Given y=2x^2-10x. Find all real values of x for which y=-3
solve 2x^2 - 10x = 3
2x^2 - 10x - 3 = 0
which methods have you learned to solve a quadratic?
I have a copy error, should have been
solve 2x^2 - 10x = -3
2x^2 - 10x + 3 = 0
To find the real values of x for which y = -3, we can set the equation for y equal to -3 and solve for x.
Given y = 2x^2 - 10x, we can rewrite the equation as:
2x^2 - 10x = -3
Now, let's move the -3 to the left-hand side by adding it to both sides of the equation:
2x^2 - 10x + 3 = 0
We have converted the equation to a quadratic equation in the standard form, ax^2 + bx + c = 0, where a = 2, b = -10, and c = 3.
To solve this equation, we can factor it or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Substituting the values a = 2, b = -10, and c = 3 into the formula, we get:
x = (-(-10) ± √((-10)^2 - 4(2)(3))) / (2(2))
Simplifying further:
x = (10 ± √(100 - 24)) / 4
x = (10 ± √76) / 4
Now, let's simplify the expression under the square root:
x = (10 ± √(4 × 19)) / 4
x = (10 ± 2√19) / 4
We can simplify it further by dividing both the numerator and denominator by 2:
x = 5 ± √19 / 2
So, the two possible solutions for x are:
1. x = (5 + √19) / 2
2. x = (5 - √19) / 2
These are the real values of x for which y = -3 for the equation y = 2x^2 - 10x.