Question
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Which of the following is a root for the function \( f(x)=2 x^{3}-\sqrt{x-\frac{1}{2}-\frac{23}{4}} \)

Ask by Valdez Carrillo. in Saudi Arabia
Nov 27,2024

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Answer

لحساب الجذر للدالة \( f(x) = 2x^3 - \sqrt{x - \frac{25}{4}} \)، يجب تبسيط التعبير تحت الجذر، ثم حل المعادلة \( 2x^3 - \sqrt{x - \frac{25}{4}} = 0 \) لمعرفة الجذر الصحيح.

Solution

لحساب الجذر للدالة \( f(x) = 2x^3 - \sqrt{x - \frac{1}{2} - \frac{23}{4}} \)، يجب علينا أولاً تبسيط التعبير تحت الجذر. نبدأ بتبسيط الجزء تحت الجذر: \[ x - \frac{1}{2} - \frac{23}{4} = x - \frac{2}{4} - \frac{23}{4} = x - \frac{25}{4} \] لذا يمكننا إعادة كتابة الدالة كالتالي: \[ f(x) = 2x^3 - \sqrt{x - \frac{25}{4}} \] لإيجاد الجذر، نحتاج إلى حل المعادلة \( f(x) = 0 \): \[ 2x^3 - \sqrt{x - \frac{25}{4}} = 0 \] هذا يعني أن: \[ \sqrt{x - \frac{25}{4}} = 2x^3 \] لإيجاد الجذر، يمكننا تربيع كلا الجانبين: \[ x - \frac{25}{4} = (2x^3)^2 \] \[ x - \frac{25}{4} = 4x^6 \] نقوم بنقل جميع الحدود إلى جهة واحدة: \[ 4x^6 - x + \frac{25}{4} = 0 \] الآن، يمكننا استخدام طرق عددية أو تحليلية لإيجاد الجذور. لكن إذا كان لديك خيارات معينة للجذور، يمكنك اختبارها في المعادلة الأصلية. إذا كان لديك خيارات معينة، يرجى تقديمها وسأساعدك في تحديد الجذر الصحيح.

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To solve for roots of the function \( f(x) = 2x^3 - \sqrt{x - \frac{1}{2} - \frac{23}{4}} \), we can start by simplifying the expression under the square root. The constant term simplifies as follows: \[ -\frac{1}{2} - \frac{23}{4} = -\frac{2}{4} - \frac{23}{4} = -\frac{25}{4}. \] Thus, the expression inside the square root becomes: \[ \sqrt{x + \frac{25}{4}}. \] Setting \( f(x) = 0 \), we want to find \( x \) such that: \[ 2x^3 = \sqrt{x + \frac{25}{4}}. \] Squaring both sides will lead us to a polynomial equation that we can solve for \( x \). In case of solving for roots, graphing the function or using numerical methods can also be effective. Now, carefully check for values of \( x \) that make the squared left-hand side equal to the right-hand side. Remember, being careful with squaring equations is crucial, as it may introduce extraneous roots! Good luck with your calculations!

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