if p and q be the roots of the equation x^2+ax+b=0,prove that p/q is a root of the equation bx^2+(2b-a^2)+b=0?

Question

sumi29 · Answer

Alright...we know that p and q are roots to the given equation. That means p+q = a, and p.q = b. Now substitute these new expressions in the problem, and you will get a new quadratic equation, again in x. This time, put x = p/q, and equate both sides. You will prove that it's a root of the given equation.

raden · Answer

p+q = -a

whpalmer4 · Answer

The problem is incorrect as written above.  For example, take p = -2, q = -3, which are the roots of the equation $$x^2+5x+6 =0$$so $a = 5, b = 6$.

Our new equation would be $$bx^2+(2b-a^2)+b = 0$$$$6x^2+(2*6-5^2) + 6= 0$$$$6x^2-7 = 0$$It isn't hard to see that $$6(2/3)^2 - 7 \ne 0$$

Now, if the equation is $$bx^2+(2b-a^2)x+b = 0$$ that becomes $$6x^2+(12-25)x+6 =0$$$$6x^2-13x+6=0$$$$6(2/3)^2-13(2/3)+6 = 0$$$$6*\frac{4}{9}-13*\frac{2}{3}+6 = 0$$$$24-78+54=0\checkmark$$

@sumi29's approach works (after you apply @RadEn's correction) with the corrected equation, but @digitalmonk please make sure what you post is correct.

anonymous · Answer

@whpalmer4 : what i posted was from a very popular mathematics text book in my country. now it is now possible for me to judge whether a sum given in a text books is correct or not just by looking at the sum.

whpalmer4 · Answer

Well, there are at least two possibilities: the book, however popular, has an error, or you made a mistake when typing.  If the error is with the book, my apologies.

sumi29 · Answer

@whpalmer4 : Yeah the equation is incorrect, I was only assuming the poster knew that he made a mistake in writing down the equation...lol

anonymous · Answer

bx^2+(2b-a^2)x+b=0  should have been the equation, i missed out the x, sorry for the typo. I eventually got the answer, that to all those who helped !!

sumi29 · Answer

@digitalmonk : You are welcome. :-)