
How do you solve $x=\sqrt{2x+3}$?
Answer
568.8k+ views
Hint: We have to solve the above given question by using certain transformations and substitutions. Right from the start you can clearly observe from the question that you can use only positive values of $x$, since taking the square root of a real positive number will always produce a positive value. This means that you need to have $x\ge 0$
Complete step by step answer:
From the question given, equation is $x=\sqrt{2x+3}$
By squaring on both sides of the equation we get,
${{x}^{2}}={{\left( \sqrt{2x+3} \right)}^{2}}$
$\Rightarrow {{x}^{2}}=2x+3$
Rearrange the above equation by moving all terms on one side,
By rearranging, we get the below equation, ${{x}^{2}}-2x-3=0$
You can find the solutions to this quadratic equation by using the quadratic formula.
We know that for a general form of quadratic equation $a{{x}^{2}}+bx+c=0$, the quadratic formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
The equation we got by rearranging is ${{x}^{2}}-2x-3=0$
By comparing the coefficients of general form of quadratic equation and above we get the values as
$\begin{align}
& a=1 \\
& b=-2 \\
& c=-3 \\
\end{align}$
Now substitute the above values in the quadratic formula,
By substituting the above values in quadratic formula we get,
$x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -3 \right)}}{2\left( 1 \right)}$
$\Rightarrow x=\dfrac{2\pm \sqrt{4-\left( -12 \right)}}{2}$
$\Rightarrow x=\dfrac{2\pm \sqrt{16}}{2}$
$\Rightarrow x=\dfrac{2\pm 4}{2}$
$\Rightarrow x=3\text{ and x=-1}$
The value $x=-1$ does not satisfy the equation as it is a negative value and square root of positive value will always be a positive value.
So the original equation will have only one solution that is\[x=3\]
Therefore $x=3$ is the solution for the given equation.
Note:
We should be well aware of the quadratic formula and its usage. We should use certain transformations and substitutions according to the question given to make the given question simplified. We should be very careful while doing the calculation of the quadratic formula. We should be well aware of the positive square roots and negative square roots. If we do not know the concept of positive square roots then we may write both the solutions here but that is a wrong conclusion.
Complete step by step answer:
From the question given, equation is $x=\sqrt{2x+3}$
By squaring on both sides of the equation we get,
${{x}^{2}}={{\left( \sqrt{2x+3} \right)}^{2}}$
$\Rightarrow {{x}^{2}}=2x+3$
Rearrange the above equation by moving all terms on one side,
By rearranging, we get the below equation, ${{x}^{2}}-2x-3=0$
You can find the solutions to this quadratic equation by using the quadratic formula.
We know that for a general form of quadratic equation $a{{x}^{2}}+bx+c=0$, the quadratic formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
The equation we got by rearranging is ${{x}^{2}}-2x-3=0$
By comparing the coefficients of general form of quadratic equation and above we get the values as
$\begin{align}
& a=1 \\
& b=-2 \\
& c=-3 \\
\end{align}$
Now substitute the above values in the quadratic formula,
By substituting the above values in quadratic formula we get,
$x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -3 \right)}}{2\left( 1 \right)}$
$\Rightarrow x=\dfrac{2\pm \sqrt{4-\left( -12 \right)}}{2}$
$\Rightarrow x=\dfrac{2\pm \sqrt{16}}{2}$
$\Rightarrow x=\dfrac{2\pm 4}{2}$
$\Rightarrow x=3\text{ and x=-1}$
The value $x=-1$ does not satisfy the equation as it is a negative value and square root of positive value will always be a positive value.
So the original equation will have only one solution that is\[x=3\]
Therefore $x=3$ is the solution for the given equation.
Note:
We should be well aware of the quadratic formula and its usage. We should use certain transformations and substitutions according to the question given to make the given question simplified. We should be very careful while doing the calculation of the quadratic formula. We should be well aware of the positive square roots and negative square roots. If we do not know the concept of positive square roots then we may write both the solutions here but that is a wrong conclusion.
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