Triangle $b$ is isosceles. Two triangles are cut on so that the remaining pentagon has five equal sides of length $x$ . The value of $x$ can be found using this equation. $\newline$ $(2b-a)x^{2}+(4a^{2}-b^{2})(2x-a)=0$ $\newline$ a. Find $x$ when $a=10$ and $b=12$ . $\newline$ b. Can $a$ be equal to $2b$ ?

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Write and solve direct variation equations

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Q. Triangle

b

is isosceles. Two triangles are cut on so that the remaining pentagon has five equal sides of length

x

. The value of

x

can be found using this equation.

\newline

(2b-a)x^{2}+(4a^{2}-b^{2})(2x-a)=0

\newline

a. Find

x

when

a=10

and

b=12

\newline

b. Can

a

be equal to

2b

Given Equation and Values: We are given the equation $(2b-a)x^{2}+(4a^{2}-b^{2})(2x-a)=0$ and the values $a=10$ and $b=12$ . We need to find the value of $x$ .
Substitute and Simplify: Substitute $a=10$ and $b=12$ into the equation. $(2(12)-10)x^{2}+(4(10)^{2}-(12)^{2})(2x-10)=0$
Quadratic Equation: Simplify the equation. $\newline$ $(24-10)x^{2}+(400-144)(2x-10)=0$
Calculate Discriminant: Continue simplifying. $14x^2+256(2x-10)=0$
Use Quadratic Formula: Distribute the $256$ . $\newline$ $14x^{2}+512x-2560=0$
Calculate Solutions: This is a quadratic equation in the form of $ax^2 + bx + c = 0$ . We can solve for $x$ using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , but first, let's check if the equation can be factored.
Solution for $x$ : The equation $14x^{2}+512x-2560=0$ does not factor easily, so we will use the quadratic formula. $\newline$ $a = 14$ , $b = 512$ , and $c = -2560$ .
Check $a=2b$ : Calculate the discriminant $(b^2 - 4ac)$ . $\newline$ Discriminant = $512^2 - 4(14)(-2560)$
Simplify and Solve: Calculate the discriminant. $\newline$ Discriminant = $262144 + 143360$
Conclusion: Add the values to find the discriminant. Discriminant = $405504$
Conclusion: Add the values to find the discriminant. $\newline$ Discriminant = $405504$ Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula. $\newline$ $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$
Conclusion: Add the values to find the discriminant. $\newline$ Discriminant = $405504$ Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula. $\newline$ $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ Calculate the square root of the discriminant. $\newline$ $\sqrt{405504} = 636.8$ (approximately)
Conclusion: Add the values to find the discriminant. $\newline$ Discriminant = $405504$ Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula. $\newline$ $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ Calculate the square root of the discriminant. $\newline$ $\sqrt{405504} = 636.8$ (approximately)Substitute the square root back into the quadratic formula. $\newline$ $x = \frac{-512 \pm 636.8}{28}$
Conclusion: Add the values to find the discriminant. $\newline$ Discriminant = $405504$ Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula. $\newline$ $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ Calculate the square root of the discriminant. $\newline$ $\sqrt{405504} = 636.8$ (approximately)Substitute the square root back into the quadratic formula. $\newline$ $x = \frac{-512 \pm 636.8}{28}$ Calculate the two possible values for x. $\newline$ $x_1 = \frac{-512 + 636.8}{28}$ $\newline$ $x_2 = \frac{-512 - 636.8}{28}$
Conclusion: Add the values to find the discriminant. $\newline$ Discriminant = $405504$ Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula. $\newline$ $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ Calculate the square root of the discriminant. $\newline$ $\sqrt{405504} = 636.8$ (approximately) Substitute the square root back into the quadratic formula. $\newline$ $x = \frac{-512 \pm 636.8}{28}$ Calculate the two possible values for $x$ . $\newline$ $x_1 = \frac{-512 + 636.8}{28}$ $\newline$ $x_2 = \frac{-512 - 636.8}{28}$ Solve for $x_1$ . $\newline$ $x_1 = \frac{124.8}{28}$ $\newline$ x_1 \approx $4$.$457$
Conclusion: Add the values to find the discriminant.$\newline$Discriminant = $405504$Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$Calculate the square root of the discriminant.$\newline$$\sqrt{405504} = 636.8$ (approximately)Substitute the square root back into the quadratic formula.$\newline$$x = \frac{-512 \pm 636.8}{28}$Calculate the two possible values for $x$.$\newline$$x_1 = \frac{-512 + 636.8}{28}$$\newline$$x_2 = \frac{-512 - 636.8}{28}$Solve for $x_1$.$\newline$$x_1 = \frac{124.8}{28}$$\newline$$x_1 \approx 4.457$Solve for $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$0$.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$1$$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$
Conclusion: Add the values to find the discriminant.$\newline$Discriminant = $405504$Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$Calculate the square root of the discriminant.$\newline$$\sqrt{405504} = 636.8$ (approximately)Substitute the square root back into the quadratic formula.$\newline$$x = \frac{-512 \pm 636.8}{28}$Calculate the two possible values for $x$.$\newline$$x_1 = \frac{-512 + 636.8}{28}$$\newline$$x_2 = \frac{-512 - 636.8}{28}$Solve for $x_1$.$\newline$$x_1 = \frac{124.8}{28}$$\newline$$x_1 \approx 4.457$Solve for $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$0$.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$1$$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$Since we are looking for a length, we cannot have a negative value for $x$. Therefore, $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$ is the solution.
Conclusion: Add the values to find the discriminant.$\newline$Discriminant = $405504$Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$Calculate the square root of the discriminant.$\newline$$\sqrt{405504} = 636.8$ (approximately)Substitute the square root back into the quadratic formula.$\newline$$x = \frac{-512 \pm 636.8}{28}$Calculate the two possible values for x.$\newline$$x_1 = \frac{-512 + 636.8}{28}$$\newline$$x_2 = \frac{-512 - 636.8}{28}$Solve for $x_1$.$\newline$$x_1 = \frac{124.8}{28}$$\newline$$x_1 \approx 4.457$Solve for $x_2$.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$0$$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$1$Since we are looking for a length, we cannot have a negative value for x. Therefore, $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$ is the solution.Now, let's address part b of the problem: Can $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$3$ be equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$?$\newline$If $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$, then we substitute $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$3$ with $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$ in the original equation.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$8$
Conclusion: Add the values to find the discriminant.$\newline$Discriminant = $405504$ Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ Calculate the square root of the discriminant.$\newline$$\sqrt{405504} = 636.8$ (approximately) Substitute the square root back into the quadratic formula.$\newline$$x = \frac{-512 \pm 636.8}{28}$ Calculate the two possible values for $x$.$\newline$$x_1 = \frac{-512 + 636.8}{28}$$\newline$$x_2 = \frac{-512 - 636.8}{28}$ Solve for $x_1$.$\newline$$x_1 = \frac{124.8}{28}$$\newline$$x_1 \approx 4.457$ Solve for $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$0$.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$1$$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$ Since we are looking for a length, we cannot have a negative value for $x$. Therefore, $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$ is the solution.Now, let's address part b of the problem: Can $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$ be equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$6$?$\newline$If $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$7$, then we substitute $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$ with $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$6$ in the original equation.$\newline$$\sqrt{405504} = 636.8$$0$ Simplify the equation.$\newline$$\sqrt{405504} = 636.8$$1$
Conclusion: Add the values to find the discriminant.$\newline$Discriminant = $405504$Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$Calculate the square root of the discriminant.$\newline$$\sqrt{405504} = 636.8$ (approximately)Substitute the square root back into the quadratic formula.$\newline$$x = \frac{-512 \pm 636.8}{28}$Calculate the two possible values for $x$.$\newline$$x_1 = \frac{-512 + 636.8}{28}$$\newline$$x_2 = \frac{-512 - 636.8}{28}$Solve for $x_1$.$\newline$$x_1 = \frac{124.8}{28}$$\newline$$x_1 \approx 4.457$Solve for $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$0$.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$1$$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$Since we are looking for a length, we cannot have a negative value for $x$. Therefore, $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$ is the solution.Now, let's address part b of the problem: Can $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$ be equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$6$?$\newline$If $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$7$, then we substitute $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$ with $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$6$ in the original equation.$\newline$$\sqrt{405504} = 636.8$$0$Simplify the equation.$\newline$$\sqrt{405504} = 636.8$$1$Further simplification gives us:$\newline$$\sqrt{405504} = 636.8$$2$
Conclusion: Add the values to find the discriminant.$\text{Discriminant} = 405504$Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula.$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$Calculate the square root of the discriminant.$\sqrt{405504} = 636.8$ (approximately)Substitute the square root back into the quadratic formula.$x = \frac{-512 \pm 636.8}{28}$Calculate the two possible values for x.$x_1 = \frac{-512 + 636.8}{28}$$x_2 = \frac{-512 - 636.8}{28}$Solve for $x_1$.$x_1 = \frac{124.8}{28}$$x_1 \approx 4.457$Solve for $x_2$.$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$0$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$1$Since we are looking for a length, we cannot have a negative value for x. Therefore, $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$ is the solution.Now, let's address part b of the problem: Can $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$3$ be equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$?If $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$, then we substitute $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$3$ with $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$ in the original equation.$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$8$Simplify the equation.$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$9$Further simplification gives us:$\sqrt{405504} = 636.8$$0$Since the first term is $0$, the equation simplifies to:$\sqrt{405504} = 636.8$$1$
Conclusion: Add the values to find the discriminant.$\newline$Discriminant = $405504$ Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ Calculate the square root of the discriminant.$\newline$$\sqrt{405504} = 636.8$ (approximately) Substitute the square root back into the quadratic formula.$\newline$$x = \frac{-512 \pm 636.8}{28}$ Calculate the two possible values for $x$.$\newline$$x_1 = \frac{-512 + 636.8}{28}$$\newline$$x_2 = \frac{-512 - 636.8}{28}$ Solve for $x_1$.$\newline$$x_1 = \frac{124.8}{28}$$\newline$$x_1 \approx 4.457$ Solve for $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$0$.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$1$$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$ Since we are looking for a length, we cannot have a negative value for $x$. Therefore, $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$ is the solution.Now, let's address part b of the problem: Can $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$ be equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$6$?$\newline$If $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$7$, then we substitute $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$ with $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$6$ in the original equation.$\newline$$\sqrt{405504} = 636.8$$0$ Simplify the equation.$\newline$$\sqrt{405504} = 636.8$$1$ Further simplification gives us:$\newline$$\sqrt{405504} = 636.8$$2$ Since the first term is $\sqrt{405504} = 636.8$$3$, the equation simplifies to:$\newline$$\sqrt{405504} = 636.8$$4$ For the equation to hold true, either $\sqrt{405504} = 636.8$$5$ or $\sqrt{405504} = 636.8$$6$. Since $\sqrt{405504} = 636.8$$7$ cannot be $\sqrt{405504} = 636.8$$3$ (as it is a given length), the only possibility is that $\sqrt{405504} = 636.8$$6$.
Conclusion: Add the values to find the discriminant.$\newline$Discriminant = $405504$Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$Calculate the square root of the discriminant.$\newline$$\sqrt{405504} = 636.8$ (approximately)Substitute the square root back into the quadratic formula.$\newline$$x = \frac{-512 \pm 636.8}{28}$Calculate the two possible values for x.$\newline$$x_1 = \frac{-512 + 636.8}{28}$$\newline$$x_2 = \frac{-512 - 636.8}{28}$Solve for $x_1$.$\newline$$x_1 = \frac{124.8}{28}$$\newline$$x_1 \approx 4.457$Solve for $x_2$.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$0$$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$1$Since we are looking for a length, we cannot have a negative value for $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$. Therefore, $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$3$ is the solution.Now, let's address part b of the problem: Can $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$ be equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$?$\newline$If $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$6$, then we substitute $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$4$ with $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$5$ in the original equation.$\newline$$x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$9$Simplify the equation.$\newline$$\sqrt{405504} = 636.8$$0$Further simplification gives us:$\newline$$\sqrt{405504} = 636.8$$1$Since the first term is $\sqrt{405504} = 636.8$$2$, the equation simplifies to:$\newline$$\sqrt{405504} = 636.8$$3$For the equation to hold true, either $\sqrt{405504} = 636.8$$4$ or $\sqrt{405504} = 636.8$$5$. Since $\sqrt{405504} = 636.8$$6$ cannot be $\sqrt{405504} = 636.8$$2$ (as it is a given length), the only possibility is that $\sqrt{405504} = 636.8$$5$.Solve for $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$$2$ when $\sqrt{405504} = 636.8$$5$.$\newline$$x = \frac{-512 \pm 636.8}{28}$$1$$\newline$x = b
Conclusion: Add the values to find the discriminant. $\newline$ Discriminant = $405504$ Since the discriminant is positive, there are two real solutions. Now, calculate the solutions using the quadratic formula. $\newline$ $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ Calculate the square root of the discriminant. $\newline$ $\sqrt{405504} = 636.8$ (approximately) Substitute the square root back into the quadratic formula. $\newline$ $x = \frac{-512 \pm 636.8}{28}$ Calculate the two possible values for $x$ . $\newline$ $x_1 = \frac{-512 + 636.8}{28}$ $\newline$ $x_2 = \frac{-512 - 636.8}{28}$ Solve for $x_1$ . $\newline$ $x_1 = \frac{124.8}{28}$ $\newline$ $x_1 \approx 4.457$ Solve for $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $0$ . $\newline$ $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $1$ $\newline$ $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $2$ Since we are looking for a length, we cannot have a negative value for $x$ . Therefore, $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $4$ is the solution.Now, let's address part b of the problem: Can $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $5$ be equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $6$ ? $\newline$ If $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $7$ , then we substitute $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $5$ with $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $6$ in the original equation. $\newline$ $\sqrt{405504} = 636.8$ $0$ Simplify the equation. $\newline$ $\sqrt{405504} = 636.8$ $1$ Further simplification gives us: $\newline$ $\sqrt{405504} = 636.8$ $2$ Since the first term is $\sqrt{405504} = 636.8$ $3$ , the equation simplifies to: $\newline$ $\sqrt{405504} = 636.8$ $4$ For the equation to hold true, either $\sqrt{405504} = 636.8$ $5$ or $\sqrt{405504} = 636.8$ $6$ . Since $\sqrt{405504} = 636.8$ $7$ cannot be $\sqrt{405504} = 636.8$ $3$ (as it is a given length), the only possibility is that $\sqrt{405504} = 636.8$ $6$ .Solve for $x$ when $\sqrt{405504} = 636.8$ $6$ . $\newline$ $x = \frac{-512 \pm 636.8}{28}$ $2$ $\newline$ $x = \frac{-512 \pm 636.8}{28}$ $3$ This means that if $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $5$ is equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $6$ , then $x$ must be equal to $\sqrt{405504} = 636.8$ $7$ . There is no contradiction here, so $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $5$ can be equal to $x = \frac{-512 \pm \sqrt{405504}}{2 \times 14}$ $6$ .