IIT JEE , AIEEE Forum - Mathematics , Algebra

alisha_gupta_27

1)For what value of a the equation x^2 - x(1-a) - (a+2) = 0 has integral roots.Find the roots.

2)If a<b<c<d prove that the equation (x-a)(x-c) + k(x-b)(x-d) = 0 has real solutions for all k(real nos.).

karthik2007

for question one, I got the value of a = -2.

Roots are 3, 0.

Discriminant of the equation = a^2 + 2a + 9.

roots will be -b +/- sqrt (d) / 2a.

I know my solution looks ugly, but first tell me if my answer is right.

lazycol

d = (1-a)²+ 4(a+2)

= a² +2a +9

= (a+1)²+8

4 integral sols d shud b perfect sq.

d & (a+1)²r perfect sq ie diff bw 2 perfect sq is 8

dis works only with 1 & 9

so (a+1)²= 1

a + 1 =

1

a=0 ; a= -2;

karthik2007

@lazycol

If d is a perfect square, then it is not necessary that the roots themselves will be integral. It cud even end up as a fraction, don't you think?

for example, if d is a perfect square, like 25, then its sqrt will be 5. But then it is not necessary that -b +/- sqrt d / (2a) needs to be integral...

lazycol

well indis case roots r (a-1)

^{[ ]}

d/2

d& a-1 r odd as per my explanation(d=1,a=-2,0)

so roots r integers

lazycol

let f(x) = (x-a)(x-c) + k(x-b)(x-d)

f(b) =(b-a)(b-c) { -ve no:}

f(d) =(d-a)(d-c) { +ve no:}

ie f(x) changes its sign in (b,d)

ie f(x) has a real root in (b,d)

hence roots of f(x) r real { if one root is real,other also has 2 b so}

karthik2007

I didnt quite follow your solution. can u elaborate?

lazycol

see, f(b)<0 ,f(d)>0

so f(x) changes its sign(initially -ve & later +ve) in the interval (b,d)

f(x) being a continous fnt before it becomes +ve it shud pass through 0

ie f(x) has a root in (b,d)

karthik2007

We can even do this for the interval (a, c) right?

lazycol

yes u can.....

again u get 1 root in (a,c)

i just gave 1 interval here

u can go 4 any of 2

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