roots of an equation from another equation

Question

The question is this:

Let the roots of $ax^2+ bx + c = 0$ be $r$ and $s$. The equation with roots $ar + b$ and $ as + b $ is:

$$ (A) x^2 − bx − ac = 0

(B) x^2 - bx + ac = 0

(C) x^2+ 3bx + ca + 2b^2= 0

(D) x^2+ 3bx - ca + 2b^2= 0

(E) x^2+ bx(2 - a) + a^2c + b^2(a + 1) = 0$$

What I did was change $ ar+b$ and $as+b$ to $ -as$ and $-ar$ using Vieta's Formula. Then I defined $x^2+px+q$ with $p$ being $as + ar$ and $q$ being $a^2sr$. This isn't one of the answer choices. Am I doing something wrong?

Answer 1

Change ar+b and as+b to −as and −ar using Vieta's Formula. Define x2+px+q with p being as+ar and q being a^2sr. Change what I defined as p and q as -b and ac making it answer choice B.

Answer 2

Hint:

Use Transformation of equations

Let $y=ax+b$

If $x=r, y=?$ and if $x=s, y=?$

$\implies x=\dfrac{y-b}a$

As $x$ is a root of the given equation,

$$a\left(\dfrac{y-b}a\right)^2+b\left(\dfrac{y-b}a\right)+c=0$$

Now simplify form the required quadratic equation

Answer 3

Hint:

As we know

$$r+s=?, rs=?$$

Use $ar+b+as+b=a(r+s)+2b=?$

and $(ar+b)(as+b)=a^2(rs)+ab(r+s)+b^2=?$

Answer 4

Working from the "vertex form" of the quadratic polynomial, the combination of "horizontal stretch" and "horizontal shift" transforms the vertex from $ \ (h \ , \ k ) \ $ to $ \ (h' \ = \ ah + b \ , \ k ) \ \ . $ We are then looking for the transformed coefficients for which $$ ax^2 \ + \ bx \ + \ c \ \ = \ \ a·(x + h)^2 \ + \ k \ \ \rightarrow \ \ a'x^2 \ + \ b'x \ + \ c' \ \ = \ \ a'·(x + h')^2 \ + \ k \ \ . $$

The $ \ x-$coordinate of the vertex is transformed as $ \ h \ = \ -\frac{b}{2a} \ \rightarrow \ h' \ = \ \left(-\frac{b}{2a} · a \right) \ + \ b \ = \ \frac{b}{2} \ \ , $ Since the $ \ y-$coordinate is unaltered, we have $$ k \ \ = \ \ c \ - \ \frac{b^2}{4a} \ \ = \ \ c' \ - \ \frac{b'^2}{4a'} \ \ . $$ If we say that the zeroes of the polynomial lie at a distance $ \ D \ $ "to either side" of the symmetry axis of the parabola representing the polynomial, the transformed zeroes are $ \ aD \ $ from the transformed axis. So we have $$ \ a·(D)^2 \ + \ k \ = \ 0 \ \rightarrow \ a'·(aD)^2 \ + \ k \ = \ 0 \ \ \Rightarrow \ \ a·D^2 \ \ = \ \ a'·a^2·D^2 \ \ \Rightarrow \ \ a' \ \ = \ \ \frac{1}{a} \ \ . $$ The vertex transformation tells us that $$ h' \ \ = \ \ -\frac{b'}{2a'} \ \ = \ \ \frac{b}{2} \ \ \Rightarrow \ \ b' \ \ = \ \ -b·a' \ \ = \ \ -\frac{b}{a} $$ and $$ c' \ \ = \ \ c \ - \ \frac{b^2}{4a} \ + \ \frac{b'^2}{4a'} \ \ = \ \ c \ - \ \frac{b^2}{4a} \ + \ \frac{(-b \ / \ a)^2}{4 \ / \ a} \ \ = \ \ c \ - \ \frac{b^2}{4a} \ + \ \frac{b^2}{4a} \ \ = \ \ c \ \ . $$ [The $ \ y-$intercept of the parabola is unchanged, as is expected since this transformation should not affect $ \ y-$coordinates.]

Hence, the transformed quadratic polynomial is $ \ \frac{1}{a} · x^2 \ -\frac{b}{a}·x \ + \ c \ \ . $ Since it does not alter the zeroes to multiply the polynomial by a constant $ \ a \neq 0 \ $ to make it monic, we have the transformed quadratic equation $ \ x^2 \ - \ bx \ + \ ac \ = \ 0 \ \ $ [choice $ \ \mathbf{(B)} \ ] \ \ . $