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

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

[tex]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}[/tex]

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:

[tex]-r^2-2r+14=-8r^2+6[/tex]

So we need to arrange this equation in the form:

[tex]ax^2+bx+c=[/tex]

Thus:

[tex]-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[/tex]

So the discriminant is:

[tex]\Delta=(-2)^2-4(7)(8) \\ \\ \Delta=4-224 \\ \\ \boxed{\Delta=-220}[/tex]

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 [tex]c[/tex] such that:

[tex]x^2+24x+c=0[/tex]

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

[tex]a^2x^2\pm 2axb+b^2[/tex]

and can be expressed in squared-binomial form as:

[tex](ax\pm b)^2[/tex]

So we can write our quadratic equation as follows:

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

Finally, the value of [tex]c[/tex] that completes the square is 144 because:

[tex]x^2+24x+144=(x+12)^2[/tex]

14. Value that completes the square.

Answer: C) [tex]\frac{121}{4}[/tex]

What we need to find is the value of [tex]c[/tex] such that:

[tex]z^2+11z+c=0[/tex]

So we can write our quadratic equation as follows:

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

Finally, the value of [tex]c[/tex] that completes the square is [tex]\frac{121}{4}[/tex] because:

[tex]z^2+11z+\frac{121}{4}=(x+\frac{11}{2})^2[/tex]

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:

[tex]A=L\times W[/tex]

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:

[tex]L=2W-1[/tex]

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

[tex]A=L\times W \\ \\ \\ Where: \\ \\ W:Width \\ \\ L:Length[/tex]

[tex]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}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ W_{1}=\frac{-(-1)+ \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{1}=\frac{1+ \sqrt{1+360}}{4} \therefore W_{1}=5 \\ \\ W_{2}=\frac{-(-1)- \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{2}=\frac{1- \sqrt{1+360}}{4} \therefore W_{2}=-\frac{9}{2}[/tex]

The only valid option is [tex]W_{1}[/tex] because is greater than zero. Recall that we can't have a negative value of the width. For the length we have:

[tex]L=2(5)-1 \\ \\ L=9[/tex]

Finally:

[tex]The \ length \ is \ 9 \ inches \\ \\ The \ width \ is \ 5 \ inches[/tex]

16. Satellite

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

[tex]d=-9t^2+776[/tex]

where [tex]t[/tex] 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, [tex]d=452[/tex]:

[tex]d=-9t^2+776 \\ \\ 452=-9t^2+776 \\ \\ 9t^2=776-452 \\ \\ 9t^2=324 \\ \\ t^2=\frac{324}{9} \\ \\ t^2=36 \\ \\ t=\sqrt{36} \\ \\ \boxed{t=6h}[/tex]

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