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mathdad
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Check the last term.greg1313 said:Well, let's see...
Is $8-2\sqrt7$ a perfect square number?
$$(a+b\sqrt7)^2=a^2+2ab\sqrt7+7b$$
topsquark said:Check the last term.
greg1313 said:Well, let's see...
Is $8-2\sqrt7$ a perfect square number?
$$(a+b\sqrt7)^2=a^2+2ab\sqrt7+7b^2$$
$$a^2+7b^2=8$$
$$2ab=-2$$
$$\implies a=-1,b=1$$
Hence $8-2\sqrt7=(\sqrt7-1)^2$
$$\sqrt7-\sqrt{8-2\sqrt7}=\sqrt7-(\sqrt7-1)=\sqrt7-\sqrt7+1=1$$
A radical equation is an equation that contains a radical, or a square root symbol. It is used to solve for a variable that is under the radical.
To solve a radical equation, you must first isolate the radical on one side of the equation. Then, square both sides of the equation to eliminate the radical. Finally, solve for the variable and check your solution by plugging it back into the original equation.
Extraneous solutions are solutions that arise during the process of solving a radical equation but do not actually satisfy the original equation. They can occur when squaring both sides of an equation to eliminate a radical, as this can introduce additional solutions that do not work in the original equation.
A radical equation contains a radical, while a quadratic equation contains a squared variable. Both equations can be solved using similar methods, but the solutions for a radical equation may need to be checked for extraneous solutions. Additionally, a quadratic equation can have multiple solutions, while a radical equation typically only has one.
Yes, radical equations can have imaginary solutions. This can occur when taking the square root of a negative number, which results in an imaginary number. Imaginary solutions are represented by the letter "i" and can be used to solve equations, but they are not real numbers.