Question

Sqrt(x+10)-sqrt(x+3)=sqrt(4x-23)

Answer 1

ОДЗ: x+10≥0       x+3≥0       4x-23≥0
           x≥-10           x≥-3          x≥5,75
x∈[5,75; + $\infty$)
$\sqrt{x+10}-\sqrt{x+3}=\sqrt{4x-23} \\ (\sqrt{x+10}-\sqrt{x+3})^2=(\sqrt{4x-23} )^2 \\ x+10-2\sqrt{x+10}*\sqrt{x+3}+x+3=4x-23 \\ -2\sqrt{(x+10)(x+3)}=4x-23-x-10-x-3 \\ -2\sqrt{x^2+13x+30}=2x-36 \\ \sqrt{x^2+13x+30}=18-x \\ (\sqrt{x^2+13x+30})^2=(18-x)^2 \\ x^2+13x+30=324-36x+x^2 \\ 49x=294 \\ x=294:49 \\ x=6$

Answer 2

I'm sorry, I could not find any web search results for your query. It seems like you are asking me to solve a mathematical equation. I can try to help you with that, but I cannot guarantee the accuracy of my answer. Here is how I would approach the problem:

First, I would isolate the square root terms on one side of the equation and the non-square root term on the other side. This gives me:

$$\sqrt{x+10}-\sqrt{x+3}=\sqrt{4x-23}$$

Next, I would square both sides of the equation to eliminate the square roots. This gives me:

$$(\sqrt{x+10}-\sqrt{x+3})^2=(\sqrt{4x-23})^2$$ $$x+10-2\sqrt{x+10}\sqrt{x+3}+x+3=4x-23$$

Then, I would simplify and rearrange the equation to get a quadratic equation. This gives me:

$$-2\sqrt{x+10}\sqrt{x+3}=2x-36$$ $$\sqrt{x+10}\sqrt{x+3}=-x+18$$

Next, I would square both sides of the equation again to eliminate the remaining square roots. This gives me:

$$(\sqrt{x+10}\sqrt{x+3})^2=(-x+18)^2$$ $$(x+10)(x+3)=x^2-36x+324$$ $$x^2+13x+30=x^2-36x+324$$

Then, I would simplify and rearrange the equation to get a linear equation. This gives me:

$$49x=-294$$ $$x=-6$$

Finally, I would check if the solution satisfies the original equation. This gives me:

$$\sqrt{-6+10}-\sqrt{-6+3}=\sqrt{4(-6)-23}$$ $$\sqrt{4}-\sqrt{-3}=\sqrt{-47}$$

Since the square roots of negative numbers are not real numbers, the solution is not valid. Therefore, the original equation has no real solutions.Ответь подробно. Sqrt(x+10)-sqrt(x+3)=sqrt(4x-23)