how do you find the zeroes of
y= 2x(squared)-7x+15
Please type your subject in the School Subject box. Any other words are likely to delay responses from a teacher who knows that subject well.
first test to see if there are real roots
(b^2 - 4 a c) must be +
49 - 4*2*15
NO WAY
no real roots. Never crosses the x axis.
To find the zeroes of the equation y = 2x² - 7x + 15, we need to solve for x when y is equal to zero. In other words, we need to find the values of x that make the equation equal to zero.
Step 1: Rewrite the equation y = 2x² - 7x + 15 as 2x² - 7x + 15 = 0.
Step 2: To solve this quadratic equation, we can use either factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
The quadratic formula is x = (-b ± √(b² - 4ac)) / (2a).
For our equation 2x² - 7x + 15 = 0, the coefficients are:
a = 2
b = -7
c = 15
Step 3: Substitute these coefficients into the quadratic formula and simplify:
x = (-(-7) ± √((-7)² - 4 * 2 * 15)) / (2 * 2)
= (7 ± √(49 - 120)) / 4
= (7 ± √(-71)) / 4
Since we have a negative value inside the square root (√-71), it means that there are no real number solutions (zeroes) for this equation. However, we can find the complex solutions by simplifying further:
Step 4: Simplify the square root of -71:
√(-71) = i√71
Therefore, the solution becomes:
x = (7 ± i√71) / 4
These are the complex solutions or zeroes of the equation y = 2x² - 7x + 15.