Solve Let P(x) = x ^ 4 - 20x ^ 3 + 108x ^ 2 - ax + 9; alpha \in I There is a unique quadruple (a, , ) satisfying the polynomial such that beta - sqrt(gamma + sqrt(3)) is a root of dot P (x). then find the value of |alpha + beta - gamma + delta| Answer: 01 02 04 05 06 07 08 09 0306 07 08 09 Select your answer from radio buttons.

Question

$$\left. \begin{array} { l } { x = x ^ { 2 } - 202 x + 102 ^ { 2 } - 9 x + 9 } \\ { \text { b ) ( a b e n g a b e n g p e n y e n d s e f o r a n d e r a n d e n g a n d e n g o f a l e n d e n \right.$$

Solve for x

$x=\sqrt{823}+106\approx 134.687976576$
$x=106-\sqrt{823}\approx 77.312023424$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Calculate $102$ to the power of $2$ and get $10404$.

$$x=x^{2}-202x+10404-9x+9$$

Combine $-202x$ and $-9x$ to get $-211x$.

$$x=x^{2}-211x+10404+9$$

Add $10404$ and $9$ to get $10413$.

$$x=x^{2}-211x+10413$$

Subtract $x^{2}$ from both sides.

$$x-x^{2}=-211x+10413$$

Add $211x$ to both sides.

$$x-x^{2}+211x=10413$$

Combine $x$ and $211x$ to get $212x$.

$$212x-x^{2}=10413$$

Subtract $10413$ from both sides.

$$212x-x^{2}-10413=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-x^{2}+212x-10413=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $212$ for $b$, and $-10413$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-212±\sqrt{212^{2}-4\left(-1\right)\left(-10413\right)}}{2\left(-1\right)}$$

Square $212$.

$$x=\frac{-212±\sqrt{44944-4\left(-1\right)\left(-10413\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-212±\sqrt{44944+4\left(-10413\right)}}{2\left(-1\right)}$$

Multiply $4$ times $-10413$.

$$x=\frac{-212±\sqrt{44944-41652}}{2\left(-1\right)}$$

Add $44944$ to $-41652$.

$$x=\frac{-212±\sqrt{3292}}{2\left(-1\right)}$$

Take the square root of $3292$.

$$x=\frac{-212±2\sqrt{823}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{-212±2\sqrt{823}}{-2}$$

Now solve the equation $x=\frac{-212±2\sqrt{823}}{-2}$ when $±$ is plus. Add $-212$ to $2\sqrt{823}$.

$$x=\frac{2\sqrt{823}-212}{-2}$$

Divide $-212+2\sqrt{823}$ by $-2$.

$$x=106-\sqrt{823}$$

Now solve the equation $x=\frac{-212±2\sqrt{823}}{-2}$ when $±$ is minus. Subtract $2\sqrt{823}$ from $-212$.

$$x=\frac{-2\sqrt{823}-212}{-2}$$

Divide $-212-2\sqrt{823}$ by $-2$.

$$x=\sqrt{823}+106$$

The equation is now solved.

$$x=106-\sqrt{823}$$ $$x=\sqrt{823}+106$$

Steps for Completing the Square

Calculate $102$ to the power of $2$ and get $10404$.

$$x=x^{2}-202x+10404-9x+9$$

Combine $-202x$ and $-9x$ to get $-211x$.

$$x=x^{2}-211x+10404+9$$

Add $10404$ and $9$ to get $10413$.

$$x=x^{2}-211x+10413$$

Subtract $x^{2}$ from both sides.

$$x-x^{2}=-211x+10413$$

Add $211x$ to both sides.

$$x-x^{2}+211x=10413$$

Combine $x$ and $211x$ to get $212x$.

$$212x-x^{2}=10413$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-x^{2}+212x=10413$$

Divide both sides by $-1$.

$$\frac{-x^{2}+212x}{-1}=\frac{10413}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\frac{212}{-1}x=\frac{10413}{-1}$$

Divide $212$ by $-1$.

$$x^{2}-212x=\frac{10413}{-1}$$

Divide $10413$ by $-1$.

$$x^{2}-212x=-10413$$

Divide $-212$, the coefficient of the $x$ term, by $2$ to get $-106$. Then add the square of $-106$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-212x+\left(-106\right)^{2}=-10413+\left(-106\right)^{2}$$

Square $-106$.

$$x^{2}-212x+11236=-10413+11236$$

Add $-10413$ to $11236$.

$$x^{2}-212x+11236=823$$

Factor $x^{2}-212x+11236$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-106\right)^{2}=823$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-106\right)^{2}}=\sqrt{823}$$

Simplify.

$$x-106=\sqrt{823}$$ $$x-106=-\sqrt{823}$$

Add $106$ to both sides of the equation.

$$x=\sqrt{823}+106$$ $$x=106-\sqrt{823}$$

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Steps Using the Quadratic Formula

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Steps Using the Quadratic Formula

Steps for Completing the Square