A quadratic equation ax2+bx+c=0 has equal roots at a=2c How could we find the sum...

minikim38

minikim38

Answered

2022-01-20

A quadratic equation ax2+bx+c=0 has equal roots at a=2c
How could we find the sum of reciprocals of the roots of this equation?

Answer & Explanation

Tyrn7i

Tyrn7i

Expert

2022-01-21Added 11 answers

Step 1
Since the roots are equal, you must have
b2=4ac=8c2
Also, the roots satisfy
0=x2+bax+ca
=x2+ba+12
so the product of the roots is 12; that is, 4c2=12. Therefore, b2=2(4c2)=1
The sum of the roots is ba; but it also is equal to 2a. So
2a=ba
This gives
2a2=b,
and we know
|b|=1,
so we must have
b=1
So the sum of the reciprocals is
4ab=4a
Now, a2=12
so either
a=22
or
a=22
Thus, the equation is either
22x2x+24
or
22x2x24
Both satisfy the desired conditions: they have a double root at a=2c. In one case, the sum of the reciprocals is 22, in the other it is 22.
Jordyn Horne

Jordyn Horne

Expert

2022-01-22Added 16 answers

Step 1
Also, just for the fun of it, since you have a root with multiplicity two:
ax2+bx+c=(ax+c)2
Then the root is
x=±ca
and from a=2c, we get
x=±12
Hence,
1x1+1x2=2x=±22
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

Step 1This shall answer your question. A polynomial is of the form,P(x)=(xr1)(xr2)(xrn)where r1, r2,  rn are the roots of the polynomial. Then the first derivative of the polynomial will be,P(x)=(xr1)(xr1)P(x)+(xr2)(xr2)P(x)+(xr1)(xr1)P(x) (xrn)(xrn)P(x)since (xk)=1, we have,P(x)P(x)=1(xr1)+1(xr2)+1(xr3)+1(xrn)for x=0P(0)P(0)=1(r1)+1(r2)+1(r3)+1(rn)P(0)k=1r1+1r2+1r3+1rnwhere k is the constant term of the polynomial.Therefore the sum of the reciprocals of the roots of any polynomial is equal to minus of the first derivative of the polynomial at x=0 divided by the constant term of the polynomial.

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