118223
If the roots of the equation \({a x^2}^2+b x+c=0\) are real and distinct then
1 both roots are greater than \(-\frac{b}{2 a}\)
2 both roots are less than \(-\frac{b}{2 a}\)
3 one of the roots exceeds \(-\frac{b}{2 a}\)
4 None of the above
Explanation:
C Given equation \(a x^2+b x+c=0\) \(\because\) Roots are real and distinct \(\therefore \quad \mathrm{D}>0\) \(\mathrm{~b}^2-4 \mathrm{a}>0\) Now, roots of given equation are - \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}, \frac{-b+\sqrt{b^2-4 a c}}{2 a}\) \(\frac{-b-\sqrt{b^2-4 a c}}{2 a}\lt \frac{-b}{2 a}\lt \frac{-b+\sqrt{b^2-4 a c}}{2 a}\)So, we can say that one of the roots exceeds \(\frac{-b}{2 a}\)
UPSEE-2010
Complex Numbers and Quadratic Equation
118224
The roots of the equation \(x^4-2 x^3+x=380\) are:
1 \(5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
2 \(-5,4,-\frac{1 \pm 5 \sqrt{-3}}{2}\)
3 \(5,4, \frac{-1 \pm 5 \sqrt{-3}}{2}\)
4 \(-5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
Explanation:
A Given, \(x^4-2 x^3+x=380\) Then, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) Now, \(\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) \((\mathrm{x}+4)(\mathrm{x}-5)=0\) \(\mathrm{x}=5,-4\) And, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)=0\) \(\mathrm{x} =\frac{1 \pm \sqrt{1^2-4 \times 19}}{2}\) \(=\frac{1 \pm \sqrt{-75}}{2}\) \(=\frac{1 \pm 5 \sqrt{-3}}{2}\)Thus, roots of the equation are \(5,-4\) and \(\frac{1 \pm 5 \sqrt{-3}}{2}\).
UPSEE-2004]
Complex Numbers and Quadratic Equation
118225
Find the sum of the real roots of the equation \(x^2\) \(+5|\mathbf{x}|+6=0\).
1 5
2 10
3 -5
4 None of these
Explanation:
D Given \(x^2+5|x|+6=0\) \(\text { Now, }\) \(\text { Case I : when } x \text { is positive }\) \(\quad x^2+5 x+6=0\) \(\quad(x+2)(x+3)=0\) Now, Case I : when \(x\) is positive \(x=-2,-3\) is not possible when \(x>0\) Case II: when \(\mathrm{x}\) is negative \(x^2-5 x+6=0\) \((x-2)(x-3)=0\) \(x=2,3\) Similarly when \(\mathrm{x}\lt 0\) then \(\mathrm{x}=2,3\) is not possible So, no real roots are possible.
JCECE-2012
Complex Numbers and Quadratic Equation
118226
If the roots of the equation \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) are equal in magnitude but opposite in sign, then their product is
1 \(\frac{1}{2}\left(a^2+b^2\right)\)
2 \(-\frac{1}{2}\left(a^2+b^2\right)\)
3 \(\frac{1}{2} \mathrm{ab}\)
4 \(-\frac{1}{2} \mathrm{ab}\)
Explanation:
B Given, \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) \(\frac{(x+a)+(x+b)}{(x+a)(x+b)}=\frac{1}{c}\) \((2 x+a+b) c=x^2+a x+b x+a b\) \(2 x c+a c+b c=x^2+(a+b) x+a b\) \(x^2+(a+b-2 c) x+a b-a c-b c=0\) Let roots are \(\alpha\) and \(-\alpha\) \(\alpha+(-\alpha)=\frac{-(a+b-2 c)}{1}\) \(a+b-2 c=0\) \(a+b=2 c\) \(c=\frac{a+b}{2}\) Now, product of roots \(=a b-a c-b c\) \(=a b-c(a+b)\) \(=a b-\frac{(a+b)(a+b)}{2}=a b-\frac{(a+b)^2}{2}\) \(=\frac{2 a b-(a+b)^2}{2}=\frac{-1}{2}\left(a^2+b^2\right)\)
JCECE-2011
Complex Numbers and Quadratic Equation
118227
If \(\alpha, \beta\) are the roots of equation \(8 x^2-3 x+27=\) 0 , then the value of \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{7}{2}\)
4 4
Explanation:
B Given, \(8 x^2-3 x+27=0\) and roots are \(\alpha\) and \(\beta\) then, \(\alpha+\beta=\frac{3}{8}\) \(\alpha . \beta=\frac{27}{8}\) Now, \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) \(=\frac{\alpha^{2 / 3}}{\beta^{1 / 3}}+\frac{\beta^{2 / 3}}{\alpha^{1 / 3}}=\frac{\alpha+\beta}{(\alpha \beta)^{1 / 3}}\) \(=\frac{\frac{3}{8}}{\left(\frac{27}{8}\right)^{1 / 3}}=\frac{\frac{3}{8}}{\frac{3}{2}}=\frac{1}{4}\)
118223
If the roots of the equation \({a x^2}^2+b x+c=0\) are real and distinct then
1 both roots are greater than \(-\frac{b}{2 a}\)
2 both roots are less than \(-\frac{b}{2 a}\)
3 one of the roots exceeds \(-\frac{b}{2 a}\)
4 None of the above
Explanation:
C Given equation \(a x^2+b x+c=0\) \(\because\) Roots are real and distinct \(\therefore \quad \mathrm{D}>0\) \(\mathrm{~b}^2-4 \mathrm{a}>0\) Now, roots of given equation are - \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}, \frac{-b+\sqrt{b^2-4 a c}}{2 a}\) \(\frac{-b-\sqrt{b^2-4 a c}}{2 a}\lt \frac{-b}{2 a}\lt \frac{-b+\sqrt{b^2-4 a c}}{2 a}\)So, we can say that one of the roots exceeds \(\frac{-b}{2 a}\)
UPSEE-2010
Complex Numbers and Quadratic Equation
118224
The roots of the equation \(x^4-2 x^3+x=380\) are:
1 \(5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
2 \(-5,4,-\frac{1 \pm 5 \sqrt{-3}}{2}\)
3 \(5,4, \frac{-1 \pm 5 \sqrt{-3}}{2}\)
4 \(-5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
Explanation:
A Given, \(x^4-2 x^3+x=380\) Then, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) Now, \(\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) \((\mathrm{x}+4)(\mathrm{x}-5)=0\) \(\mathrm{x}=5,-4\) And, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)=0\) \(\mathrm{x} =\frac{1 \pm \sqrt{1^2-4 \times 19}}{2}\) \(=\frac{1 \pm \sqrt{-75}}{2}\) \(=\frac{1 \pm 5 \sqrt{-3}}{2}\)Thus, roots of the equation are \(5,-4\) and \(\frac{1 \pm 5 \sqrt{-3}}{2}\).
UPSEE-2004]
Complex Numbers and Quadratic Equation
118225
Find the sum of the real roots of the equation \(x^2\) \(+5|\mathbf{x}|+6=0\).
1 5
2 10
3 -5
4 None of these
Explanation:
D Given \(x^2+5|x|+6=0\) \(\text { Now, }\) \(\text { Case I : when } x \text { is positive }\) \(\quad x^2+5 x+6=0\) \(\quad(x+2)(x+3)=0\) Now, Case I : when \(x\) is positive \(x=-2,-3\) is not possible when \(x>0\) Case II: when \(\mathrm{x}\) is negative \(x^2-5 x+6=0\) \((x-2)(x-3)=0\) \(x=2,3\) Similarly when \(\mathrm{x}\lt 0\) then \(\mathrm{x}=2,3\) is not possible So, no real roots are possible.
JCECE-2012
Complex Numbers and Quadratic Equation
118226
If the roots of the equation \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) are equal in magnitude but opposite in sign, then their product is
1 \(\frac{1}{2}\left(a^2+b^2\right)\)
2 \(-\frac{1}{2}\left(a^2+b^2\right)\)
3 \(\frac{1}{2} \mathrm{ab}\)
4 \(-\frac{1}{2} \mathrm{ab}\)
Explanation:
B Given, \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) \(\frac{(x+a)+(x+b)}{(x+a)(x+b)}=\frac{1}{c}\) \((2 x+a+b) c=x^2+a x+b x+a b\) \(2 x c+a c+b c=x^2+(a+b) x+a b\) \(x^2+(a+b-2 c) x+a b-a c-b c=0\) Let roots are \(\alpha\) and \(-\alpha\) \(\alpha+(-\alpha)=\frac{-(a+b-2 c)}{1}\) \(a+b-2 c=0\) \(a+b=2 c\) \(c=\frac{a+b}{2}\) Now, product of roots \(=a b-a c-b c\) \(=a b-c(a+b)\) \(=a b-\frac{(a+b)(a+b)}{2}=a b-\frac{(a+b)^2}{2}\) \(=\frac{2 a b-(a+b)^2}{2}=\frac{-1}{2}\left(a^2+b^2\right)\)
JCECE-2011
Complex Numbers and Quadratic Equation
118227
If \(\alpha, \beta\) are the roots of equation \(8 x^2-3 x+27=\) 0 , then the value of \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{7}{2}\)
4 4
Explanation:
B Given, \(8 x^2-3 x+27=0\) and roots are \(\alpha\) and \(\beta\) then, \(\alpha+\beta=\frac{3}{8}\) \(\alpha . \beta=\frac{27}{8}\) Now, \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) \(=\frac{\alpha^{2 / 3}}{\beta^{1 / 3}}+\frac{\beta^{2 / 3}}{\alpha^{1 / 3}}=\frac{\alpha+\beta}{(\alpha \beta)^{1 / 3}}\) \(=\frac{\frac{3}{8}}{\left(\frac{27}{8}\right)^{1 / 3}}=\frac{\frac{3}{8}}{\frac{3}{2}}=\frac{1}{4}\)
118223
If the roots of the equation \({a x^2}^2+b x+c=0\) are real and distinct then
1 both roots are greater than \(-\frac{b}{2 a}\)
2 both roots are less than \(-\frac{b}{2 a}\)
3 one of the roots exceeds \(-\frac{b}{2 a}\)
4 None of the above
Explanation:
C Given equation \(a x^2+b x+c=0\) \(\because\) Roots are real and distinct \(\therefore \quad \mathrm{D}>0\) \(\mathrm{~b}^2-4 \mathrm{a}>0\) Now, roots of given equation are - \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}, \frac{-b+\sqrt{b^2-4 a c}}{2 a}\) \(\frac{-b-\sqrt{b^2-4 a c}}{2 a}\lt \frac{-b}{2 a}\lt \frac{-b+\sqrt{b^2-4 a c}}{2 a}\)So, we can say that one of the roots exceeds \(\frac{-b}{2 a}\)
UPSEE-2010
Complex Numbers and Quadratic Equation
118224
The roots of the equation \(x^4-2 x^3+x=380\) are:
1 \(5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
2 \(-5,4,-\frac{1 \pm 5 \sqrt{-3}}{2}\)
3 \(5,4, \frac{-1 \pm 5 \sqrt{-3}}{2}\)
4 \(-5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
Explanation:
A Given, \(x^4-2 x^3+x=380\) Then, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) Now, \(\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) \((\mathrm{x}+4)(\mathrm{x}-5)=0\) \(\mathrm{x}=5,-4\) And, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)=0\) \(\mathrm{x} =\frac{1 \pm \sqrt{1^2-4 \times 19}}{2}\) \(=\frac{1 \pm \sqrt{-75}}{2}\) \(=\frac{1 \pm 5 \sqrt{-3}}{2}\)Thus, roots of the equation are \(5,-4\) and \(\frac{1 \pm 5 \sqrt{-3}}{2}\).
UPSEE-2004]
Complex Numbers and Quadratic Equation
118225
Find the sum of the real roots of the equation \(x^2\) \(+5|\mathbf{x}|+6=0\).
1 5
2 10
3 -5
4 None of these
Explanation:
D Given \(x^2+5|x|+6=0\) \(\text { Now, }\) \(\text { Case I : when } x \text { is positive }\) \(\quad x^2+5 x+6=0\) \(\quad(x+2)(x+3)=0\) Now, Case I : when \(x\) is positive \(x=-2,-3\) is not possible when \(x>0\) Case II: when \(\mathrm{x}\) is negative \(x^2-5 x+6=0\) \((x-2)(x-3)=0\) \(x=2,3\) Similarly when \(\mathrm{x}\lt 0\) then \(\mathrm{x}=2,3\) is not possible So, no real roots are possible.
JCECE-2012
Complex Numbers and Quadratic Equation
118226
If the roots of the equation \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) are equal in magnitude but opposite in sign, then their product is
1 \(\frac{1}{2}\left(a^2+b^2\right)\)
2 \(-\frac{1}{2}\left(a^2+b^2\right)\)
3 \(\frac{1}{2} \mathrm{ab}\)
4 \(-\frac{1}{2} \mathrm{ab}\)
Explanation:
B Given, \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) \(\frac{(x+a)+(x+b)}{(x+a)(x+b)}=\frac{1}{c}\) \((2 x+a+b) c=x^2+a x+b x+a b\) \(2 x c+a c+b c=x^2+(a+b) x+a b\) \(x^2+(a+b-2 c) x+a b-a c-b c=0\) Let roots are \(\alpha\) and \(-\alpha\) \(\alpha+(-\alpha)=\frac{-(a+b-2 c)}{1}\) \(a+b-2 c=0\) \(a+b=2 c\) \(c=\frac{a+b}{2}\) Now, product of roots \(=a b-a c-b c\) \(=a b-c(a+b)\) \(=a b-\frac{(a+b)(a+b)}{2}=a b-\frac{(a+b)^2}{2}\) \(=\frac{2 a b-(a+b)^2}{2}=\frac{-1}{2}\left(a^2+b^2\right)\)
JCECE-2011
Complex Numbers and Quadratic Equation
118227
If \(\alpha, \beta\) are the roots of equation \(8 x^2-3 x+27=\) 0 , then the value of \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{7}{2}\)
4 4
Explanation:
B Given, \(8 x^2-3 x+27=0\) and roots are \(\alpha\) and \(\beta\) then, \(\alpha+\beta=\frac{3}{8}\) \(\alpha . \beta=\frac{27}{8}\) Now, \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) \(=\frac{\alpha^{2 / 3}}{\beta^{1 / 3}}+\frac{\beta^{2 / 3}}{\alpha^{1 / 3}}=\frac{\alpha+\beta}{(\alpha \beta)^{1 / 3}}\) \(=\frac{\frac{3}{8}}{\left(\frac{27}{8}\right)^{1 / 3}}=\frac{\frac{3}{8}}{\frac{3}{2}}=\frac{1}{4}\)
118223
If the roots of the equation \({a x^2}^2+b x+c=0\) are real and distinct then
1 both roots are greater than \(-\frac{b}{2 a}\)
2 both roots are less than \(-\frac{b}{2 a}\)
3 one of the roots exceeds \(-\frac{b}{2 a}\)
4 None of the above
Explanation:
C Given equation \(a x^2+b x+c=0\) \(\because\) Roots are real and distinct \(\therefore \quad \mathrm{D}>0\) \(\mathrm{~b}^2-4 \mathrm{a}>0\) Now, roots of given equation are - \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}, \frac{-b+\sqrt{b^2-4 a c}}{2 a}\) \(\frac{-b-\sqrt{b^2-4 a c}}{2 a}\lt \frac{-b}{2 a}\lt \frac{-b+\sqrt{b^2-4 a c}}{2 a}\)So, we can say that one of the roots exceeds \(\frac{-b}{2 a}\)
UPSEE-2010
Complex Numbers and Quadratic Equation
118224
The roots of the equation \(x^4-2 x^3+x=380\) are:
1 \(5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
2 \(-5,4,-\frac{1 \pm 5 \sqrt{-3}}{2}\)
3 \(5,4, \frac{-1 \pm 5 \sqrt{-3}}{2}\)
4 \(-5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
Explanation:
A Given, \(x^4-2 x^3+x=380\) Then, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) Now, \(\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) \((\mathrm{x}+4)(\mathrm{x}-5)=0\) \(\mathrm{x}=5,-4\) And, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)=0\) \(\mathrm{x} =\frac{1 \pm \sqrt{1^2-4 \times 19}}{2}\) \(=\frac{1 \pm \sqrt{-75}}{2}\) \(=\frac{1 \pm 5 \sqrt{-3}}{2}\)Thus, roots of the equation are \(5,-4\) and \(\frac{1 \pm 5 \sqrt{-3}}{2}\).
UPSEE-2004]
Complex Numbers and Quadratic Equation
118225
Find the sum of the real roots of the equation \(x^2\) \(+5|\mathbf{x}|+6=0\).
1 5
2 10
3 -5
4 None of these
Explanation:
D Given \(x^2+5|x|+6=0\) \(\text { Now, }\) \(\text { Case I : when } x \text { is positive }\) \(\quad x^2+5 x+6=0\) \(\quad(x+2)(x+3)=0\) Now, Case I : when \(x\) is positive \(x=-2,-3\) is not possible when \(x>0\) Case II: when \(\mathrm{x}\) is negative \(x^2-5 x+6=0\) \((x-2)(x-3)=0\) \(x=2,3\) Similarly when \(\mathrm{x}\lt 0\) then \(\mathrm{x}=2,3\) is not possible So, no real roots are possible.
JCECE-2012
Complex Numbers and Quadratic Equation
118226
If the roots of the equation \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) are equal in magnitude but opposite in sign, then their product is
1 \(\frac{1}{2}\left(a^2+b^2\right)\)
2 \(-\frac{1}{2}\left(a^2+b^2\right)\)
3 \(\frac{1}{2} \mathrm{ab}\)
4 \(-\frac{1}{2} \mathrm{ab}\)
Explanation:
B Given, \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) \(\frac{(x+a)+(x+b)}{(x+a)(x+b)}=\frac{1}{c}\) \((2 x+a+b) c=x^2+a x+b x+a b\) \(2 x c+a c+b c=x^2+(a+b) x+a b\) \(x^2+(a+b-2 c) x+a b-a c-b c=0\) Let roots are \(\alpha\) and \(-\alpha\) \(\alpha+(-\alpha)=\frac{-(a+b-2 c)}{1}\) \(a+b-2 c=0\) \(a+b=2 c\) \(c=\frac{a+b}{2}\) Now, product of roots \(=a b-a c-b c\) \(=a b-c(a+b)\) \(=a b-\frac{(a+b)(a+b)}{2}=a b-\frac{(a+b)^2}{2}\) \(=\frac{2 a b-(a+b)^2}{2}=\frac{-1}{2}\left(a^2+b^2\right)\)
JCECE-2011
Complex Numbers and Quadratic Equation
118227
If \(\alpha, \beta\) are the roots of equation \(8 x^2-3 x+27=\) 0 , then the value of \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{7}{2}\)
4 4
Explanation:
B Given, \(8 x^2-3 x+27=0\) and roots are \(\alpha\) and \(\beta\) then, \(\alpha+\beta=\frac{3}{8}\) \(\alpha . \beta=\frac{27}{8}\) Now, \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) \(=\frac{\alpha^{2 / 3}}{\beta^{1 / 3}}+\frac{\beta^{2 / 3}}{\alpha^{1 / 3}}=\frac{\alpha+\beta}{(\alpha \beta)^{1 / 3}}\) \(=\frac{\frac{3}{8}}{\left(\frac{27}{8}\right)^{1 / 3}}=\frac{\frac{3}{8}}{\frac{3}{2}}=\frac{1}{4}\)
118223
If the roots of the equation \({a x^2}^2+b x+c=0\) are real and distinct then
1 both roots are greater than \(-\frac{b}{2 a}\)
2 both roots are less than \(-\frac{b}{2 a}\)
3 one of the roots exceeds \(-\frac{b}{2 a}\)
4 None of the above
Explanation:
C Given equation \(a x^2+b x+c=0\) \(\because\) Roots are real and distinct \(\therefore \quad \mathrm{D}>0\) \(\mathrm{~b}^2-4 \mathrm{a}>0\) Now, roots of given equation are - \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}, \frac{-b+\sqrt{b^2-4 a c}}{2 a}\) \(\frac{-b-\sqrt{b^2-4 a c}}{2 a}\lt \frac{-b}{2 a}\lt \frac{-b+\sqrt{b^2-4 a c}}{2 a}\)So, we can say that one of the roots exceeds \(\frac{-b}{2 a}\)
UPSEE-2010
Complex Numbers and Quadratic Equation
118224
The roots of the equation \(x^4-2 x^3+x=380\) are:
1 \(5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
2 \(-5,4,-\frac{1 \pm 5 \sqrt{-3}}{2}\)
3 \(5,4, \frac{-1 \pm 5 \sqrt{-3}}{2}\)
4 \(-5,-4, \frac{1 \pm 5 \sqrt{-3}}{2}\)
Explanation:
A Given, \(x^4-2 x^3+x=380\) Then, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) Now, \(\left(\mathrm{x}^2-\mathrm{x}-20\right)=0\) \((\mathrm{x}+4)(\mathrm{x}-5)=0\) \(\mathrm{x}=5,-4\) And, \(\left(\mathrm{x}^2-\mathrm{x}+19\right)=0\) \(\mathrm{x} =\frac{1 \pm \sqrt{1^2-4 \times 19}}{2}\) \(=\frac{1 \pm \sqrt{-75}}{2}\) \(=\frac{1 \pm 5 \sqrt{-3}}{2}\)Thus, roots of the equation are \(5,-4\) and \(\frac{1 \pm 5 \sqrt{-3}}{2}\).
UPSEE-2004]
Complex Numbers and Quadratic Equation
118225
Find the sum of the real roots of the equation \(x^2\) \(+5|\mathbf{x}|+6=0\).
1 5
2 10
3 -5
4 None of these
Explanation:
D Given \(x^2+5|x|+6=0\) \(\text { Now, }\) \(\text { Case I : when } x \text { is positive }\) \(\quad x^2+5 x+6=0\) \(\quad(x+2)(x+3)=0\) Now, Case I : when \(x\) is positive \(x=-2,-3\) is not possible when \(x>0\) Case II: when \(\mathrm{x}\) is negative \(x^2-5 x+6=0\) \((x-2)(x-3)=0\) \(x=2,3\) Similarly when \(\mathrm{x}\lt 0\) then \(\mathrm{x}=2,3\) is not possible So, no real roots are possible.
JCECE-2012
Complex Numbers and Quadratic Equation
118226
If the roots of the equation \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) are equal in magnitude but opposite in sign, then their product is
1 \(\frac{1}{2}\left(a^2+b^2\right)\)
2 \(-\frac{1}{2}\left(a^2+b^2\right)\)
3 \(\frac{1}{2} \mathrm{ab}\)
4 \(-\frac{1}{2} \mathrm{ab}\)
Explanation:
B Given, \(\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}\) \(\frac{(x+a)+(x+b)}{(x+a)(x+b)}=\frac{1}{c}\) \((2 x+a+b) c=x^2+a x+b x+a b\) \(2 x c+a c+b c=x^2+(a+b) x+a b\) \(x^2+(a+b-2 c) x+a b-a c-b c=0\) Let roots are \(\alpha\) and \(-\alpha\) \(\alpha+(-\alpha)=\frac{-(a+b-2 c)}{1}\) \(a+b-2 c=0\) \(a+b=2 c\) \(c=\frac{a+b}{2}\) Now, product of roots \(=a b-a c-b c\) \(=a b-c(a+b)\) \(=a b-\frac{(a+b)(a+b)}{2}=a b-\frac{(a+b)^2}{2}\) \(=\frac{2 a b-(a+b)^2}{2}=\frac{-1}{2}\left(a^2+b^2\right)\)
JCECE-2011
Complex Numbers and Quadratic Equation
118227
If \(\alpha, \beta\) are the roots of equation \(8 x^2-3 x+27=\) 0 , then the value of \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{7}{2}\)
4 4
Explanation:
B Given, \(8 x^2-3 x+27=0\) and roots are \(\alpha\) and \(\beta\) then, \(\alpha+\beta=\frac{3}{8}\) \(\alpha . \beta=\frac{27}{8}\) Now, \(\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3}\) \(=\frac{\alpha^{2 / 3}}{\beta^{1 / 3}}+\frac{\beta^{2 / 3}}{\alpha^{1 / 3}}=\frac{\alpha+\beta}{(\alpha \beta)^{1 / 3}}\) \(=\frac{\frac{3}{8}}{\left(\frac{27}{8}\right)^{1 / 3}}=\frac{\frac{3}{8}}{\frac{3}{2}}=\frac{1}{4}\)
