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:,
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:,
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:,
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:,