1
$\begingroup$

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.

$\endgroup$
  • 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
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $8x+9.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.