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The equation $x^2-a=\sqrt{x+a}$ has real or

imaginary roots depending on the values of $a.$

Then range of $a$ for which the equation.

$(a)\;\; $ No real roots

$(b)\;\; $ One real root

$(c)\;\;$ Exactly two real roots

$(d)\;\;$ At least two real roots

what i try

$x^2-a=\sqrt{x+a}\Rightarrow (x^2-a)^2=x+a$

$x^4+a^2-2ax^2=x+a\Rightarrow a^2-(2x^2+1)a+x^4-x=0$

$$a=\frac{(2x^2+1)\pm \sqrt{(2x^2+1)^2-4(x^4-x)}}{2}$$

$$a=\frac{2x^2+1\pm (2x+1)}{2}$$

$$a=x^2-x,a=x^2+x+1$$

$(a)$ For no real roots

$x^2-x-a=0$ and $x^2+x+1-a=0$ has no real roots

So $1+4a>0$ and $\displaystyle 1-4(1-a)>0\Rightarrow a>-\frac{1}{4}.$

How do i solve other parts Help me please

Community
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jacky
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    For example: if $;x+a<0;$ then there is no equation, so it must be $;x\ge-a;$, and then $;\sqrt{x+a}\ge0;$, and then it can't be $;x^2-a<0;$ ... So it must be that both sides of that equation are non-negative for you have any chance at all to begin to work, and then tou can rise to the second power (as then you've made sure both sides are non-negative) and etc. – DonAntonio Nov 18 '19 at 10:18
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    Related https://math.stackexchange.com/questions/417098/how-find-a-such-that-x2-sqrta-x-a-has-exactly-two-real-solutions/417102#417102 – Fallen_Prince Nov 18 '19 at 11:17

1 Answers1

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First of all, a quadratic equation has no real roots when its discriminant is negative. So you have $1+4a<0$ and similarly for the other inequality. However, these are not necessarily all the cases. The equation is equivalent to the system $$\begin{align} (x^2-a)^2 &= x+a \\ x^2-a &\geq0 \Leftrightarrow x^2\geq a \end{align}$$ Notice that we don't need $x+a\geq 0$ because that's guaranteed to be non-negative since it's equal to a square. So we're interested in the number of solutions to the system, not just the equation. It might happen that the equation has real roots but they don't satisfy the inequality which would imply the original equation has no real roots. Continuing your work, the equation has roots (by solving the $2$ quadratics) $$x_{1,2}= \frac{1\pm\sqrt{1+4a}}{2}\\ x_{3,4}=\frac{-1\pm\sqrt{4a-3}}{2}$$ Now you need to plug in each of those in $x^2\geq a$ and solve for $a$. The values for $a$ tell you when the root you plugged in is actually a real root of the original equation. After that, you can determine the answers for (a) to (d).

bjorn93
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  • There's also a geometrical approach to this problem which doesn't involve as much computation. See here: https://math.stackexchange.com/a/417239 – bjorn93 Nov 19 '19 at 02:58
  • The condition that $x^2 \ge a$ appears to be incorrect. Thus, consider, for example that $a= \frac{1}{4}$, so that the initial equation is simplified to $x^2-\frac{1}{4}=\sqrt{x+\frac{1}{4}}$. Plain to see that $x=\frac{1-\sqrt{2}}{2}$ is a solution thereto $$\left(\frac{1-\sqrt{2}}{2} \right) ^2-\frac{1}{4}=\frac{1-\sqrt{2}}{2}=\sqrt{\frac{1-\sqrt{2}}{2}+\frac{1}{4}}=\sqrt{\frac{2-2\sqrt{2}+1}{4}}=\frac{\sqrt{(1-\sqrt{2}})^2}{2},$$ and yet, $$\left(\frac{1-\sqrt{2}}{2} \right) ^2<\frac{1}{4}$$ – USIKPA Jul 02 '23 at 16:55
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    @USIKPA The condition $x^2\ge a$ is necessary. For your values of $a$ and $x,$ $x$ is not a solution because $x^2-a=-\sqrt{x+a}\ne\sqrt{x+a}.$ – Anne Bauval Jul 02 '23 at 22:24
  • Yes, I forgot what a principal root is – USIKPA Jul 04 '23 at 17:59