
How do you solve the quadratic equation ${{x}^{2}}+11x=7$?
Answer
552k+ views
Hint: In this problem we need to solve the given quadratic equation i.e., we need to calculate the values of $x$ where the given equation is satisfied. For solving a quadratic equation, we have several methods. But in the problem, we are going to use the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.To use the above formula we need to have the quadratic equation in standard form which is $a{{x}^{2}}+bx+c=0$, so we will simplify the given equation and convert it into the standard form. After that we will compare the given equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0$ and write the values of $a$, $b$, $c$. Now we will substitute those values in the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and simplify the obtained equation to get the required result.
Complete step by step solution:
Given equation ${{x}^{2}}+11x=7$.
Shifting the constant which is right hand side to left hand side, then we will get
$\Rightarrow {{x}^{2}}+11x-7=0$
Comparing the above quadratic equation with standard quadratic equation $a{{x}^{2}}+bx+c=0$, then we will get the values of $a$, $b$, $c$ as
$a=1$, $b=11$, $c=-7$.
We have the quadratic formula for the solution as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of $a$, $b$, $c$ in the above equation, then we will get
$\Rightarrow x=\dfrac{-\left( 11 \right)\pm \sqrt{{{\left( 11 \right)}^{2}}-4\left( 1 \right)\left( -7 \right)}}{2\left( 1 \right)}$
We know that when we multiplied a negative sign with a positive sign, then we will get a negative sign. Applying the above rule and simplifying the above equation, then we will get
$\begin{align}
& \Rightarrow x=\dfrac{-11\pm \sqrt{121+28}}{2} \\
& \Rightarrow x=\dfrac{-11\pm \sqrt{149}}{2} \\
\end{align}$
In the above equation we have the value $\sqrt{149}$. We can observe that the number $149$ is a prime number, so we can’t simplify the value of $\sqrt{149}$.
Hence the solution of the given quadratic equation ${{x}^{2}}+11x=7$ is $\dfrac{-11\pm \sqrt{149}}{2}$.
Note: There are several methods to solve the quadratic equation. But the easiest one is the using quadratic formula. There are less chances of making mistakes in this method. So, when they don’t mention the method to use, we can freely use this method and solve the equation.
Complete step by step solution:
Given equation ${{x}^{2}}+11x=7$.
Shifting the constant which is right hand side to left hand side, then we will get
$\Rightarrow {{x}^{2}}+11x-7=0$
Comparing the above quadratic equation with standard quadratic equation $a{{x}^{2}}+bx+c=0$, then we will get the values of $a$, $b$, $c$ as
$a=1$, $b=11$, $c=-7$.
We have the quadratic formula for the solution as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of $a$, $b$, $c$ in the above equation, then we will get
$\Rightarrow x=\dfrac{-\left( 11 \right)\pm \sqrt{{{\left( 11 \right)}^{2}}-4\left( 1 \right)\left( -7 \right)}}{2\left( 1 \right)}$
We know that when we multiplied a negative sign with a positive sign, then we will get a negative sign. Applying the above rule and simplifying the above equation, then we will get
$\begin{align}
& \Rightarrow x=\dfrac{-11\pm \sqrt{121+28}}{2} \\
& \Rightarrow x=\dfrac{-11\pm \sqrt{149}}{2} \\
\end{align}$
In the above equation we have the value $\sqrt{149}$. We can observe that the number $149$ is a prime number, so we can’t simplify the value of $\sqrt{149}$.
Hence the solution of the given quadratic equation ${{x}^{2}}+11x=7$ is $\dfrac{-11\pm \sqrt{149}}{2}$.
Note: There are several methods to solve the quadratic equation. But the easiest one is the using quadratic formula. There are less chances of making mistakes in this method. So, when they don’t mention the method to use, we can freely use this method and solve the equation.
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