Quadratic equation with roots

Question

The quadratic equation $2x^2 + 7x + 5 = 0$ has a roots $α$ and $β$. Form a quadratic equation with roots $3α$ and $3β$.

Answer 1

Hint:

By Vieta's rules The sum of roots of new quadratic must be 3 times of sum of roots original one while the product of roots of new quadratic must be 9 times product of roots of original quadratic. Hence new quadratic is $$x^2+\frac {21}{2}x+\frac {45}{2}=0$$

Answer 2

As far as I understand your question you want to get a new equation out of the old one. By using the $p$-$q$-formula with $p=\frac72$ and $q=\frac52$we get

$$x_{1,2}~=~-\frac{7}{4}\pm\sqrt{\frac{49}{16}-\frac52}=-\frac{7}{4}\pm\sqrt{\frac{9}{16}}$$

and therefore $x_1=-1,x_2=-\frac52$. Lets set $x_1=\alpha$ and $x_2=\beta$. So $3\alpha=-3$ and $3\beta=-\frac{15}2$. Now by Vieta's rule that you can write a polynomial using the roots as follows

$$x^2+px+q=(x-x_1)(x-x_2)$$

we obtain

$$(x+3)\left(x+\frac{15}2\right)=x^2+\frac{21}2x+\frac{45}2$$

Answer 3

If $f(x) = 0$ has a solution $x = \alpha$, then $f\left(\frac{x}{3}\right) = 0$ will have a solution $x = 3\alpha$. We can thus write the desired quadratic equation as:

$$2 \left(\frac{x}{3}\right)^2 + 7 \left(\frac{x}{3}\right) + 5 = 0$$

Which can be simplified to:

$$2 x^2 + 21 x + 45 = 0$$

Answer 4

If $\alpha$ and $\beta$ are roots of a quadratic equation we have $$x^2-(\alpha+\beta)x+\alpha\beta=0$$therefore a quadratic equation with $3\alpha$ and $3\beta$ as its roots is $$x^2-3(\alpha+\beta)x+9\alpha\beta=0$$in our question we have $$\alpha+\beta=-3.5\\\alpha\beta=2.5$$so the requested equation is $$x^2+3\times3.5x+9\times 2.5=0$$or $$2x^2+21x+45=0$$