Answer:
[tex]t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)[/tex]
Step-by-step explanation:
Rearranging your quadratic to standard form, you get ...
16t^2 -4t +d = 0
For the quadratic formula, you have ...
a = 16
b = -4
c = d
so the solution is ...
[tex]t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(16)(d)}}{2(16)}=\dfrac{4\pm 4\sqrt{1-4d}}{32}\\\\t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)[/tex]
Answer:
C. [tex]t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}[/tex]
Step-by-step explanation:
We have been given an equation [tex]d=-16t^2+4t[/tex]. We are asked to solve for t.
To solve for t we will rewrite our given equation in general form of equation [tex](ax^2+bx+c=0)[/tex].
[tex]d+16t^2-4t=0[/tex]
[tex]16t^2-4t+d=0[/tex]
Since our given equation is quadratic, so we will use quadratic formula to solve for t.
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Upon comparing our given equation with general form of equation we can see,
[tex]a=16[/tex]
[tex]b=-4[/tex]
[tex]c=d[/tex]
Upon substituting our given values in quadratic formula we will get,
[tex]t=\frac{--4\pm\sqrt{(-4)^2-4\cdot 16\cdot d}}{2\cdot 16}[/tex]
[tex]t=\frac{4\pm\sqrt{16-64\cdot d}}{32}[/tex]
Upon factoring out 16 inside the square root we will get,
[tex]t=\frac{4\pm\sqrt{16(1-4\cdot d)}}{32}[/tex]
[tex]t=\frac{4\pm 4\sqrt{(1-4\cdot d)}}{32}[/tex]
[tex]t=\frac{4}{32}\pm\frac{4\sqrt{(1-4\cdot d)}}{32}[/tex]
[tex]t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}[/tex]
Therefore, the solutions of t are[tex]t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}[/tex] and option C is the correct choice.
