Question

Det
If a, b, c are three positive roots of the equation x3 – px2 + qx – 48 = 0 (p,q> 0) then find
2
the minimum value of 2
3
b
+
+
3)
a​

Answer 1

Explanation:

the minimum value of 2 is 50

Answer 2

Given cubic equation is x3+px2+qx+r=0

Also given that roots are in consecutive positive integers. so lets the roots be y,y+2,y+4.

{Also for cubic equation ax3+bx2+cx+d=0 .Let p,q and r be its roots then:

Root expression:

p+q+r=-b/a……(i)

pq+qr+pr=c/a…..(ii)

pqr=-d/a……(iii)

}

so according to our problem we have a=1,b=p,c=q and d=r.

Also for finding (q) we need only root expression (iii).

=>y∗(y+2)+(y+2)∗(y+4)+(y+4)∗y=q

After aolving we get,

=>3y2+12y+8=q

=>y2+4y+(8−q)/3=0

Since the cubic equation reduces to quadratic form and the roots are in the same condition.

{ For quadratic equation a2x+bx+c=0 having roots p and q. Root expression are:

p+q=-b/a….(i)

pq=c/a….(ii)

}

So lets take roots be mandm+2 as given in the question that roots are in consecutive positive integers.

So, y2+4y+(8−q)/3, a=1,b=4 and c=(8-q)/3

=>m+(m+2)=−4 from (i)

=>2m+2=−4

=>m=−3

and

m(m+2)=(8−q)/3 from (ii)

putting the value of m in (ii), we get

=>−3(−3+2)=(8−q)/3

=>3∗3=8−q

=>q=−1ans.

Thanks.

Hope it helps.