Solve for z
z=4
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\left(\sqrt{4z+9}\right)^{2}=\left(z+1\right)^{2}
Square both sides of the equation.
4z+9=\left(z+1\right)^{2}
Calculate \sqrt{4z+9} to the power of 2 and get 4z+9.
4z+9=z^{2}+2z+1
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z+1\right)^{2}.
4z+9-z^{2}=2z+1
Subtract z^{2} from both sides.
4z+9-z^{2}-2z=1
Subtract 2z from both sides.
2z+9-z^{2}=1
Combine 4z and -2z to get 2z.
2z+9-z^{2}-1=0
Subtract 1 from both sides.
2z+8-z^{2}=0
Subtract 1 from 9 to get 8.
-z^{2}+2z+8=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=2 ab=-8=-8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -z^{2}+az+bz+8. To find a and b, set up a system to be solved.
-1,8 -2,4
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.
-1+8=7 -2+4=2
Calculate the sum for each pair.
a=4 b=-2
The solution is the pair that gives sum 2.
\left(-z^{2}+4z\right)+\left(-2z+8\right)
Rewrite -z^{2}+2z+8 as \left(-z^{2}+4z\right)+\left(-2z+8\right).
-z\left(z-4\right)-2\left(z-4\right)
Factor out -z in the first and -2 in the second group.
\left(z-4\right)\left(-z-2\right)
Factor out common term z-4 by using distributive property.
z=4 z=-2
To find equation solutions, solve z-4=0 and -z-2=0.
\sqrt{4\times 4+9}=4+1
Substitute 4 for z in the equation \sqrt{4z+9}=z+1.
5=5
Simplify. The value z=4 satisfies the equation.
\sqrt{4\left(-2\right)+9}=-2+1
Substitute -2 for z in the equation \sqrt{4z+9}=z+1.
1=-1
Simplify. The value z=-2 does not satisfy the equation because the left and the right hand side have opposite signs.
z=4
Equation \sqrt{4z+9}=z+1 has a unique solution.
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Simultaneous equation
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\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
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Limits
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