Question

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Answer 1

Answer:

5. The quadratic function has a bigger positive solution is f(x)=2x^2-32.

6. It will take approximately 2.5 seconds for the screwdiver to reach the ground.

7. The value of c so that -9 and 9 are both solutions of x^2+c=103 is c=22.

Step-by-step explanation:

5. f(x)=2x^2-32

Solution:

$f(x)=0\\ 2x^{2}-32=0$

Solving for x: Adding 32 both sides of the equation:

$2x^{2} -32+32=0+32\\ 2x^{2} =32$

Dividing both sides of the equation by 2:

$\frac{2x^{2} }{2}=\frac{32}{2}\\ x^{2}=16$

Square root both sides of the equation:

$\sqrt{x^{2} } =+-\sqrt{16}$

x=±4

Solution: x=-4 and x=4

g(x)=12x^2-48

Solution:

$g(x)=0\\ 12x^2-48=0$

Solving for x: Adding 48 both sides of the equation:

$12x^{2}-48+48=0+48\\12x^{2} =48$

Dividing both sides of the equation by 12:

$\frac{12x^{2} }{12}=\frac{48}{12}\\x^{2} =4$

Square root both sides of the equation:

$\sqrt{x^{2} } =+-\sqrt{4}$

x=±2

Solution: x=-2 and x=2

h(x)=100x^2

Solution:

$h(x)=0\\ 100x^{2} =0$

Solving for x: Dividing both sides of the equation by 100:

$\frac{100x^{2} }{100}=\frac{0}{100}\\ x^{2} =0$

Square root both sides of the equation:

$\sqrt{x^{2} } =\sqrt{0}\\ x=0$

Solution: x=0

Answer: The quadratic function has a bigger positive solution is f(x)=2x^2-32.

6. h=-16t^2+98

How long will it take for the screwdiver to reach the ground?

In the ground h=0, then:

$-16t^2+98=0$

Solving for t: Subtracting 98 from both sides of the equation:

$-16t^2+98-98=0-98\\ -16t^2=-98$

Dividing both sides of the equation by -16:

$\frac{-16t^2}{-16}=\frac{-98}{-16}\\t^2=6.125$

Square root both sides of the equation, taking only the positive value, because the time must be a positive number:

$\sqrt{t^2}=\sqrt{6.125}\\t=2.474873734$

Rounding to the nearest tenth:

t=2.5 seconds

Answer: It will take approximately 2.5 seconds for the screwdiver to reach the ground.

7. What is the value of c so that -9 and 9 are both solutions of x^2+c=103?

x=-9 or x=9 are solutions. Replacing the values in the equation:

$\left \{ {{(-9)^{2}+c =103} \atop {(9)^{2}+c=103}} \right.$

In both case we get:

$81+c=103$

Solving for c: Subtracting 81 from both sides of the equation:

$81+c-81=103-81\\ c=22$

Answer: The value of c so that -9 and 9 are both solutions of x^2+c=103 is c=22

Answer 2

QUESTION 5a

The given function is $f(x)=2x^2-32$

We factor to obtain;  $f(x)=2(x^2-16)$.

This implies that;

$f(x)=2(x^2-4^2)$.

We apply difference of two squares to obtain;

$f(x)=2(x-4)(x+4))$.

To find the solutions of the function, we solve $f(x)=0$.

$2(x-4)(x+4)=0$

$(x-4)(x+4)=0$

$\Rightarrow (x-4)=0,(x+4)=0$.

$\Rightarrow x=4,x=-4$

The positive solution is 4

QUESTION 5b.

The given function is

$g(x)=12x^2-48$.

To find the solution, we solve $g(x)=0$.

This implies that;

$12x^2-48=0$.

We divide through by 12 to get;

$x^2-4=0$

$x^2=4$

$x=\pm \sqrt{4}$

$x=\pm 2$

The positive solution is $x=2$.

QUESTION 5c

The given function is $h(x)=100x^2$.

To find the solution of this function, we solve the  equation;

$100x^2=0$

This implies that;

$x^2=0$

$x=0$

Therefore the quadratic function with the biggest positive solution is ;

$f(x)=2x^2-32$.

QUESTION 6.

The height of the screwdriver is modeled by:

$h=-16t^2+98$

When the screwdriver reach the ground its height will be zero.

This implies that;

$-16t^2+98=0$

This implies that;

$-16t^2=-98$

We divide through by -16 to get;

$t^2=\frac{-98}{-16}$

$t^2=\frac{49}{8}$

$t=\pm \sqrt{\frac{49}{8}}$

$t=\pm \frac{7}{\sqrt{8} }$

$t=\pm 2.475$

Time is always positive, so we discard the negative value to get;

$t=2.5$ seconds to the nearest tenth.

QUESTION 7

The given equation is

$x^2+c=103$

This implies that

$x^2=103-c$

$x=\pm \sqrt{103-c}$

$x=-\sqrt{103-c}$

or

$x=\sqrt{103-c}...eqn(1)$

We substitute $x=9\:or-9$ into  either equation (1) or (2) to get

$9=\sqrt{103-c}...eqn(2)$

$9^2=103-c$

$81=103-c$

$81-103=-c$

$c=22$

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