Immylesidn
Answered

IDNStudy.com, ang iyong mapagkakatiwalaang mapagkukunan para sa maaasahang mga sagot. Tuklasin ang mga maaasahang impormasyon sa anumang paksa sa pamamagitan ng aming network ng bihasang mga propesyonal.

4. Find the solutions of each of the following QE by factoring. Explain how you arrived at your answer.

a. (x+3)^2 = 25 c. (2t - 3)^2 = 2t^2 + 5t - 26

b. (s+4)^2 = -2s d. 3(x+2)^2 = 2x^2 + 3x - 8

5. Do you agree that x^2 + 5x - 14 = 0 and 14 - 5x - x^2 = 0 have the same solutions? Justify your answer.
6. Show that the equation (x - 4)^2 = 9 can be solved both by factoring and extracting square roots.

7. A computer manufacturing company would like to come up with a new
laptop computer such that its monitor is 80 sq inches smaller than the
present ones. Suppose the length of the monitor of the larger computer
is 5 inches longer than its width and the area of the smaller computer
is 70 sq inches. What are the dimensions of the monitor of the larger
computer?
THANK YOU PO :)

Sagot :

Quenjolangceidn
(x-3)^2 = 25
x^2 + 9 - 6x = 25
x^2 - 6x + 9 - 25 = 0
x^2 - 6x - 16 = 0
x^2 - 8x + 2x - 16 = 0
x(x - 8) + 2(x - 8) = 0
(x + 2)(x - 8) = 0
x = -2 , 8


Rodriguezmina94idn
4. [tex](x+3)^2=25[/tex]
[tex] \sqrt{(x+3)^2}= + or-\sqrt{25} [/tex]
[tex]x+3=+or-5[/tex]
[tex]x=5-3[/tex]
[tex]x=2[/tex]
[tex]x=-5-3[/tex]
[tex]x=-8[/tex]
[tex] \left \{ {{x=2} \atop {x=-8}} \right. [/tex]

[tex](s+4)^2=-2s[/tex]
[tex]s^2+8s+16+2s=0[/tex]
[tex]s^2+10s+16=0[/tex]
[tex]s^2+10s=-16[/tex]
[tex]s^2+10s+(5)^2=-16+(5)^2[/tex]
[tex] \sqrt{(s+5)^2}=+or- \sqrt{9} [/tex]
[tex]s+5=+or-3[/tex]
[tex]s=3-5[/tex]
[tex]s=-2[/tex]
[tex]s=-3-5[/tex]
[tex]s=-8[/tex]
[tex] \left \{ {{x=-2} \atop {x=-8}} \right. [/tex]

[tex](2t-3)^2=2t^2+5t-26[/tex]
[tex]4t^2-12t+9=2t^2+5t-26[/tex]
[tex]4t^2-2t^2-12t-5t+9+26=0[/tex]
[tex]2t^2-17t+35=0[/tex]
[tex]2t^2-17t=-35[/tex]
[tex] \frac{2t^2-17t}{2} = \frac{-35}{2} [/tex]
[tex]t^2- \frac{17}{2}t=- \frac{35}{2} [/tex]
[tex]t^2- \frac{17}{2} t+( \frac{17}{4})^2=- \frac{35}{2}+( \frac{17}{4})^2 [/tex]
[tex] \sqrt{(t- \frac{17}{4}) } =+or- \sqrt{ \frac{289}{16} } [/tex]
[tex]t- \frac{17}{4}=+or- \frac{17}{4} [/tex]
[tex]t= \frac{17}{4}+ \frac{17}{4} [/tex]
[tex]t= \frac{34}{4} or \frac{17}{2} [/tex]
[tex]t=- \frac{17}{4}+ \frac{17}{4} [/tex]
[tex]t=0[/tex]
[tex] \left \{ {{t= \frac{17}{2} } \atop {x=0}} \right. [/tex]

[tex]3(x+2)^2=2x^2+3x-8[/tex]
[tex]x^2+4x+4+3=2x^2+3x-8[/tex]
[tex] x^{2} -2 x^{2} +4x-3x+7+8=0[/tex]
[tex]- x^{2} -x+15=0[/tex]
[tex] \frac{- x^{2} -x}{-1} = \frac{-15}{-1} [/tex]
[tex] x^{2} +x+( \frac{1}{2})^2 =15+( \frac{1}{2})^2 [/tex]
[tex] \sqrt{(x+ \frac{1}{2})^2 } =+ or - \sqrt{ \frac{61}{4} } [/tex]
[tex]x+ \frac{1}{2} = +or- \sqrt{ \frac{61}{2} } [/tex] (When you write it on your paper don't include 2 in the square root sign)
[tex]x= \sqrt{ \frac{61-1}{2} } [/tex] (Likewise here, don't include -1/2 in the square root sign)
[tex]x=- \sqrt{ \frac{61-1}{2} } [/tex] (Here too)

5.Yes,
[tex]x^2+5x-14=0[/tex]
[tex](x+7)(x-2)=0[/tex]
[tex]x+7=0[/tex]
[tex]x-2=0[/tex]
[tex]x=-7[/tex]
[tex]x=2[/tex]

[tex]14-5x-x^2=0[/tex]
[tex]-1[14-5x-x^2=0][/tex]
Use distributive property then it will be equal to x²+5x-14=0.

[tex](x-4)^2=9[/tex]
[tex] \sqrt{(x-4)^2} = \sqrt{9} [/tex]
[tex]x-4=+or-3[/tex]
[tex]x=3+4[/tex]
[tex]x=7[/tex]
[tex]x=-3+4[/tex]
[tex]x=1[/tex]

[tex](x+4)^2=9[/tex]
[tex] x^{2} -8x+16-9=0[/tex]
[tex] x^{2} -8x+7=0[/tex]
[tex](x-7)(x-1)=0[/tex]
[tex]x-7=0[/tex]
[tex]x=7[/tex]
[tex]x-1=0[/tex]
[tex]x=1[/tex]

P.S. Done at last... Sorry for the late answers.... Solving it was time consuming... Sorry also because I can't quite get number 7.... I hope this helps you though... :(