How to tell if a quadratic equation has a real solution?

Mathematics

2 answers:

Dennis_Churaev [7]2 years ago

7 0

The easiest way to determine if a quadratic equation has real solutions is by using the discriminant.

Start by finding a,b and c. You can find a by looking at the coefficient of x^2. You can find b by looking at the coefficient of x. You can find c by looking at the constant at the end.

Now use the discriminant formula.

b^2 - 4ac

After plugging in and solving, check the number. If it is negative, there are no real solutions. If it is positive there are 2 solutions. If it is 0, there is one real answer and it is a double root.

attashe74 [19]2 years ago

4 0

If the discriminant(b^2-4ac) is rational, then the quadratic is rational.

~ThePirc

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Step-by-step explanation:

Expand both factors and collect like term

Using Pascal' triangle with n = 6 to obtain the coefficients

1 6 15 20 15 6 1

Decreasing powers of 1 from $1^{6}$ to $1^{0}$

Increasing powers of 3x from $(3x)^{0}$ to $(3x)^{6}$

$1+3x)^{6}$

= 1. $1^{6}$ $(3x)^{0}$ + 6. $1^{5}$ $(3x)^{1}$ + 15. $1^{4}$ $(3x)^{2}$ + 20. $1^{3}$ $(3x)^{3}$ + 15.1² $(3x)^{4}$ + 6. $1^{1}$ $(3x)^{5}$ + 1. $1^{0}$ $(3x)^{6}$

= 1 + 18x + 135x² + 540x³ + 1215 $x^{4}$ + 1458 $x^{5}$ + 729 $x^{6}$

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$(1-3x)^{6}$

= 1. $1^{6}$ $(-3x)^{0}$ + 6. $1^{5}$ $(-3x)^{1}$ + 15. $1^{4}$ $(-3x)^{2}$ + 20. $1^{3}$ $(-3x)^{3}$ + 15.1² $(-3x)^{4}$ + 6. $1^{1}$ $(-3x)^{5}$ + 1. $1^{0}$ $(-3x)^{6}$