Question

name
teacher
hour
solving quadratic functions
unit 3 test: b
solve the quadratic equation by taking square roots.

1. ( 4 x ^ { 2 } = 49 )
2. on mercury, the distance ( d ) (in feet) that an object falls in time ( t ) (in seconds) can be

modeled by the function ( d ( t ) = 6 t ^ { 2 } ). how long would it take an object to fall
96 feet on mercury?

1. solve the equation by completing the square. state whether the solutions are real or non-

real.
( x ^ { 2 } - 10 x + 17 = 0 )
solution:
circle one: real non-real

1. write the following function in vertex form.

( g ( x ) = x ^ { 2 } - 4 x + 7 )
( g ( x ) = )

Explanation:

1. Solve \(4x^{2}=49\)

Step1: Isolate \(x^{2}\)

Divide both sides of the equation \(4x^{2}=49\) by \(4\).
\(x^{2}=\frac{49}{4}\)

Step2: Take square roots

Take the square root of both sides. Remember that if \(x^{2}=a\) (\(a\geq0\)), then \(x = \pm\sqrt{a}\).
\(x=\pm\sqrt{\frac{49}{4}}=\pm\frac{7}{2}\)

Step1: Isolate \(t^{2}\)

Divide both sides of the equation \(6t^{2}=96\) by \(6\).
\(t^{2}=\frac{96}{6}=16\)

Step2: Take square roots

Take the square root of both sides. Since \(t\) represents time, \(t\geq0\).
\(t=\sqrt{16} = 4\) (we discard \(t = - 4\) because time cannot be negative)

Step1: Rewrite the equation

\(x^{2}-10x=-17\)

Step2: Complete the square

The coefficient of \(x\) is \(-10\). Half of it is \(-5\), and \((-5)^{2}=25\). Add \(25\) to both sides of the equation:
\(x^{2}-10x + 25=-17 + 25\)
\((x - 5)^{2}=8\)

Step3: Take square roots

\(x-5=\pm\sqrt{8}=\pm2\sqrt{2}\)

Step4: Solve for \(x\)

\(x=5\pm2\sqrt{2}\)
Since \(2\sqrt{2}\approx2.83\) is a real - number, the solutions are real.

Answer:

\(x = \frac{7}{2}\) or \(x=-\frac{7}{2}\)

2. Find \(t\) when \(d(t) = 6t^{2}=96\)