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kogti [31]
2 years ago
10

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

Mathematics
2 answers:
Stella [2.4K]2 years ago
6 0
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.

kirill115 [55]2 years ago
5 0

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}.

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