Show that $-1$ is a root of the equation $$(a+b-2c)x^2+(2a-b-c)x+(c+a-2b)=0$$ Find the other root.

My Attempt:

Since replacing $x=-1$ satisfies the given equation, it is the root. But how to find the other root.

  • 2
    $\begingroup$ Hint: product of the roots is... $\endgroup$ – dxiv Aug 17 '17 at 23:24
  • $\begingroup$ Vieta's relations. $\endgroup$ – Bernard Aug 17 '17 at 23:25
  • $\begingroup$ ... and the sum of the roots is ... $\endgroup$ – Henry Aug 17 '17 at 23:25
  • $\begingroup$ Given a quadratic equation $px^2+qx+r=0$, the sum of the roots is $-q/p$ and the product is $r/p$. $\endgroup$ – Oscar Lanzi Aug 17 '17 at 23:25
  • $\begingroup$ @dxiv, Sum of the roots is $\dfrac {b+c-2a}{a+b-2c}$. $\endgroup$ – pi-π Aug 17 '17 at 23:26

A few people in the comments are citing Vieta's rlations, and that's a good strategy. There's also this: If $-1$ is a root, then $(x+1)$ is a factor. Use polynomial long division to divide the given polynomial by $(x+1)$, set the linear quotient equal to zero, and solve.


Suppose you have something like $8x^2 + 17x+9,$ and you plug in a number, such as $-1,$ and you find that $8(-1)^2 + 17(-1) + 9 =0.$ If you plug a number into a polynomial and get $0$, that means $x$ minus that number is a factor, so in this case you have $x+1$ is a factor: $$ 8x^2 + 17x+9 = (x+1)(\cdots\cdots\cdots). $$ If all else fails, you can find the other factor by long division: In this case, divide $8x^2+17x+9$ by $x+1,$ getting $8x+9.$


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