Solution of Quadratic and Higher Degree Equations
Complex Numbers and Quadratic Equation

118088 If the equation \(z^2+(p+i q) z+r+i s=0\), where \(p, q r\) and s are real and non-zero roots, then

1 \(\mathrm{pqr}=\mathrm{r}^2+\mathrm{p}^2 \mathrm{~s}\)
2 \(p r s=q^2+r^2 p\)
3 \(q r s=p^2+s^2 q\)
4 \(\mathrm{pqs}=\mathrm{s}^2+\mathrm{q}^2 \mathrm{r}\)
Complex Numbers and Quadratic Equation

118089 If the equation
\(\left(a^2+4 a+3\right) x^2+\left(a^2-a-2\right) x+a(a+1)=0 \text { has }\)
\(\text { more than two roots, then value of } a \text { is }\)
#[Qdiff: Hard, QCat: Numerical Based, examname: JCECE-2005], Now, So, standard quadratic equation is, \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\), we know that, it any quadratic equation has more than, two roots, then it is identity and \(a=b=c=0\), \(\quad a^2+4 a+3=0\), \(a^2+3 a+a+3=0\), \((a+1)(a+3)=0\), \(\mathrm{a}=-1,-3\), \(\mathrm{a}^2-\mathrm{a}-2=0\), \(\mathrm{a}^2-2 \mathrm{a}+\mathrm{a}-2=0\), \((a-2)(a+1)=0 \Rightarrow a=2,-1\), and \(a(a+1)=0\), \(a=-1\), Therefore value of \(a=-1\), 638. If \(\alpha\) and \(\beta\) are the solutions of the quadratic equation \(a x^2+b x+c=0\) such that \(\beta=\alpha^{1 / 3}\), then:,

1 0
2 1
3 -1
4 None of these
Complex Numbers and Quadratic Equation

118090 If \(x+\frac{1}{x}=2 \sin \alpha, y+\frac{1}{y}=2 \cos \beta\)
then
\(x^3 y^3+\frac{1}{x^3 y^3}\) is:

1 \(2 \cos 3(\beta-\alpha)\)
2 \(2 \cos 3(\beta+\alpha)\)
3 \(2 \sin 3(\beta-\alpha)\)
4 \(2 \sin 3(\beta+\alpha)\)
Complex Numbers and Quadratic Equation

118091 If \(x^2-3 x+2\) be a factor of \(x^4-p x^2+q\), then (p,
q) is equal to:

1 \((3,4)\)
2 \((4,5)\)
3 \((4,3)\)
4 \((5,4)\)
Complex Numbers and Quadratic Equation

118088 If the equation \(z^2+(p+i q) z+r+i s=0\), where \(p, q r\) and s are real and non-zero roots, then

1 \(\mathrm{pqr}=\mathrm{r}^2+\mathrm{p}^2 \mathrm{~s}\)
2 \(p r s=q^2+r^2 p\)
3 \(q r s=p^2+s^2 q\)
4 \(\mathrm{pqs}=\mathrm{s}^2+\mathrm{q}^2 \mathrm{r}\)
Complex Numbers and Quadratic Equation

118089 If the equation
\(\left(a^2+4 a+3\right) x^2+\left(a^2-a-2\right) x+a(a+1)=0 \text { has }\)
\(\text { more than two roots, then value of } a \text { is }\)
#[Qdiff: Hard, QCat: Numerical Based, examname: JCECE-2005], Now, So, standard quadratic equation is, \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\), we know that, it any quadratic equation has more than, two roots, then it is identity and \(a=b=c=0\), \(\quad a^2+4 a+3=0\), \(a^2+3 a+a+3=0\), \((a+1)(a+3)=0\), \(\mathrm{a}=-1,-3\), \(\mathrm{a}^2-\mathrm{a}-2=0\), \(\mathrm{a}^2-2 \mathrm{a}+\mathrm{a}-2=0\), \((a-2)(a+1)=0 \Rightarrow a=2,-1\), and \(a(a+1)=0\), \(a=-1\), Therefore value of \(a=-1\), 638. If \(\alpha\) and \(\beta\) are the solutions of the quadratic equation \(a x^2+b x+c=0\) such that \(\beta=\alpha^{1 / 3}\), then:,

1 0
2 1
3 -1
4 None of these
Complex Numbers and Quadratic Equation

118090 If \(x+\frac{1}{x}=2 \sin \alpha, y+\frac{1}{y}=2 \cos \beta\)
then
\(x^3 y^3+\frac{1}{x^3 y^3}\) is:

1 \(2 \cos 3(\beta-\alpha)\)
2 \(2 \cos 3(\beta+\alpha)\)
3 \(2 \sin 3(\beta-\alpha)\)
4 \(2 \sin 3(\beta+\alpha)\)
Complex Numbers and Quadratic Equation

118091 If \(x^2-3 x+2\) be a factor of \(x^4-p x^2+q\), then (p,
q) is equal to:

1 \((3,4)\)
2 \((4,5)\)
3 \((4,3)\)
4 \((5,4)\)
Complex Numbers and Quadratic Equation

118088 If the equation \(z^2+(p+i q) z+r+i s=0\), where \(p, q r\) and s are real and non-zero roots, then

1 \(\mathrm{pqr}=\mathrm{r}^2+\mathrm{p}^2 \mathrm{~s}\)
2 \(p r s=q^2+r^2 p\)
3 \(q r s=p^2+s^2 q\)
4 \(\mathrm{pqs}=\mathrm{s}^2+\mathrm{q}^2 \mathrm{r}\)
Complex Numbers and Quadratic Equation

118089 If the equation
\(\left(a^2+4 a+3\right) x^2+\left(a^2-a-2\right) x+a(a+1)=0 \text { has }\)
\(\text { more than two roots, then value of } a \text { is }\)
#[Qdiff: Hard, QCat: Numerical Based, examname: JCECE-2005], Now, So, standard quadratic equation is, \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\), we know that, it any quadratic equation has more than, two roots, then it is identity and \(a=b=c=0\), \(\quad a^2+4 a+3=0\), \(a^2+3 a+a+3=0\), \((a+1)(a+3)=0\), \(\mathrm{a}=-1,-3\), \(\mathrm{a}^2-\mathrm{a}-2=0\), \(\mathrm{a}^2-2 \mathrm{a}+\mathrm{a}-2=0\), \((a-2)(a+1)=0 \Rightarrow a=2,-1\), and \(a(a+1)=0\), \(a=-1\), Therefore value of \(a=-1\), 638. If \(\alpha\) and \(\beta\) are the solutions of the quadratic equation \(a x^2+b x+c=0\) such that \(\beta=\alpha^{1 / 3}\), then:,

1 0
2 1
3 -1
4 None of these
Complex Numbers and Quadratic Equation

118090 If \(x+\frac{1}{x}=2 \sin \alpha, y+\frac{1}{y}=2 \cos \beta\)
then
\(x^3 y^3+\frac{1}{x^3 y^3}\) is:

1 \(2 \cos 3(\beta-\alpha)\)
2 \(2 \cos 3(\beta+\alpha)\)
3 \(2 \sin 3(\beta-\alpha)\)
4 \(2 \sin 3(\beta+\alpha)\)
Complex Numbers and Quadratic Equation

118091 If \(x^2-3 x+2\) be a factor of \(x^4-p x^2+q\), then (p,
q) is equal to:

1 \((3,4)\)
2 \((4,5)\)
3 \((4,3)\)
4 \((5,4)\)
Complex Numbers and Quadratic Equation

118088 If the equation \(z^2+(p+i q) z+r+i s=0\), where \(p, q r\) and s are real and non-zero roots, then

1 \(\mathrm{pqr}=\mathrm{r}^2+\mathrm{p}^2 \mathrm{~s}\)
2 \(p r s=q^2+r^2 p\)
3 \(q r s=p^2+s^2 q\)
4 \(\mathrm{pqs}=\mathrm{s}^2+\mathrm{q}^2 \mathrm{r}\)
Complex Numbers and Quadratic Equation

118089 If the equation
\(\left(a^2+4 a+3\right) x^2+\left(a^2-a-2\right) x+a(a+1)=0 \text { has }\)
\(\text { more than two roots, then value of } a \text { is }\)
#[Qdiff: Hard, QCat: Numerical Based, examname: JCECE-2005], Now, So, standard quadratic equation is, \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\), we know that, it any quadratic equation has more than, two roots, then it is identity and \(a=b=c=0\), \(\quad a^2+4 a+3=0\), \(a^2+3 a+a+3=0\), \((a+1)(a+3)=0\), \(\mathrm{a}=-1,-3\), \(\mathrm{a}^2-\mathrm{a}-2=0\), \(\mathrm{a}^2-2 \mathrm{a}+\mathrm{a}-2=0\), \((a-2)(a+1)=0 \Rightarrow a=2,-1\), and \(a(a+1)=0\), \(a=-1\), Therefore value of \(a=-1\), 638. If \(\alpha\) and \(\beta\) are the solutions of the quadratic equation \(a x^2+b x+c=0\) such that \(\beta=\alpha^{1 / 3}\), then:,

1 0
2 1
3 -1
4 None of these
Complex Numbers and Quadratic Equation

118090 If \(x+\frac{1}{x}=2 \sin \alpha, y+\frac{1}{y}=2 \cos \beta\)
then
\(x^3 y^3+\frac{1}{x^3 y^3}\) is:

1 \(2 \cos 3(\beta-\alpha)\)
2 \(2 \cos 3(\beta+\alpha)\)
3 \(2 \sin 3(\beta-\alpha)\)
4 \(2 \sin 3(\beta+\alpha)\)
Complex Numbers and Quadratic Equation

118091 If \(x^2-3 x+2\) be a factor of \(x^4-p x^2+q\), then (p,
q) is equal to:

1 \((3,4)\)
2 \((4,5)\)
3 \((4,3)\)
4 \((5,4)\)