Multiple choice

If $\alpha,\beta$ are the roots of the equation $\left( x-a \right) \left( x-b \right) +c=0\left( c\neq 0 \right) ,$ then the roots of the equation $\left( x-c-\alpha \right) \left( x-c-\beta \right) =c$ are

  1. $a$ and $b+c$
  2. $a^2+c^2$ and $b^2+c^2$
  3. $a+c$ and $a-c$
  4. $a+c$ and $b+c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The roots of (x-a)(x-b) + c = 0 are alpha and beta. This means x^2 - (a+b)x + ab + c = 0. The roots of (x-c-alpha)(x-c-beta) = c are x = c+alpha and x = c+beta. Since alpha and beta are roots of the original equation, adding c to them shifts the roots by c. Thus, the new roots are a+c and b+c.

AI explanation

Expand the first equation to x squared minus (a + b)x plus (ab + c) equals 0, meaning the sum of the roots alpha plus beta equals a + b and the product alpha beta equals ab + c. Expand the second equation to get x squared minus (c + alpha + beta + c)x plus (c + alpha)(c + beta) equals c, and substitute the sum and product of the first roots into this expanded form. This changes the equation to x squared minus (a + b + 2c)x plus (c squared + c(a + b) + ab + c) equals c, which simplifies to x squared minus (a + b + 2c)x plus (c squared + ac + bc + ab) equals 0. Factoring this resulting quadratic gives (x minus (a + c))(x minus (b + c)) equals 0, so the roots of the new equation are a + c and b + c.

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