Question

3x^2+kx=-3 What is the value of K will result in exactly one solution to the equation?

Answer 1

Answer:

For k = 6 or k = -6, the equation will have exactly one solution.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

If $\bigtriangleup = 0$, the equation has only one solution.

In this problem, we have that:

$3x^{2} + kx + 3 = 0$

So

$a = 3, b = k, c = 3$

$\bigtriangleup = b^{2} - 4ac$

$\bigtriangleup = k^{2} - 4*3*3$

$\bigtriangleup = k^{2} - 36$

We will only have one solution if $\bigtriangleup = 0$. So

$\bigtriangleup = 0$

$k^{2} - 36 = 0$

$k^{2} = 36$

$k = \pm \sqrt{36}$

$k = \pm 6$

For k = 6 or k = -6, the equation will have exactly one solution.