Question

Find the value of $a$ if $x^3-3x+=0$ has three distinct real roots ?

• A. $-2,a,2$
• B. $(-\infty ,-2)$
• C. $(2,\infty )$
• D. $(0,\infty )$

Answer 1

The correct option is A $-2,a,2$

$f\left(x\right)=x^3-3x+a=0$

$f^{\prime }\left(x\right)=3x^2-3=0$

$x=\pm 1$

for 3 real roots

$f\left(x_1\right)+f\left(x_2\right)<0$

$f\left(1\right)f(-1)<0$

$(1-3+a)(-1+3+a)$

$(a-2)(a+2)<0$

$a\Rightarrow -2<a<2.$