118376
The real values of a for which the quadratic equation \(2 x^2-\left(a^3-8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite signs are given by
1 \(a>5\)
2 \(0\lt \) a \(\lt 4\)
3 \(a>0\)
4 \(a>7\)
Explanation:
B \(: \text { Let, } \alpha \text { and } \beta \text { are the roots of } 2 \mathrm{x}^2-\left(\mathrm{a}^3-8 \mathrm{a}-\right.\) \(\text { 1)x } \mathrm{x}+\mathrm{a}^2-4 \mathrm{a}=0\) \(\because \text { Roots are of opposite sign. }\) \(\therefore \text { Product of roots }\lt 0\) \(\alpha \beta\lt 0\) \(\Rightarrow \frac{\mathrm{a}^2-4 \mathrm{a}}{2}\lt 0\) \(\Rightarrow \mathrm{a}(\mathrm{a}-4)\lt 0\) \(\Rightarrow 0\lt \mathrm{a}\lt 4\)\(\because\) Roots are of opposite sign.
SRM JEEE-2008
Complex Numbers and Quadratic Equation
118377
Let \(\alpha, \beta\) be the roots of the equation \(a^2+b x+c\) \(=0\). A root of the equation \(\mathbf{a}^3 \mathbf{x}^2+a b c x+c^3=0\) is
1 \(\alpha+\beta\)
2 \(\alpha^2+\beta\)
3 \(a^2-\beta\)
4 \(\alpha^2 \beta\)
Explanation:
D Given \( a^3 x^2+a b c x+c^3=0\) \(x^2+\frac{b}{a} \cdot \frac{c}{a} x+\left(\frac{c}{a}\right)^3=0\) \(x^2-(\alpha+\beta) \alpha \beta x+\alpha^3 \beta^3=0\) \(\Rightarrow x=\alpha^2 \beta, \alpha \beta^2 .\)
COMEDK-2017
Complex Numbers and Quadratic Equation
118379
If \(x^2+a x+10=0\) and \(x^2+b x-10=0\) have common roots, then \(a^2-b^2\) is equal to
1 10
2 20
3 30
4 40
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{ax}+10=0\) and \(\mathrm{x}^2+\mathrm{bx}-10=0\) Let, \(\alpha\) be a common root of both the equations. So, \(\alpha^2+\mathrm{a} \alpha+10=0\) and \(\alpha^2+b \alpha-10=0\) or \(\alpha^2=10-b \alpha\) Putting (ii) in (i), we get - \((10-\mathrm{b} \alpha)+\mathrm{a} \alpha+10=0\) \(\Rightarrow \alpha(\mathrm{a}-\mathrm{b})=-20 \text { or } \alpha=\frac{20}{\mathrm{~b}-\mathrm{a}}\) \(\text { On substituting in (i), } \frac{400}{(b-a)^2}+\frac{20 a}{(b-a)}+10=0\) \(\Rightarrow 10(\mathrm{~b}-\mathrm{a})^2+20 \mathrm{a}(\mathrm{b}-\mathrm{a})+400=0\) \(\Rightarrow(\mathrm{b}-\mathrm{a})^2+2 \mathrm{a}(\mathrm{b}-\mathrm{a})+40=0\) \(\Rightarrow \mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab}+2 \mathrm{ab}-2 \mathrm{a}^2+40=0\) \(\Rightarrow \mathrm{b}^2-\mathrm{a}^2+40=0 \Rightarrow \mathrm{a}^2-\mathrm{b}^2=40 \text {. }\)
118376
The real values of a for which the quadratic equation \(2 x^2-\left(a^3-8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite signs are given by
1 \(a>5\)
2 \(0\lt \) a \(\lt 4\)
3 \(a>0\)
4 \(a>7\)
Explanation:
B \(: \text { Let, } \alpha \text { and } \beta \text { are the roots of } 2 \mathrm{x}^2-\left(\mathrm{a}^3-8 \mathrm{a}-\right.\) \(\text { 1)x } \mathrm{x}+\mathrm{a}^2-4 \mathrm{a}=0\) \(\because \text { Roots are of opposite sign. }\) \(\therefore \text { Product of roots }\lt 0\) \(\alpha \beta\lt 0\) \(\Rightarrow \frac{\mathrm{a}^2-4 \mathrm{a}}{2}\lt 0\) \(\Rightarrow \mathrm{a}(\mathrm{a}-4)\lt 0\) \(\Rightarrow 0\lt \mathrm{a}\lt 4\)\(\because\) Roots are of opposite sign.
SRM JEEE-2008
Complex Numbers and Quadratic Equation
118377
Let \(\alpha, \beta\) be the roots of the equation \(a^2+b x+c\) \(=0\). A root of the equation \(\mathbf{a}^3 \mathbf{x}^2+a b c x+c^3=0\) is
1 \(\alpha+\beta\)
2 \(\alpha^2+\beta\)
3 \(a^2-\beta\)
4 \(\alpha^2 \beta\)
Explanation:
D Given \( a^3 x^2+a b c x+c^3=0\) \(x^2+\frac{b}{a} \cdot \frac{c}{a} x+\left(\frac{c}{a}\right)^3=0\) \(x^2-(\alpha+\beta) \alpha \beta x+\alpha^3 \beta^3=0\) \(\Rightarrow x=\alpha^2 \beta, \alpha \beta^2 .\)
COMEDK-2017
Complex Numbers and Quadratic Equation
118379
If \(x^2+a x+10=0\) and \(x^2+b x-10=0\) have common roots, then \(a^2-b^2\) is equal to
1 10
2 20
3 30
4 40
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{ax}+10=0\) and \(\mathrm{x}^2+\mathrm{bx}-10=0\) Let, \(\alpha\) be a common root of both the equations. So, \(\alpha^2+\mathrm{a} \alpha+10=0\) and \(\alpha^2+b \alpha-10=0\) or \(\alpha^2=10-b \alpha\) Putting (ii) in (i), we get - \((10-\mathrm{b} \alpha)+\mathrm{a} \alpha+10=0\) \(\Rightarrow \alpha(\mathrm{a}-\mathrm{b})=-20 \text { or } \alpha=\frac{20}{\mathrm{~b}-\mathrm{a}}\) \(\text { On substituting in (i), } \frac{400}{(b-a)^2}+\frac{20 a}{(b-a)}+10=0\) \(\Rightarrow 10(\mathrm{~b}-\mathrm{a})^2+20 \mathrm{a}(\mathrm{b}-\mathrm{a})+400=0\) \(\Rightarrow(\mathrm{b}-\mathrm{a})^2+2 \mathrm{a}(\mathrm{b}-\mathrm{a})+40=0\) \(\Rightarrow \mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab}+2 \mathrm{ab}-2 \mathrm{a}^2+40=0\) \(\Rightarrow \mathrm{b}^2-\mathrm{a}^2+40=0 \Rightarrow \mathrm{a}^2-\mathrm{b}^2=40 \text {. }\)
118376
The real values of a for which the quadratic equation \(2 x^2-\left(a^3-8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite signs are given by
1 \(a>5\)
2 \(0\lt \) a \(\lt 4\)
3 \(a>0\)
4 \(a>7\)
Explanation:
B \(: \text { Let, } \alpha \text { and } \beta \text { are the roots of } 2 \mathrm{x}^2-\left(\mathrm{a}^3-8 \mathrm{a}-\right.\) \(\text { 1)x } \mathrm{x}+\mathrm{a}^2-4 \mathrm{a}=0\) \(\because \text { Roots are of opposite sign. }\) \(\therefore \text { Product of roots }\lt 0\) \(\alpha \beta\lt 0\) \(\Rightarrow \frac{\mathrm{a}^2-4 \mathrm{a}}{2}\lt 0\) \(\Rightarrow \mathrm{a}(\mathrm{a}-4)\lt 0\) \(\Rightarrow 0\lt \mathrm{a}\lt 4\)\(\because\) Roots are of opposite sign.
SRM JEEE-2008
Complex Numbers and Quadratic Equation
118377
Let \(\alpha, \beta\) be the roots of the equation \(a^2+b x+c\) \(=0\). A root of the equation \(\mathbf{a}^3 \mathbf{x}^2+a b c x+c^3=0\) is
1 \(\alpha+\beta\)
2 \(\alpha^2+\beta\)
3 \(a^2-\beta\)
4 \(\alpha^2 \beta\)
Explanation:
D Given \( a^3 x^2+a b c x+c^3=0\) \(x^2+\frac{b}{a} \cdot \frac{c}{a} x+\left(\frac{c}{a}\right)^3=0\) \(x^2-(\alpha+\beta) \alpha \beta x+\alpha^3 \beta^3=0\) \(\Rightarrow x=\alpha^2 \beta, \alpha \beta^2 .\)
COMEDK-2017
Complex Numbers and Quadratic Equation
118379
If \(x^2+a x+10=0\) and \(x^2+b x-10=0\) have common roots, then \(a^2-b^2\) is equal to
1 10
2 20
3 30
4 40
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{ax}+10=0\) and \(\mathrm{x}^2+\mathrm{bx}-10=0\) Let, \(\alpha\) be a common root of both the equations. So, \(\alpha^2+\mathrm{a} \alpha+10=0\) and \(\alpha^2+b \alpha-10=0\) or \(\alpha^2=10-b \alpha\) Putting (ii) in (i), we get - \((10-\mathrm{b} \alpha)+\mathrm{a} \alpha+10=0\) \(\Rightarrow \alpha(\mathrm{a}-\mathrm{b})=-20 \text { or } \alpha=\frac{20}{\mathrm{~b}-\mathrm{a}}\) \(\text { On substituting in (i), } \frac{400}{(b-a)^2}+\frac{20 a}{(b-a)}+10=0\) \(\Rightarrow 10(\mathrm{~b}-\mathrm{a})^2+20 \mathrm{a}(\mathrm{b}-\mathrm{a})+400=0\) \(\Rightarrow(\mathrm{b}-\mathrm{a})^2+2 \mathrm{a}(\mathrm{b}-\mathrm{a})+40=0\) \(\Rightarrow \mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab}+2 \mathrm{ab}-2 \mathrm{a}^2+40=0\) \(\Rightarrow \mathrm{b}^2-\mathrm{a}^2+40=0 \Rightarrow \mathrm{a}^2-\mathrm{b}^2=40 \text {. }\)
118376
The real values of a for which the quadratic equation \(2 x^2-\left(a^3-8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite signs are given by
1 \(a>5\)
2 \(0\lt \) a \(\lt 4\)
3 \(a>0\)
4 \(a>7\)
Explanation:
B \(: \text { Let, } \alpha \text { and } \beta \text { are the roots of } 2 \mathrm{x}^2-\left(\mathrm{a}^3-8 \mathrm{a}-\right.\) \(\text { 1)x } \mathrm{x}+\mathrm{a}^2-4 \mathrm{a}=0\) \(\because \text { Roots are of opposite sign. }\) \(\therefore \text { Product of roots }\lt 0\) \(\alpha \beta\lt 0\) \(\Rightarrow \frac{\mathrm{a}^2-4 \mathrm{a}}{2}\lt 0\) \(\Rightarrow \mathrm{a}(\mathrm{a}-4)\lt 0\) \(\Rightarrow 0\lt \mathrm{a}\lt 4\)\(\because\) Roots are of opposite sign.
SRM JEEE-2008
Complex Numbers and Quadratic Equation
118377
Let \(\alpha, \beta\) be the roots of the equation \(a^2+b x+c\) \(=0\). A root of the equation \(\mathbf{a}^3 \mathbf{x}^2+a b c x+c^3=0\) is
1 \(\alpha+\beta\)
2 \(\alpha^2+\beta\)
3 \(a^2-\beta\)
4 \(\alpha^2 \beta\)
Explanation:
D Given \( a^3 x^2+a b c x+c^3=0\) \(x^2+\frac{b}{a} \cdot \frac{c}{a} x+\left(\frac{c}{a}\right)^3=0\) \(x^2-(\alpha+\beta) \alpha \beta x+\alpha^3 \beta^3=0\) \(\Rightarrow x=\alpha^2 \beta, \alpha \beta^2 .\)
COMEDK-2017
Complex Numbers and Quadratic Equation
118379
If \(x^2+a x+10=0\) and \(x^2+b x-10=0\) have common roots, then \(a^2-b^2\) is equal to
1 10
2 20
3 30
4 40
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{ax}+10=0\) and \(\mathrm{x}^2+\mathrm{bx}-10=0\) Let, \(\alpha\) be a common root of both the equations. So, \(\alpha^2+\mathrm{a} \alpha+10=0\) and \(\alpha^2+b \alpha-10=0\) or \(\alpha^2=10-b \alpha\) Putting (ii) in (i), we get - \((10-\mathrm{b} \alpha)+\mathrm{a} \alpha+10=0\) \(\Rightarrow \alpha(\mathrm{a}-\mathrm{b})=-20 \text { or } \alpha=\frac{20}{\mathrm{~b}-\mathrm{a}}\) \(\text { On substituting in (i), } \frac{400}{(b-a)^2}+\frac{20 a}{(b-a)}+10=0\) \(\Rightarrow 10(\mathrm{~b}-\mathrm{a})^2+20 \mathrm{a}(\mathrm{b}-\mathrm{a})+400=0\) \(\Rightarrow(\mathrm{b}-\mathrm{a})^2+2 \mathrm{a}(\mathrm{b}-\mathrm{a})+40=0\) \(\Rightarrow \mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab}+2 \mathrm{ab}-2 \mathrm{a}^2+40=0\) \(\Rightarrow \mathrm{b}^2-\mathrm{a}^2+40=0 \Rightarrow \mathrm{a}^2-\mathrm{b}^2=40 \text {. }\)
118376
The real values of a for which the quadratic equation \(2 x^2-\left(a^3-8 a-1\right) x+a^2-4 a=0\) possesses roots of opposite signs are given by
1 \(a>5\)
2 \(0\lt \) a \(\lt 4\)
3 \(a>0\)
4 \(a>7\)
Explanation:
B \(: \text { Let, } \alpha \text { and } \beta \text { are the roots of } 2 \mathrm{x}^2-\left(\mathrm{a}^3-8 \mathrm{a}-\right.\) \(\text { 1)x } \mathrm{x}+\mathrm{a}^2-4 \mathrm{a}=0\) \(\because \text { Roots are of opposite sign. }\) \(\therefore \text { Product of roots }\lt 0\) \(\alpha \beta\lt 0\) \(\Rightarrow \frac{\mathrm{a}^2-4 \mathrm{a}}{2}\lt 0\) \(\Rightarrow \mathrm{a}(\mathrm{a}-4)\lt 0\) \(\Rightarrow 0\lt \mathrm{a}\lt 4\)\(\because\) Roots are of opposite sign.
SRM JEEE-2008
Complex Numbers and Quadratic Equation
118377
Let \(\alpha, \beta\) be the roots of the equation \(a^2+b x+c\) \(=0\). A root of the equation \(\mathbf{a}^3 \mathbf{x}^2+a b c x+c^3=0\) is
1 \(\alpha+\beta\)
2 \(\alpha^2+\beta\)
3 \(a^2-\beta\)
4 \(\alpha^2 \beta\)
Explanation:
D Given \( a^3 x^2+a b c x+c^3=0\) \(x^2+\frac{b}{a} \cdot \frac{c}{a} x+\left(\frac{c}{a}\right)^3=0\) \(x^2-(\alpha+\beta) \alpha \beta x+\alpha^3 \beta^3=0\) \(\Rightarrow x=\alpha^2 \beta, \alpha \beta^2 .\)
COMEDK-2017
Complex Numbers and Quadratic Equation
118379
If \(x^2+a x+10=0\) and \(x^2+b x-10=0\) have common roots, then \(a^2-b^2\) is equal to
1 10
2 20
3 30
4 40
Explanation:
D Given, \(\mathrm{x}^2+\mathrm{ax}+10=0\) and \(\mathrm{x}^2+\mathrm{bx}-10=0\) Let, \(\alpha\) be a common root of both the equations. So, \(\alpha^2+\mathrm{a} \alpha+10=0\) and \(\alpha^2+b \alpha-10=0\) or \(\alpha^2=10-b \alpha\) Putting (ii) in (i), we get - \((10-\mathrm{b} \alpha)+\mathrm{a} \alpha+10=0\) \(\Rightarrow \alpha(\mathrm{a}-\mathrm{b})=-20 \text { or } \alpha=\frac{20}{\mathrm{~b}-\mathrm{a}}\) \(\text { On substituting in (i), } \frac{400}{(b-a)^2}+\frac{20 a}{(b-a)}+10=0\) \(\Rightarrow 10(\mathrm{~b}-\mathrm{a})^2+20 \mathrm{a}(\mathrm{b}-\mathrm{a})+400=0\) \(\Rightarrow(\mathrm{b}-\mathrm{a})^2+2 \mathrm{a}(\mathrm{b}-\mathrm{a})+40=0\) \(\Rightarrow \mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab}+2 \mathrm{ab}-2 \mathrm{a}^2+40=0\) \(\Rightarrow \mathrm{b}^2-\mathrm{a}^2+40=0 \Rightarrow \mathrm{a}^2-\mathrm{b}^2=40 \text {. }\)