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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 :

quenjolangce
(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


rodriguezmina94
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... :(

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