Question

Use the quadratic formula to find the solution to the quadratic equation given below.
x^2-5x+25/4=0

Answer 1

Is this the equation?

\( x^2 - 5x + \dfrac{25}{4} = 0\)

Answer 2

yes

Answer 3

\[x ^{2}-5 x+25/4=0,or 4 x ^{2}-20 x+25=0\]
a=4,b=-20,c=25
\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{2a }\]
solve

Answer 4

Here you have a = 1, b = -5, and c = 25/4.

The quadratic formula is:

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

We substitute our values of a, b, and c, and then we simplify.

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\(x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)\frac{25}{4}}}{2(1)} \)

\(x = \dfrac{5 \pm \sqrt{25 - 25}}{2} \)

\(x = \dfrac{5 \pm \sqrt{0}}{2} \)

\(x = \dfrac{5}{2} \)

Answer 5

it should be ( -20 )^{2}=400
& not -20^{2}

Answer 6

If you prefer, you can do it this way:

\(4x^2 - 20x + 25 = 0\)

with a = 4, b = -20, c = 25

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \dfrac{20 \pm \sqrt{400 - 4(4)(25)}}{2(4)}\)

\(x = \dfrac{20}{8} = \dfrac{5}{2}\)

The result is the same.