How do you find the discriminant for $2x^2-11x+10=0$ and determine the number and type of solutions?

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Answer 1 · 2017-07-18T13:23:27

$Delta = sqrt41$ , so $f$ has two real zeroes.

We know that, for any quadratic polynomial of the form $ax^2+bx+c$ , the discriminant, $Delta$ , is given by $Delta= b^2-4ac$ .

If $∆ > 0$ , then the polynomial will have two real solutions.

If $∆=0$ , then the polynomial will have one real solution - a double root.

And if $∆ < 0$ , then the polynomial will have no real roots and two complex roots.

And if $∆$ is a perfect square, then the polynomial will have rational roots.

For the given polynomial, $∆=(11^2-4xx2xx10)=41$ . Since $∆ > 0$ , but $sqrt∆ notin QQ$ , the polynomial will have two real roots.

How do you find the discriminant for 2x^2-11x+10=02x^2-11x+10=0 and determine the number and type of solutions?