Solve the following equations.
\dfrac{x^{2}+2x-3}{x-4}=0
\dfrac{P}{Q}=0 if and only if P=0
In our problem, that means:
x^{2}+2x-3=0
Applying the quadratic formula:
x=\dfrac{-2-\sqrt{\left(2\right)^{2}-4\left(1\right)\left(-3\right)}}{2\left(1\right)}=\dfrac{-2-\sqrt{16}}{2}=-3
Or x=\dfrac{-2+\sqrt{\left(2\right)^{2}-4\left(1\right)\left(-3\right)}}{2\left(1\right)}=\dfrac{-2+\sqrt{16}}{2}=1
The denominator of the original equation cannot equal zero:
x-4\neq0
Which is equivalent to:
x\neq4
The solutions to the equation \dfrac{x^{2}+2x-3}{x-4}=0 are x=1 and x=-3.
\dfrac{4x^2+5x-10}{3x+5}+\dfrac{2x^2-4x-5}{3x+5}=0
Adding the fractions, we get:
\dfrac{6x^2+x-15}{3x+5}=0
\dfrac{P}{Q}=0 if and only if P=0
That means:
6x^{2}+x-15=0
Applying the quadratic formula:
x=\dfrac{-1-\sqrt{\left(1\right)^{2}-4\left(6\right)\left(-15\right)}}{2\left(6\right)}=\dfrac{-1-\sqrt{361}}{12}=-\dfrac{20}{12}=-\dfrac{5}{3} or x=\dfrac{-1+\sqrt{\left(1\right)^{2}-4\left(6\right)\left(-15\right)}}{2\left(6\right)}=\dfrac{-1+\sqrt{361}}{12}=\dfrac{18}{12}=\dfrac{3}{2}
The denominator of the original equation cannot equal zero:
3x+5\neq0
Which is equivalent to:
x\neq-\dfrac{5}{3}
That means x=-\dfrac{5}{3} is not a solution to the equation.
The solution to the equation \dfrac{4x^2+5x-10}{3x+5}+\dfrac{2x^2-4x-5}{3x+5}=0 is x=\dfrac{3}{2}.
\dfrac{2x^2+8x-10}{x-1}=3
Subtracting 3 from both sides gives us:
\dfrac{2x^2+8x-10}{x-1}-3=0
Since the common denominator is \left(x-1\right), we write:
\dfrac{2x^2+8x-10}{x-1}-\dfrac{3\left(x-1\right)}{x-1}=0
Subtracting the fractions, we have:
\dfrac{2x^{2}+5x-7}{x-1}=0
Which is equivalent to:
2x^{2}+5x-7=0
Applying the quadratic formula:
x=\dfrac{-5-\sqrt{\left(5\right)^{2}-4\left(2\right)\left(-7\right)}}{2\left(2\right)}=\dfrac{-5-\sqrt{81}}{4}=-\dfrac{7}{2} or x=\dfrac{-5+\sqrt{\left(5\right)^{2}-4\left(2\right)\left(-7\right)}}{2\left(2\right)}=\dfrac{-5+\sqrt{81}}{4}=1
The denominator of the original equation cannot equal zero:
x-1\neq0
Which is equivalent to:
x\neq1
That means x=1 is not a solution to the equation.
The solution to the equation \dfrac{2x^2+8x-10}{x-1}=3 is x=-\dfrac{7}{2}.
\dfrac{2x^2-5x-25}{2x+5}=0
\dfrac{P}{Q}=0 if and only if P=0.
In our problem, that means:
2x^{2}-5x-25=0
Applying the quadratic formula:
x=\dfrac{-\left(-5\right)-\sqrt{\left(-5\right)^{2}-4\left(2\right)\left(-25\right)}}{2\left(2\right)}=\dfrac{5-\sqrt{225}}{4}=-\dfrac{5}{2} or x=\dfrac{-\left(-5\right)+\sqrt{\left(-5\right)^{2}-4\left(2\right)\left(-25\right)}}{2\left(2\right)}=\dfrac{5+\sqrt{225}}{4}=5
The denominator of the original equation cannot equal zero:
2x+5\neq0
Which is equivalent to:
x\neq-\dfrac{5}{2}
That means x=-\dfrac{5}{2} is not a solution to the equation.
The solution to the equation \dfrac{2x^2-5x-25}{2x+5}=0 is x=5.
\dfrac{7x^2+2x-1}{4x-5}+\dfrac{x^2-16x+6}{4x-5}=0
Adding the fractions, we get:
\dfrac{8x^2-14x+5}{4x-5}=0
\dfrac{P}{Q}=0 if and only if P=0
That means:
8x^{2}-14x+5=0
Applying the quadratic formula:
x=\dfrac{-\left(-14\right)-\sqrt{\left(-14\right)^{2}-4\left(8\right)\left(5\right)}}{2\left(8\right)}=\dfrac{14-\sqrt{36}}{16}=\dfrac{1}{2} or x=\dfrac{-\left(-14\right)+\sqrt{\left(-14\right)^{2}-4\left(8\right)\left(5\right)}}{2\left(8\right)}=\dfrac{14+\sqrt{36}}{16}=\dfrac{5}{4}
The denominator of the original equation cannot equal zero:
4x-5\neq0
Which is equivalent to:
x\neq\dfrac{5}{4}
That means x=\dfrac{5}{4} is not a solution to the equation.
The solution to the equation \dfrac{7x^2+2x-1}{4x-5}+\dfrac{x^2-16x+6}{4x-5}=0 is x=\dfrac{1}{2}.
\dfrac{4x^2+4x-24}{x+3}=12
Subtracting 12 from both sides gives us:
\dfrac{4x^2+4x-24}{x+3}-12=0
Since the common denominator is \left(x+3\right), we write:
\dfrac{4x^2+4x-24}{x+3}-\dfrac{12\left(x+3\right)}{x+3}=0
Subtracting the fractions, we have:
\dfrac{4x^{2}-8x-60}{x+3}=0
Which is equivalent to:
4x^{2}-8x-60=0
Dividing the equation by 4, we have:
x^{2}-2x-15=0
Applying the quadratic formula:
x=\dfrac{-\left(-2\right)-\sqrt{\left(-2\right)^{2}-4\left(1\right)\left(-15\right)}}{2\left(1\right)}=\dfrac{2-\sqrt{64}}{2}=-3 or x=\dfrac{-\left(-2\right)+\sqrt{\left(-2\right)^{2}-4\left(1\right)\left(-15\right)}}{2\left(1\right)}=\dfrac{2+\sqrt{64}}{2}=5
The denominator of the original equation cannot equal zero:
x+3\neq0
Which is equivalent to:
x\neq-3
That means x=-3 is not a solution to the equation.
The solution to the equation \dfrac{4x^2+4x-24}{x+3}=12 is x=5.
\dfrac{6x^2+11x-7}{3x+7}=0
\dfrac{P}{Q}=0 if and only if P=0.
In our problem, that means:
6x^{2}+11x-7=0
Applying the quadratic formula:
x=\dfrac{-11-\sqrt{\left(11\right)^{2}-4\left(6\right)\left(-7\right)}}{2\left(6\right)}=\dfrac{-11-\sqrt{289}}{12}=-\dfrac{7}{3} or x=\dfrac{-11+\sqrt{\left(11\right)^{2}-4\left(6\right)\left(-7\right)}}{2\left(6\right)}=\dfrac{-11+\sqrt{289}}{12}=\dfrac{1}{2}
The denominator of the original equation cannot equal zero:
3x+7\neq0
Which is equivalent to:
x\neq-\dfrac{7}{3}
That means x=-\dfrac{7}{3} is not a solution to the equation.
The solution to the equation \dfrac{6x^2+11x-7}{3x+7}=0 is x=\dfrac{1}{2}.