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let, equation is ax2+bx+c=0

so, its solution will be

x= (-b-sqrt(b*b-4ac))/2a

x= (-b+sqrt(b*b-4ac))/2a

it is generalized equation for finding roots of Quadratic eq.

Q: How do you solve quadratic equation by competing the square which is not a perfect square?

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Rational.

There are many ways: one is to factorise. If the quadratic is written as ax2 + bx + c then, if b2 = 4ac, the quadratic is a perfect square. It is (x - b/2a)2

By using the quadratic equation formula

By using the quadratic equation formula or by completing the square

Square

Related questions

Rational.

Yes. You can calculate the two roots of a quadratic equation by using the quadratic formula, and because there are square roots on the quadratic formula, and if the radicand is not a perfect square, so the answer to that equation has decimal.

The two solutions are coincident.

There are many ways: one is to factorise. If the quadratic is written as ax2 + bx + c then, if b2 = 4ac, the quadratic is a perfect square. It is (x - b/2a)2

square

A quadratic equation.

Yes, it won't be exact, but you can round the number to get a close estimate.

By using the quadratic equation formula

By using the quadratic equation formula or by completing the square

In that case, the discriminant is not a perfect square.

Square

The discriminant must be a perfect square or a square of a rational number.