Question

Given y = 2x2 - 10x. Find all real values of x for which y = -3.

Answer 1

First, plug in -3 for the y value.
-3=2x²-10x
Then, add 3 to both sides.
0=2x²-10x+3
Use the quadratic equation to solve
(-b+-√b²-4ac)/2a
You get 4.68 or 0.32
You still have to check your answer to see if there is an excluded value.
Both values check.

Answer 2

Answer:  The required values of x are $\dfrac{-5\pm\sqrt{19}}{2}.$

Step-by-step explanation:  We are given the following function of x :

$y=2x^2-10x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to find all real values of x such that y = -3.

To find the required values of x, we must substitute y = -3 in equation (i).

Therefore, from equation (i), we have

$y=-3\\\\\Rightarrow 2x^2-10x=-3\\\\\Rightarrow 2x^2-10x+3=0\\\\\Rightarrow x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times2\times3}}{2\times2}\\\\\\\Rightarrow x=\dfrac{-10\pm\sqrt{100-24}}{4}\\\\\\\Rightarrow x=\dfrac{-10\pm\sqrt{76}}{4}\\\\\\\Rightarrow x=\dfrac{-10\pm2\sqrt{19}}{4}\\\\\\\Rightarrow x=\dfrac{-5\pm\sqrt{19}}{2}.$

We notice that both the values of x are real numbers.

Thus, the required values of x are $\dfrac{-5\pm\sqrt{19}}{2}.$