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There is the following equation.

$$x^{2}+2(m-a)x+3am-2=0$$

a) Find $a$ such that the equation has real roots, $\forall m\in \mathbb{R}$

b) Find $m$ such that the equation has real roots, $\forall a\in \mathbb{R}$

The discriminant is $m^{2}-5am+a^{2}+2\geq 0$ and $a^{2}-5am+m^{2}+2\geq 0$

I found that $a,m\in (-\infty , -\sqrt{\frac{8}{21}}]\cup [\sqrt{\frac{8}{21}}, +\infty)$ but I don't know why $|a|\leqslant \sqrt{\frac{8}{21}}$and $|m|\leqslant \sqrt{\frac{8}{21}}$

Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in $a$ is $\le0$. For radicals I know that it should be $\ge0$.

nonuser
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Vali RO
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  • Yes, I saw that question but I don't understand why in that answer the discriminant of the new quadratic in a is <=0.For radicals I know that it should be >=0 – Vali RO Jan 18 '19 at 18:56
  • You could ask this first at the original question before posting a duplicate. – Dietrich Burde Jan 18 '19 at 18:59
  • $|a|\leq \sqrt{\frac{8}{21}}$ because for $m^2-5am + a^2 +2 \geq 0$, the discriminant of this equation i.e. $D=(25a^2 - 4(a^2+2))$ must be less than equal to $0$. – Sauhard Sharma Jan 18 '19 at 19:07

1 Answers1

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So you have a discriminant $$\Delta_1(m,a)=m^2-5am+a^2+2$$ and you want it to be greater than zero for all $m$ values. This expression in $m$ with $a$ as parameter is a quadratic. If it is always positive, it means that for $m=0$ you have $\Delta_1(0,a)>0$ (which is true). But you want to be true for all $m$ values. If this expression would have real roots, there are some values where the discriminant is negative, so the original equation does not have real solutions. So $$\Delta_2(a)=25 a^2-4(a^2+2)<0$$ This will ensure that $\Delta_1(m,a)>0$ for any $m$.

Andrei
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  • I tried to understand and analyze but I don't get too far.So I have a quadratic with a as parameter: m^2-5am+a^2+2.This expression need to be >=0.I need to find all m values such this this expression to be true.For m=0 I understood, it's true."If this expression would have real roots, there are some values where the discriminant is negative" this part I don't really understand.I know that if this expression has real roots the discriminant is >=0 otherwise the roots are complex. – Vali RO Jan 18 '19 at 19:27
  • So if I would rewrite this by changing $m$ to $x$: find values of $a$ such as $x^2-5ax+(a^2+2)$ is always greater than $0$. So you want a parabola, above the $x$ axis, for all $x$. It means that the equation $x^2-5ax+(a^2+2)=0$ does not have any real roots (either the function is always above or always below), and you have a point that is above. That means that the discriminant of this equation is $\le 0$ (to be one one side of the axis), and I choose the $x=0$ point to calculate if it's above or below. – Andrei Jan 18 '19 at 19:36
  • I finally understood.Thank you so much for your help! – Vali RO Jan 18 '19 at 19:41