How do I know what integer the irrational number that was squared is closest to?

Mathematics

1 answer:

djverab [1.8K]3 years ago

3 0

Answer:

Approximate Irrational Numbers (solutions, examples, videos ...www.onlinemathlearning.com

Step-by-step explanation:try this it might help

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Please help!!! Select the quadratic equation that has no real solution.

Gnom [1K]

Answer:

$25x^2-10x+4$

Step-by-step explanation:

You can use the quadratic formula to determine if a quadratic equation has real or imaginary solutions. The quadratic formula is: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . What really matters in this case is the discriminant, which is the stuff under the radical which is: $b^2-4ac$ . This is because the solutions are only imaginary, if the discriminant is negative, because then you would be taking the square root of a negative number. So let's look through each example:

$25x^2+20x-4$ ; a=25, b=20, c-4: $20^2-4(25)(-4) = 800$ . By examining this one example, it's important to note, if you have one negative number as a or c, then it cancels out the negative sign in the -4, and it becomes positive. So let's look at examples where a or c doesn't equal a negative number OR both a and c equal negative, that way they cancel out and over -4ac is still negative.

$25x^2-10x+4$ ; a=25, b=-10, c=4. In this case both a and c are positive so -4ac will remain negative. This gives you: $(-10)^2-4(25)(4) = 100-400 = -300$ . So this has a negative discriminant meaning it will have no real solution but rather imaginary solutions