Solve the following equations.
\left(2x+3\right)^{-1/2} = 2
Raise both sides to the \left(-2\right) power:
\left( \left(2x+3\right)^{-\frac{1}{2}} \right)^{-2}=2^{-2}
Which is equivalent to:
2x+3=\dfrac{1}{4}
2x=-\dfrac{11}{4}
x=-\dfrac{11}{8}
The solution of the equation is x=-\dfrac{11}{8}.
\left(4x-3\right)^{-1/4} = 5
Raise both sides to the \left(-4\right) power:
\left( \left(4x-3\right)^{-\frac{1}{4}} \right)^{-4}=5^{-4}
Which is equivalent to:
4x-3=\dfrac{1}{625}
4x=\dfrac{1\ 876}{625}
x=\dfrac{469}{625}.
The solution of the equation is x=\dfrac{469}{625}.
\left(5x+8\right)^{-2/3} = 4
Raise both sides to the \left( -\dfrac{3}{2} \right) power:
\left( \left(5x+8\right)^{-\frac{2}{3}} \right)^{-\frac{3}{2}}=4^{-\frac{3}{2}}
Which is equivalent to:
5x+8=\dfrac{1}{8}
5x=-\dfrac{63}{8}
x=-\dfrac{63}{40}.
The solution of the equation is x=-\dfrac{63}{40}.
\left(3x-5\right)^{1/2} = 7
Raise both sides to the second power:
\left( \left(3x-5\right)^{\frac{1}{2}} \right)^{2}=7^{2}
Which is equivalent to:
3x-5=49
3x=54
x=18.
The solution of the equation is x=18.
\sqrt{2x^{2}+5x-3}=x-3
Raise both sides to the second power:
\left( \sqrt{2x^{2}+5x-3} \right)^{2}=\left(x-3\right)^{2}
Which is equivalent to:
2x^{2}+5x-3=x^{2}-6x+9
x^{2}+11x-12=0
Apply the quadratic formula:
x=\dfrac{-11-\sqrt{\left(11\right)^{2}-4\left(1\right)\left(-12\right)}}{2\left(1\right)}=\dfrac{-11-\sqrt{169}}{2}=-12
Or
x=\dfrac{-11+\sqrt{\left(11\right)^{2}-4\left(1\right)\left(-12\right)}}{2\left(1\right)}=\dfrac{-11+\sqrt{169}}{2}=1
Test if any of the solutions are extraneous by plugging them into the initial equation:
\sqrt{2\left(-12\right)^{2}+5\left(-12\right)-3}=\left(-12\right)-3
\sqrt{225}=-15 which is not true, so x=-12 is an extraneous solution.
\sqrt{2\left(1\right)^{2}+5\left(1\right)-3}=1-3
\sqrt{4}=-2 which is not true, so x=1 is an extraneous solution as well.
The equation \sqrt{2x^{2}+5x-3}=x-3 has no solution.
\sqrt{x^{2}+11x-3}=x-4
Raise both sides to the second power:
\left( \sqrt{x^{2}+11x-3} \right)^{2}=\left(x-4\right)^{2}
Which is equivalent to:
x^{2}+11x-3=x^{2}-8x+16
19x=19
x=1
Test if the solution is extraneous by plugging it in the initial equation:
\sqrt{\left(1\right)^{2}+11\left(1\right)-3}=\left(1\right)-4
\sqrt{9}=-3 which is not true, so x=1 is an extraneous solution.
The equation \sqrt{x^{2}+11x-3}=x-4 has no solution.
\sqrt{13x+43}=x+5
Raise both sides to the second power:
\left( \sqrt{13x+43} \right)^{2}=\left(x+5\right)^{2}
Which is equivalent to:
13x+43=x^{2}+10x+25
x^{2}-3x-18=0
Apply the quadratic formula:
x=\dfrac{-\left(-3\right)-\sqrt{\left(-3\right)^{2}-4\left(1\right)\left(-18\right)}}{2\left(1\right)}=\dfrac{3-\sqrt{81}}{2}=-3
Or
x=\dfrac{-\left(-3\right)+\sqrt{\left(-3\right)^{2}-4\left(1\right)\left(-18\right)}}{2\left(1\right)}=\dfrac{3+\sqrt{81}}{2}=6
Test if any of the solutions are extraneous by plugging them in the initial equation:
\sqrt{13\left(-3\right)+43}=\left(-3\right)+5
\sqrt{4}=2 which is true, so x=-3 is a viable solution.
\sqrt{13\left(6\right)+43}=\left(6\right)+5
\sqrt{121}=11 which is true, so x=6 is a viable solution as well.
The solutions of the equation \sqrt{13x+43}=x+5 are x=-3 and x=6.