MathematicsMedium244×since 2002Q4470∣x+1xxxx+λxxxx+λ2∣=98(103x+81)\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)x+1xxxx+λxxxx+λ2=89(103x+81), then λ,λ3\lambda, \frac{\lambda}{3}λ,3λ are the roots of the equation :A4x2+24x−27=04 x^{2}+24 x-27=04x2+24x−27=0B4x2−24x+27=04 x^{2}-24 x+27=04x2−24x+27=0C4x2−24x−27=04 x^{2}-24 x-27=04x2−24x−27=0D4x2+24x+27=04 x^{2}+24 x+27=04x2+24x+27=0Check answerSkip