Question

Equation $12x^4-56x^3+89x^2-56x+12=0$ has

• A. four real and roots
• B. two irrational roots
• C. one integer roots
• D. two imaginary roots

Answer 1

Answer: (A)

four real and roots

Given $12x^4-56x^3+89x^2-56x+12=0$

Lets, divide by $x^2$,

$12x^2-56x+89-\frac{56}{x}+\frac{12}{x^2}=0$

Rearranging

$12\left(\right(x+\frac{1}{x}{)}^2-2)-56(x+\frac{1}{x})+89=0$

let $y=x+\frac{1}{x}$

$12(y^2-2)-56y+89=0$

$12y^2-56y+65=0$

Evaluating , we get

$y=\frac{29}{6}$, $y=\frac{9}{2}$

Substituting, $y=x+\frac{1}{x}$

we get

$6x^2-29x+6=0$ ( 1 )

$2x^2-9x+2=0$ ( 2 )

for ( 1 ) $b^2-4ac={29}^2-4\left(6\right)\left(6\right)=697>0$

So ( 1 ) has two real roots

for ( 2 )$b^2-4ac=9^2-4\left(2\right)\left(2\right)=65>0$

So ( 2 ) has two real roots

So given polynomial has 4 real roots

$optionAiscorrect$