Nope. Consider x2+0x+1=0. This means x2+1=0. This has two solutions, but they are complex numbers: +i and -i, where i is the squareroot of -1.
How about x2+0x+0=0? This means x2=0. This has two solutions, sure, but they aren't distinct. In this case, x=0 for both solutions, so we just consider them one solution.
Plug 'a', 'b', and 'c' from the equation into the formula. When you do that, the formula becomes a pair of numbers ... one number when you pick the 'plus' sign, and another number when you pick the 'minus' sign. Those two numbers are the 'solutions' to the quadratic equation you started with.
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I am assuming that "an algebra equation" refers to a quadratic equation and not a higher polynomial.For a quadratic equation of the form y = ax^2 + bx + c, where a, b and c are real numbers and a is non-zero, the discriminant is b^2 – 4ac.
In that case, the discriminant is not a perfect square.
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x2 - 4x + 4 or (x - 2)2
0x2 + 1x - 7 = 0
Plug 'a', 'b', and 'c' from the equation into the formula. When you do that, the formula becomes a pair of numbers ... one number when you pick the 'plus' sign, and another number when you pick the 'minus' sign. Those two numbers are the 'solutions' to the quadratic equation you started with.
I don't think there is any easy way to estimate it; just use the quadratic equation to calculate the solutions. You can round some of the numbers to get an estimate; in this case you might even do most of the calculation in your head, but it's probably easier just to do the full calculation.
They are page numbers 24 and 25 . ( 24 x 25 = 600 ) The easiest way to solve this is by trial and error. Multiply two consecutive numbers; if the product is too low, try larger numbers, if it is too high, try smaller numbers. You can also write an equation and use the quadratic formula. The equation in this case is x(x+1) = 600. Re-written for use of the quadratic equation, it becomes x2 + x - 600 = 0. This will give you a positive and a negative solution; only the positive solution is sensible in this case.
The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.
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I am assuming that "an algebra equation" refers to a quadratic equation and not a higher polynomial.For a quadratic equation of the form y = ax^2 + bx + c, where a, b and c are real numbers and a is non-zero, the discriminant is b^2 – 4ac.
In that case, the discriminant is not a perfect square.
153
Ax2 + Bx + C = 0A, B, and C are constants (numbers)
The complex roots of an equation is any solution to that equation which cannot be expressed in terms of real numbers. For example, the equation 0 = x² + 5 does not have any solution in real numbers. But in complex numbers, it has solutions.