Question

I need help from question 11- 16! Please help!

Answer 1

11. Find discriminant.

Answer: D) 0, one real solution

A quadratic function is given of the form:

`ax^2+bx+c=`

We can find the roots of this equation using the quadratic formula:

`x_(12)=(-b \pm √(b^2-4ac))/(2a)`

Where
`\Delta=b^2-4ac` is named the discriminant. This gives us information about the roots without computing them. So, arranging our equation we have:

`4a^2-4a-6=-7 \\ \\ Adding \ 7 \ to \ both \ sides \ of \ the \ equation: \\ \\ 4a^2-4a-6+7=-7+7 \\ \\ 4a^2-4a+1=0 \\ \\ Then \ the \ discriminant: \\ \\ \Delta=(-4)^2-4(4)(1) \\ \\ \Delta=16-16 \\ \\ \boxed{Delta=0}`

Since the discriminant equals zero, then we just have one real solution.

12. Find discriminant.

Answer: D) -220, no real solution

In this exercise, we have the following equation:

`-r^2-2r+14=-8r^2+6`

So we need to arrange this equation in the form:

`ax^2+bx+c=`

Thus:

`-r^2-2r+14=-8r^2+6 \\ \\ Adding \ 8r^2 \ to \ both \ sides \ of \ the \ equation: \\ \\ -r^2-2r+14+8r^2=-8r^2+6+8r^2 \\ \\ Associative \ Property: \\ \\ (-r^2+8r^2)-2r+14=(-8r^2+8r^2)+6 \\ \\ 7r^2-2r+14=6 \\ \\ Subtracting \ 6 \ from \ both \ sides: \\ \\ 7r^2-2r+14-6=6-6 \\ \\ 7r^2-2r+8=0`

So the discriminant is:

`\Delta=(-2)^2-4(7)(8) \\ \\ \Delta=4-224 \\ \\ \boxed{\Delta=-220}`

Since the discriminant is less than one, then there is no any real solution

13. Value that completes the squares

Answer: C) 144

What we need to find is the value of
`c` such that:

`x^2+24x+c=0`

is a perfect square trinomial, that are given of the form:

`a^2x^2\pm 2axb+b^2`

and can be expressed in squared-binomial form as:

`(ax\pm b)^2`

So we can write our quadratic equation as follows:

`x^2+2(12)x+c \\ \\ So: \\ \\ a=1 \\ \\ b=12 \\ \\ c=b^2 \therefore c=12^2 \therefore \boxed{c=144}`

Finally, the value of
`c` that completes the square is 144 because:

`x^2+24x+144=(x+12)^2`

14. Value that completes the square.

Answer: C)
`(121)/(4)`

What we need to find is the value of
`c` such that:

`z^2+11z+c=0`

So we can write our quadratic equation as follows:

`z^2+2(11)/(2)z+c \\ \\ So: \\ \\ a=1 \\ \\ b=(11)/(2) \\ \\ c=b^2 \therefore c=\left((11)/(2)\left)^2 \therefore \boxed{c=(121)/(4)}`

Finally, the value of
`c` that completes the square is
`(121)/(4)` because:

`z^2+11z+(121)/(4)=(x+(11)/(2))^2`

15. Rectangle.

In this problem, we need to find the length and width of a rectangle. We are given the area of the rectangle, which is 45 square inches. We know that the formula of the area of a rectangle is:

`A=L* W`

From the statement we know that the length of the rectangle is is one inch less than twice the width, this can be written as:

`L=2W-1`

So we can introduce this into the equation of the area, hence:

`A=L* W \\ \\ \\ Where: \\ \\ W:Width \\ \\ L:Length`

`A=(2W-1)(W) \\ \\ But \ A=45: \\ \\ 45=(2W-1)(W) \\ \\ Distributive \ Property:\\ \\ 45=2W^2-W \\ \\ 2W^2-W-45=0 \\ \\ Quadratic \ Formula: \\ \\ x_(12)=(-b\pm √(b^2-4ac))/(2a) \\ \\ W_(1)=(-(-1)+ √((-1)^2-4(2)(-45)))/(2(2)) \\ \\ W_(1)=(1+ √(1+360))/(4) \therefore W_(1)=5 \\ \\ W_(2)=(-(-1)- √((-1)^2-4(2)(-45)))/(2(2)) \\ \\ W_(2)=(1- √(1+360))/(4) \therefore W_(2)=-(9)/(2)`

The only valid option is
`W_(1)` because is greater than zero. Recall that we can't have a negative value of the width. For the length we have:

`L=2(5)-1 \\ \\ L=9`

Finally:

`The \ length \ is \ 9 \ inches \\ \\ The \ width \ is \ 5 \ inches`

16. Satellite

The distance in miles between mars and a satellite is given by the equation:

`d=-9t^2+776`

where
`t` is the number of hours it has fallen. So we need to find when the satellite will be 452 miles away from mars, that is,
`d=452`:

`d=-9t^2+776 \\ \\ 452=-9t^2+776 \\ \\ 9t^2=776-452 \\ \\ 9t^2=324 \\ \\ t^2=(324)/(9) \\ \\ t^2=36 \\ \\ t=√(36) \\ \\ \boxed{t=6h}`

Finally, the satellite will be 452 miles away from mars in 6 hours.