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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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Discriminant—When a quadratic equation is in standard form, Ax2 + Bx + C,
= 0 the expression, B2- 4AC, that is found under the square root part of the
quadratic formula is called the discriminant. The sign of the discriminant
can reveal how many solutions there are going to be and if the solutions are
real numbers or complex numbers.

 When the Discriminant b2-4ac is Characteristic Solution Comment > 0 Two distinct real solutions As the discriminant is found under the square root, if its value is positive, then the quadratic possesses two distinct real roots = 0 One real solution As the discriminant is found under the square root, if its value is zero, then the quadratic possesses only one real root. < 0 Two distinct complex imaginary solutions As the discriminant is found under the square root, if its value is positive, then the quadratic possesses two distinct complex roots

To determine the determinant, it is necessary to convert a given trinomial
form to standard form. Then you can easily read off A, B and C.

For the following quadratics, determine the value of the discriminant and
give the number and type of roots:

(a) 2x2 – 8x + 7 discr. = 82 - 4(2)(7) = 64 - 56 = 8, >0,

(b) x2 + x - 1 discr. =

(c) x2 + 4x + 4 discr. =

(d)   Determine over the entire domain 0 ≤ θ ≤ ½π

discrim = [2cos(θ)]2 – 4( ½)sin(2θ)

=

Now from a trigonometric identity:

Note that the discriminant is a function of θ. For some values of θ, it is
positive, for others it is negative. So, we must divide the domain into 3
regions:

Now for 0 < θ < 45°, cos(θ) > sin(θ), whereas for 45° < θ <90, cos(θ) < sin (θ).

Thus for 0 < θ < 45°.

When θ = 45°, cos(45°) = sin(45 °), thus

For 45° < θ < 90° cos(θ) < sin(θ) so

(e) 3x2 – 5x + 2

(f) x2 + 4 = 0 discr. =

(g) An alternate form of the quadratic equation is given by dividing all the terms
the standard form of the quadratic equation by x2 to obtain;

This form is helpful if B2 >> 4AC where denotes much greater, in which case
the usual form of the quadratic formula can give inaccurate numerical results
for one of the roots. Show that it leads to the solution:

Solution. Let’s solve by completing the square:

Step 1: Keep the x -2 and x -1 terms on one side of the equation, and the
constant term on the other.

Step 2: Complete the square=

Inside the parentheses, take half of and square it, add result to both sides:

Step 3: Factor the perfect square (created in step 2) as a binomial squared:

Divide by C :

Step 4: Now we can solve the equation in step 3 by using the square root
method outlined previously.

subtract from both sides:

Finally, taking the inverse to recover x :

x =

Okay-eee, that should just about do it—Either you love quadratics more or
you loathe them more now.