ximenamonica
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How do you figure out 2a^2-46a+252=0?
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Algebra 1 problem
- Thread starter ximenamonica
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BigGlenntheHeavy
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\(\displaystyle 2a^2-46a+252 \ = \ 0\)
\(\displaystyle 2[a^2-23a+126] \ = \ 0, \ first, \ always \ pull \ out \ common \ factors\)
\(\displaystyle Grouping: \ (-9)(-14) \ = \ 126, \ -9-14 \ = \ -23\)
\(\displaystyle 2[a^2-9a-14a+126] \ = \ 0\)
\(\displaystyle 2[a(a-9)-14(a-9)] \ = \ 0\)
\(\displaystyle 2(a-9)(a-14) \ = \ 0\)
\(\displaystyle Hence, \ a \ = \ 9 \ or \ a \ = \ 14\)
\(\displaystyle Note: \ This \ form \ (grouping) \ works \ for \ integers, \ but \ in \ the \ real \ world, \ stick \ with \ the\)
\(\displaystyle \ quadratic \ formula, \ to \ wit: \ x \ = \ \frac{-b \ \pm\sqrt{b^2-4ac}}{2a}\)
\(\displaystyle Real \ world: \ Solve \ for \ x, \ .0253x^2-457x+832.567 \ = \ 0\)
\(\displaystyle 2[a^2-23a+126] \ = \ 0, \ first, \ always \ pull \ out \ common \ factors\)
\(\displaystyle Grouping: \ (-9)(-14) \ = \ 126, \ -9-14 \ = \ -23\)
\(\displaystyle 2[a^2-9a-14a+126] \ = \ 0\)
\(\displaystyle 2[a(a-9)-14(a-9)] \ = \ 0\)
\(\displaystyle 2(a-9)(a-14) \ = \ 0\)
\(\displaystyle Hence, \ a \ = \ 9 \ or \ a \ = \ 14\)
\(\displaystyle Note: \ This \ form \ (grouping) \ works \ for \ integers, \ but \ in \ the \ real \ world, \ stick \ with \ the\)
\(\displaystyle \ quadratic \ formula, \ to \ wit: \ x \ = \ \frac{-b \ \pm\sqrt{b^2-4ac}}{2a}\)
\(\displaystyle Real \ world: \ Solve \ for \ x, \ .0253x^2-457x+832.567 \ = \ 0\)
First, simplify by dividing by 2:ximenamonica said:How do you figure out 2a^2-46a+252=0?
a^2 - 23a + 126 = 0
Next, factor. If you have missed classes when factoring was taught, go learn here:
http://www.youtube.com/watch?v=1B7b7VnG2t0