Question

The equation whose roots exceed by $2$ than those of $2x^3+3x^2-4x+5=0$, is

• A. $2x^3+9x^2-8x+9=0$
• B. $2x^3+9x^2+8x+9=0$
• C. $2x^3-9x^2+8x+9=0$
• D. $2x^3-9x^2-8x+9=0$

Answer 1

Answer: (C)

$2x^3-9x^2+8x+9=0$

Let the roots of given equation $2x^3+3x^2-4x+5=0$ is $p,q,r$

Relation betwen roots and coefficients are

Sum of the roots$=\frac{-b}{a}$

$\therefore p+q+r=\frac{-b}{a}=\frac{-\left(3\right)}{2}=\frac{-3}{2}$

Product of the roots$=\frac{-d}{a}$

$\therefore p\times q\times r=\frac{-5}{2}$

Sum of products of the roots taken two at a time$=\frac{c}{a}$

$\therefore p\times q+p\times r+q\times r=\frac{c}{a}=\frac{-4}{2}=-2$

Now new roots are $p^{\prime }=p+2,q^{\prime }=q+2,r^{\prime }=r+2$

New equation is $x^3-(p^{\prime }+q^{\prime }+r^{\prime })x^2+\left(\right(p^{\prime }\times q^{\prime })+(q^{\prime }\times r^{\prime })+(r^{\prime }\times p^{\prime }\left)\right)x-(p^{\prime }\times q^{\prime }\times r^{\prime })=0$

$p^{\prime }+q^{\prime }+r^{\prime }=p+2+q+2+r+2=p+q+r+6=-\frac{3}{2}+6=\frac{9}{2}$

$p^{\prime }\times q^{\prime }\times r^{\prime }=(p+2)(q+2)(r+2)=pqr+2(pq+pr+qr)+4(p+q+r)+8=-\frac{5}{2}+2(-2)-4\times \frac{3}{2}+8=-\frac{9}{2}$

$(p^{\prime }\times q^{\prime })+(q^{\prime }\times r^{\prime })+(r^{\prime }\times p^{\prime })=(p+2)(q+2)+(r+2)(q+2)+(p+2)(r+2)=pq+qr+pr+12+4p+4q+4r$$=-2+12+4\times \frac{-3}{2}=4$

So, the equation is $x^3-\frac{9}{2}{x}^2+4x-(-\frac{9}{2})=0$

$\Rightarrow 2x^3-9x^2+8x+9=0$