Question

2 use the quadratic formula to determine how long it takes for a t - shirt to land back on the ground after the mascot launches it. ask yourself... do you think an exact solution or approximate solution is more appropriate for this context? 3 meredith solves the quadratic equation x² - 7x - 8 = 3. her work is shown. a identify merediths error. she didnt make her equation = to o(zero) meredith x² - 7x - 8 = 3 a = 1, b = -7, c = -8 x = \frac{-(-7)pmsqrt{(-7)^{2}-4(1)(-8)}}{2(1)} x = \frac{7pmsqrt{49 + 32}}{2} x = \frac{7pmsqrt{81}}{2} x = \frac{7pm9}{2} x = \frac{7 + 9}{2} and x = \frac{7 - 9}{2} x = \frac{16}{2} and x = \frac{-2}{2}=-1 the roots are 8 and (-1). b determine the solution to merediths quadratic equation. 4 use the quadratic formula to determine the zeros for each function. round the solutions to the nearest hundredth. a f(x)=2x² + 10x - 102 b h(x)=3x² + 11x - 2

Explanation:

3(a)

Step1: Recall quadratic - formula condition

The quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ is used for equations in the form $ax^{2}+bx + c=0$. Meredith used it on $x^{2}-7x - 8 = 3$ without first rewriting it as $x^{2}-7x-11 = 0$. She incorrectly took $c=-8$ instead of $c = - 11$.

3(b)

Step1: Rewrite the equation in standard form

Rewrite $x^{2}-7x - 8=3$ as $x^{2}-7x-11 = 0$. Here, $a = 1$, $b=-7$, and $c=-11$.

Step2: Apply the quadratic formula

$x=\frac{-(-7)\pm\sqrt{(-7)^{2}-4(1)(-11)}}{2(1)}=\frac{7\pm\sqrt{49 + 44}}{2}=\frac{7\pm\sqrt{93}}{2}$.

Step3: Calculate the two solutions

$x_1=\frac{7+\sqrt{93}}{2}\approx\frac{7 + 9.644}{2}=\frac{16.644}{2}=8.322$ and $x_2=\frac{7-\sqrt{93}}{2}\approx\frac{7 - 9.644}{2}=\frac{-2.644}{2}=-1.322$.

4(a)

Step1: Identify coefficients

For the function $f(x)=2x^{2}+10x - 102$, we have $a = 2$, $b = 10$, and $c=-102$.

Step2: Apply the quadratic formula

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-10\pm\sqrt{10^{2}-4(2)(-102)}}{2(2)}=\frac{-10\pm\sqrt{100 + 816}}{4}=\frac{-10\pm\sqrt{916}}{4}=\frac{-10\pm2\sqrt{229}}{4}=\frac{-5\pm\sqrt{229}}{2}$.

Step3: Calculate the two solutions

$\sqrt{229}\approx15.133$. So $x_1=\frac{-5 + 15.133}{2}=\frac{10.133}{2}=5.07$ and $x_2=\frac{-5-15.133}{2}=\frac{-20.133}{2}=-10.07$.

4(b)

Answer:

Step1: Identify coefficients

For the function $h(x)=3x^{2}+11x - 2$, we have $a = 3$, $b = 11$, and $c=-2$.

Step2: Apply the quadratic formula

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-11\pm\sqrt{11^{2}-4(3)(-2)}}{2(3)}=\frac{-11\pm\sqrt{121+24}}{6}=\frac{-11\pm\sqrt{145}}{6}$.

Step3: Calculate the two solutions

$\sqrt{145}\approx12.042$. So $x_1=\frac{-11 + 12.042}{6}=\frac{1.042}{6}\approx0.17$ and $x_2=\frac{-11-12.042}{6}=\frac{-23.042}{6}\approx - 3.84$.