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

mrtaurho
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Eppy Eka
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  • What exactly is your question? Do you want an answer, a hint, a solution? Please add these details right in your question and your own attempt aswell. – mrtaurho Aug 01 '18 at 17:24
  • Actually i should form an quadratic equation. But i keep messing up with the roots. And in the end i failed to form it... – Eppy Eka Aug 01 '18 at 17:26
  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Aug 01 '18 at 17:26

4 Answers4

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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:

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

Count Iblis
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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 $$x^2+\frac {21}{2}x+\frac {45}{2}=0$$

Rohan Shinde
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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

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

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

Mostafa Ayaz
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