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3 Answers

y = ax2 + bx + c
To find zeros of the quadratic function use formula:
x12 = (- b ± √D) / (2a) , D = b2 - 4ac
In the set of real numbers:
1. There are two zeros, if D > 0
2. There is one zero, if D = 0
3. There is no zeros, if D < 0
Let's find value of x if y = 0
4x2 + 6x - 2 = 0 ---> 2x2 + 3x - 1 = 0
D = 32 - 4*2*(-1) = 17 > 0
x1 = (-3 + √17) / 4
x2 = (-3 - √17) / 4

Adding to what Mark said, you can graph the equation too (either by hand or with the aid of a graphing calculator) to check that what you have solved using the Quadratic equation is correct. You'll be able to see if and where the function crosses the x-axis.

Just a hint but the x-intercept is where the graph crosses the x-axis so ask yourself what value does "y" have when this occurs.  You can directly use the Quadratic equation and solve for values of x for this special case.  If the values are real numbers then they exist.