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anonymous

  • 5 years ago

can someone help me solve 15x^4-14x^2+3=0

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  1. anonymous
    • 5 years ago
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    Robo just did for you on your previous message.

  2. anonymous
    • 5 years ago
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    i didnt see it only a conversation

  3. anonymous
    • 5 years ago
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    \[3-14 x^2+15 x^4=\left(-1+3 x^2\right) \left(-3+5 x^2\right) \] \[\left(-1+3 x^2\right)=0,\left(-3+5 x^2\right)=0 \] Solve the two equations above for x. The solutions are: \[x=\left\{-\sqrt{\frac{3}{5 }},\sqrt{\frac{3}{5}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right\} \]

  4. anonymous
    • 5 years ago
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    i keyed that answer and its saying incorrect robtobey

  5. anonymous
    • 5 years ago
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    Keyed answer. Computerized exercises are very specific. They may want the answer in a different form.

  6. anonymous
    • 5 years ago
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    equations reducible to quadratic is what i'm doing

  7. anonymous
    • 5 years ago
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    robo its still saying incorrect

  8. anonymous
    • 5 years ago
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    topdollar843: "equations reducible to quadratic is what i'm doing" First of all if I'm confident that an answer is valid, then, I will move on to other problems to be solved on this site or quit for the day. If you factor the expression on the left hand side of your problem equation, then, you will find that the quartic expression is equal to the product of the two quadratic expressions: \[\left(3 x^2-1\right)*\left(5 x^2-3\right) \] If you replace the left hand side of the problem expression with the product of the two quadratic expressions you have: \[\left(3 x^2-1\right)*\left(5 x^2-3\right)=0 \] In order for the left hand side of the equation to be equal to zero, either one of the quadratic expressions can be zero and the problem equation is valid, that is, 0 times the quantity 5x^2-3 is zero and 3x^2-1 times zero is zero. At the equation level 0=0 which is true. Now in order to find the value of x in 3x^2-1 that makes 3x^2-1 zero, you write down 3x^2-1=0 and then solve this equation for x. You then do the same thing for the other expression, set it to zero and solve the result for x. In the end there will be four unique values for x which when plugged into the original equation one at a time will make the equation true.

  9. anonymous
    • 5 years ago
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    HI ROB AFTER A FEW TRIES THE CORRECT ANSWER WILL BE GIVEN ON THE SITE I DO HOMEWORK ON AFTER THE WRONG ANSWER MORE THAN 3 TIMES IT GIVES YOU THE ANSWER AND A NEW PROBLEM

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