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If you are to solve each of the following quadratic equations, which method would you use and why? Explain your answer.
a. 9x2 = 225 d. 2x2 + x – 28 = 0

b. 4x2 – 121 = 0 e. 4x2 + 16x + 15 = 0

c. x2 + 11x + 30 = 0 f. 4x2 + 4x – 15 = 0

Sagot :

kaius
a. Factoring - since the it can be factored to (3x - 5)(3x + 5) by Binomial Theorem. b. Factoring - binomial theorem is also applied (2x - 11)(2x + 11) c. Quadratic Equation - can not be factored easily d. Factoring - can be factored (2x - 7)(x + 4) e. Factoring - can be factored (2x + 3)(2x + 5) f. Factoring - can be factored (2x - 3)(2x + 5)
riza1
[tex]a.\\ 9x^2 = 225 \\9x^2-225=0\\(3x)^2-15^2=0\\(3x-15)(3x+15)=0 \\3x-15=0\ \ or\ \ 3x+15 =0 \\3x=15 \ \ or \ \ 3x=-15 \\x= 3 \ \ or\ \ x=-3\\ \\Factoring : \ a^2-b^2=(a-b)(a+b)[/tex]

[tex]b.\\\\ 4x^2 - 121 = 0\\(2x)^2-11^2 =0\\(2x-11)(2x+11)=0 \\\\2x-11=0 \ \ or \ \ 2x+11=0 \\2x=11 \ \ or \ \ 2x=-11\\ x=\frac{11}{2} \ \ or \ \ x=-\frac{11}{2} \\ x=5.5 \ \ or \ \ x=-5.5 \\Factoring[/tex]

[tex]c.\\\\ x^2 + 11x + 30 = 0 \\a=1, \ \ b=11, \ \c=30 \\\\ \Delta =b^2-4ac = 11^2 -4\cdot1\cdot 30 = 121-120=1 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-11-\sqrt{1 }}{2 }=\frac{ -11-1}{2}=\frac{-12}{2}=-6 \\\\ x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-11+\sqrt{1 }}{2 }=\frac{ -11+1}{2}=\frac{-10}{2}=-5\\\\ Quadratic \ Equation - can \ not \ be \ easily \ decomposed \ into \ factors[/tex]
 
[tex]d.\\\\ 2x^2 + x - 28 = 0 \\a=2, \ \ b=1 , \ \c=-28 \\\\ \Delta =b^2-4ac = 1^2 -4\cdot2\cdot (-28) = 1+224=225 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-1-\sqrt{225}}{2\cdot 2 }=\frac{ -1-15}{4}=\frac{-16}{4}=-4 \\\\ x_{2}=\frac{-b+\sqrt{\Delta} }{2a} =\frac{-1+\sqrt{225}}{2\cdot 2 }=\frac{ -1+15}{4}=\frac{ 14}{4}= 3.5\\\\ Quadratic \ Equation[/tex]

[tex]e. \\\\4x^2 + 16x + 15 = 0 \\a=4, \ \ b=16 , \ \c=15 \\\\ \Delta =b^2-4ac = 16^2 -4\cdot4\cdot 15 = 256-240=16 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-16-\sqrt{16}}{2\cdot 4 }=\frac{ -16-4}{8}=\frac{-20}{8}=- \frac{5}{2}=-2.5 \\\\ x_{2}=\frac{-b+\sqrt{\Delta} }{2a} =\frac{-16+\sqrt{16}}{2\cdot 4 }=\frac{ -16+4}{8}=\frac{-12}{8}=- \frac{3}{2}=-1.5\\\\ Quadratic \ Equation[/tex]

[tex]e.\\\\ 4x^2 + 4x - 15 = 0 \\a=4, \ \ b=4, \ \c=-15 \\\\ \Delta =b^2-4ac = 4^2 -4\cdot4\cdot (- 15 )= 16+240=256 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-4-\sqrt{256}}{2\cdot 4 }=\frac{ -4-16 }{8}=\frac{-20}{8}=- \frac{5}{2}=-2.5 \\\\ x_{2}=\frac{-b+\sqrt{\Delta} }{2a} =\frac{-4+\sqrt{256}}{2\cdot 4 }=\frac{ -4+16}{8}=\frac{ 12}{8}= \frac{3}{2}= 1.5\\\\ Quadratic \ Equation[/tex]


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