(\alpha^4 \times \beta^4=s\) \(\alpha^4+\beta^4=\left(p^2-2 q\right)^2-2 q^2\) \(r=\left(p^2-2 q\right)^2-2 q^2\) \(r+2 q^2=\left(p^2-2 q\right)^2\) The roots of the equation \(x^2-4 q x+2 q^2-r=0\) \(x=\frac{4 q \pm \sqrt{(4 q)^2-4\left(2 q^2-r\right)}}{2}\) \(x=2 q \pm \sqrt{2 q^2+r}\) From equation (i), \(x=2 q \pm \sqrt{\left(p^2-2 q\right)^2}\) \(x=2 q \pm\left(p^2-2 q\right)\) \(\text { One of the root }=2 q+p^2-2 q\) \(\quad=p^2, \text { which is positive }\) \(\text { Second root is }=2 q-p^2+2 q\) \(\quad=4 q-p^2\) \(\therefore p^2-4 q>0, \text { since the first quadratic has two real }\) \(\text { roots }\) \(4 q-p^2 \text { is negative. }\) Second root is \(=2 q-p^2+2 q\) \(\therefore p^2-4 q>0\), since the first quadratic has two real roots \(4 q-p^2\) is negative.
AP EAMCET-21.04.2019
Complex Numbers and Quadratic Equation
118280
If \(\alpha, \beta\) are the roots of the equation \(x^2+2 x+4\) \(=0\), then \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\) equal to
1 \(\frac{-1}{4}\)
2 \(\frac{1}{4}\)
3 32
4 \(\frac{1}{32}\)
Explanation:
A Given, \(\alpha, \beta\) are the roots of the equations \(x^2+2 x+4=0\) \(\alpha+\beta=-2\) \(\alpha \cdot \beta=4\) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \cdot \beta}{(\alpha \cdot \beta)^2}\) \(=\frac{(-2)^2-2(4)}{(4)^2}=\frac{4-8}{16}=\frac{-1}{4}\)
AP EAMCET-05.10.2021
Complex Numbers and Quadratic Equation
118282
If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(a x^2+b x+c=0\) and \(3 b^2=16 a c\), then
1 \(\alpha=4 \beta\) or \(\beta=4 \alpha\)
2 \(\alpha=-4 \beta\) or \(\beta=4 \alpha\)
3 \(\alpha=3 \beta\) or \(\beta=3 \alpha\)
4 \(\alpha=-3 \beta\) or \(\beta=-3 \alpha\)
Explanation:
C Given, \(a x^2+b x+c=0\) Sum of roots, \(\alpha+\beta=\frac{-b}{a}\) Product of roots \(\alpha \cdot \beta=\frac{\mathrm{c}}{\mathrm{a}}\) Given, \(3 b^2=16 a c\) \(3 a^2(\alpha+\beta)^2=16 a \cdot a(\alpha \beta)\) \(3 \alpha^2+3 \beta^2+6 \alpha \beta=16 \alpha \beta\) \(3 \alpha^2-10 \alpha \beta+3 \beta^2=0\) \((\alpha-3 \beta)(3 \alpha-\beta)=0\) \(\alpha=3 \beta, \beta=3 \alpha\)
WB JEE-2013
Complex Numbers and Quadratic Equation
118283
If one root of \(x^2-x-k=0\) is square of the other, then \(k\) is equal to
1 \(2 \pm \sqrt{3}\)
2 \(3 \pm \sqrt{2}\)
3 \(2 \pm \sqrt{5}\)
4 \(5 \pm \sqrt{2}\)
Explanation:
C Given that one root of \(\mathrm{x}^2-\mathrm{x}-\mathrm{k}=0\) is square of the other then we have to find \(\mathrm{k}\) is = ? Let us consider the roots be \(\alpha, \alpha^2\) The quadratic equation is \((x-\alpha)\left(x-\alpha^2\right)=0\) \(x^2-\left(\alpha+\alpha^2\right) x+\alpha^3=0\) Comparing this with the given quadratic equation \(x^2-x-k=x^2-\left(\alpha+\alpha^2\right) x+\alpha^3\) Comparing the coefficient we get \(\alpha+\alpha^2=1 \text { and } \mathrm{k}=-\alpha^3\) \(\alpha+\alpha^2=1\) \(\left(\alpha+\alpha^2\right)^3=1\) \(\alpha^3+\alpha^6+3 \alpha^3\left(\alpha+\alpha^2\right)=1\) \(\mathrm{k}=-\alpha^3\) Substituting this in equation (i) \(-\mathrm{k}+\mathrm{k}^2+3(-\mathrm{k})(1)=1\) \(\mathrm{k}^2-4 \mathrm{k}-1=0\) \(\mathrm{k}=\frac{4 \pm \sqrt{16+4}}{2}\) \(\mathrm{k}=2 \pm \sqrt{5}\)
