Question

What is the value of m? If $$x^{2}+3x+1=0$$, then the roots of the equation $$x=\dfrac{-3\pm \sqrt{m} }{2}$$.

Answer 1

Hint: In this question it is given that the roots of the equation $$x^{2}+3x+1=0$$ is $$x=\dfrac{-3\pm \sqrt{m} }{2}$$, then we have to find the value of m. So to find the solution we have to know the quadratic formula, which states that if $$ax^{2}+bx+c=0$$ be any quadratic equation then its roots, $$x=\dfrac{-b\pm \sqrt{b^{2}-4ac} }{2a}$$
Complete step-by-step solution:
Given quadratic equation $$x^{2}+3x+1=0$$,
Now by comparing the given equation with $$ax^{2}+bx+c=0$$, we can write, a=1, b=3, c=1.
Therefore, by quadratic formula, we can write,
$$x=\dfrac{-b\pm \sqrt{b^{2}-4ac} }{2a}$$
$$\Rightarrow x=\dfrac{-3\pm \sqrt{3^{2}-4\times 1\times 1} }{2\times 1}$$
$$\Rightarrow x=\dfrac{-3\pm \sqrt{9-4} }{2}$$
$$\Rightarrow x=\dfrac{-3\pm \sqrt{5} }{2}$$
So by comparing the above solution with $$x=\dfrac{-3\pm \sqrt{m} }{2}$$, we can easily say that m=5.
Which is our required solution.
Note: So while solving we just compare the two equations to write the value of ‘m’. So in mathematics comparison method is a vital part, so for comparing you need two equations whose structures are identical and after that you can write the unknown values of one equation by comparing with the other one.