Question

How do you solve ${{x}^{2}}+6x-1=0$?

Answer 1

In this problem we have to solve and find the value of x. We can solve this quadratic equation by different methods. one is using a quadratic formula and the other is by creating a perfect square trinomial. In this problem we can solve using quadratic formula to solve and find the value of x. To solve by using the quadratic formula, we can find the value of a, b, c by comparing the given quadratic equation and the general equation and substitute them, to solve for x.

Complete step-by-step solution:
We know that the given quadratic equation is,
${{x}^{2}}+6x-1=0$
We also know that the quadratic formula for the quadratic equation form $a{{x}^{2}}+bx+c=0$ is
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now we can compare the given quadratic equation and the general quadratic equation, we get
a = 1, b = 6, c = -1.
Now we can substitute the above values in the quadratic formula, we get

& \Rightarrow x=\dfrac{-6\pm \sqrt{{{\left( 6 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{-6\pm \sqrt{36+4}}{2} \\\ & \Rightarrow x=\dfrac{-6\pm \sqrt{40}}{2} \\\ & \Rightarrow x=\dfrac{-6\pm \sqrt{10\times 4}}{2} \\\ \end{aligned}$$ Now we can cancel the square term inside the root, we get $$\Rightarrow x=\dfrac{-6\pm 2\sqrt{10}}{2}$$ Now we can take common term form the numerator, we get $$\Rightarrow x=\dfrac{2\left( -3\pm \sqrt{10} \right)}{2}$$ We can now cancel the similar terms and separate the terms individually $$\begin{aligned} & \Rightarrow x=-3\pm \sqrt{10} \\\ & \Rightarrow x=-3+\sqrt{10},-3-\sqrt{10} \\\ \end{aligned}$$ **Therefore, the value of $$x=-3+\sqrt{10},-3-\sqrt{10}$$.** **Note:** We can also solve this by using perfect square trinomial by taking the constant to the left-hand side and add a constant on both sides and then divide the coefficient of x by 2, to get a perfect square and add it both sides of the equation to solve for x. Student should know the correct formula for using quadratic formula.