Question

Does anyone know hw to find the zeros of this function h(x)=x2 - 4x + 1

Answer 1

Answer:

The zeros are $\sqrt{3} + 2\ and \sqrt-{3} + 2$

Step-by-step explanation:

$x = \frac{-b +- \sqrt{b^2 - 4ac} \\}{2a}$

Use the quadratic formula

Then plug in all the numbers

$x = \frac{4 +- \sqrt{(-4)^2 - 4(1)(1)} \\}{2(1)}$

Then solve (-4)^2

$x = \frac{4 +- \sqrt{16 - 4(1)(1)} \\}{2(1)}$

Then solve everything inside the radical the radical

$\sqrt{12} = 2\sqrt{3}$

Then divide and simplify the radical. After that, you will get your final answer.

$x = \frac{4 +- 2\sqrt{3)} \\}{2} = 2 + or - \sqrt{3}$

So your final answer is $\sqrt{3} + 2\ and \sqrt-{3} + 2$

Hope it helped! My answer is expert verified.