How do you factor 25x^2-9

2 answers:

4 0

How<span> you would factor this is,

</span>The two terms of the binomial equation 25x<span>^2 + -9 = 0 are perfect square. When factoring the difference of two perfect squares: x^2 – a^2 = (x+a)(x-a) . STEP 2: Set up and solve the equation. (5x) (5x) + (5x) (-3) + (3) (5x) + (3) (-3) = 0 With the Distributive Prop
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PLZ IF IT HELP LIKE AND RATE!!!!

:3

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hoa [83]4 years ago

3 0

If I understood the question correctly you can think of 25x^2-9 as being 25x^2+0x-9. Therefore a=25, b=0, and c=-9. Using the quadratic formula x=(-b(+or-)√(b²-(4ac)))/(2a). after solving this equation with the given a, b, and c values you should get x=3/5 and -3/5.

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I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have