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anonymous

  • 4 years ago

Can someone help me solve the following system of equations. Please explain how. 20a + b = -3 b^2 = 4ac 100a + 10b + c = 10 As you may see I'm trying to find a parabola.

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  1. jhonyy9
    • 4 years ago
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    b from the first equation b=-3-20a and substitute in second (-3-20a)^2 =4ac 9 -2(-3)(-20a)+40a^2 =4ac 9+6(-20a)+40a^2 =4ac 9-120a+40a^2 =4ac

  2. campbell_st
    • 4 years ago
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    re-write 20a + b = -3 into the form b = -20a - 3 (1) re-write 100a + 10b + c=10 as c = 10 - 100a - 10b (2) Substitute 1 into 2 c = 10 - 100a -10(-20a - 3) c = 100a - 20 (3) substitute it into the equation (1) and (3) into b^2 = 4ac then (-20a -3)^2 = 4a(100a - 20) expand, collect like terms and then solve for a

  3. anonymous
    • 4 years ago
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    The answers are not integers.\[\left\{a=\frac{9}{40},b=-\frac{15}{2},c=\frac{125}{2}\right\} \]

  4. anonymous
    • 4 years ago
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    b=-3-20a Applying in the 2nd eqn (-3-20a)^2=4ac 9+120a+400a^2=4ac Applying in the 3rd eqn 100a-30-200a+c=10 c=100a+40 applying in 9+120a+400a^2=4ac 9+120a+400a^2=4a(100a+40) 9+120a+400a^2=400a^2+160a 9=40a a=9/40 b=-3-9/2 =-15/2 c=125/2

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