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Question

Let α,β be the roots of the quadratic equation (1m)x2(23m)x3m=0,mW has distinct integer roots. Then find the equation whose roots are α+5 & β+5.

A
x2+12x35=0
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B
x212x35=0
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C
x2+12x+35=0
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D
x212x+35=0
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Solution

The correct option is D x212x+35=0
Given: (1m)x2(23m)x3m=0
On comparing with standard form of quadratic equation ax2+bx+c=0
we get, a=(1m),b=(23m),c=3m

If roots are distinct integers, then discriminant should be perfect square of an integer.
D=b24ac=((23m))24(1m)(3m)
D=4+9m212m+12m12m2
D=3m2+4
Also, D>03m2+4>0
3m24<0
(3m2)(3m+2)<0
m(23,23)

Since, mWm{0,1}
If m=0,D=3(0)2+4=4, which is a perfect square.
Similarly, if m=1,D=3(1)2+4=1 which is a perfect square.

Hence, the possible values of m are 0,1.
When m=0, the quadratic equation becomes x22x=0
α=0;β=2 (i)

When m=1, the equation becomes x3=0 which is not a quadratic equation since coefficient of x2 is 0

Thus, we have α=0;β=2

Now, we need to find the equation whose roots are α+5 & β+5
roots are 0+5=5,2+5=7
The equation is (x5)(x7)=0
x212x+35=0

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