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?

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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?

Tyrn7i · Accepted Answer

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  22x2−x+24  or  −22x2−x−24  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 · Accepted Answer

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 · Accepted Answer

Step 1This shall answer your question. A polynomial is of the form,P(x)=(x−r1)(x−r2)⋯(x−rn)where r1, r2, ⋯ rn are the roots of the polynomial. Then the first derivative of the polynomial will be,P′(x)=(x−r1)′(x−r1)P(x)+(x−r2)′(x−r2)P(x)+(x−r1)′(x−r1)P(x)⋯ (x−r−n)′(x−rn)P(x)since (x−k)′=1, we have,P′(x)P(x)=1(x−r1)+1(x−r2)+1(x−r3)+⋯1(x−rn)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.

A quadratic equation ax^{2}+bx+c=0 has equal roots at a=2c How could

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