Write the formula for the discriminant.  The discriminant is the term inside the square root portion of the quadratic formula.

$\displaystyle D=b^2-4ac$

Substitute the known coefficients of the polynomial in standard form:

$\displaystyle y=ax^2+bx+c$

$\displaystyle a=-2, b=0, c=-6$

$\displaystyle D=0^2-4(-2)(-6) =-48$

The answer is:  $\displaystyle -48$

`$\displaystyle f(x)=x^3+7\sqrt{A}x^2+2Bx$

If we were to find the roots of this equation we would first set $\displaystyle f(x)=0$ and attempt to factor:  

$\displaystyle x^3+7\sqrt{A}x^2+2Bx=0$

$\displaystyle x(x^2+7\sqrt{A}x+2B) = 0$

The trivial solution is $\displaystyle x=0$, the other possible non-zero solutions will depend on the quadratic factor. A quadratic equation may have either 2 complex roots, 2 real roots, or 1 real root. The conditions determined by a the radical term in the quadratic equation...the discriminate: 

For a quadratic equation, or quadratic factor, of the form $\displaystyle ax^2+bx+c = 0$ the discriminate is defined as: 

$\displaystyle D = b^2-4ca$

For the quadratic factor in this problem we have, $\displaystyle a = 1 \: \: \: \:b=7\sqrt{A}\: \: \: \:c=2B$

The discriminate in this case is then,  

$\displaystyle D=49A-4(2B)$

$\displaystyle D=49A-8B$

Now let's recall how the discriminate determines the number and types of roots 

$\displaystyle D>0 \: \: \: \: \rightarrow\: \: \: \:\; two \: \:real \: \:roots$

$\displaystyle D=0 \: \: \: \: \rightarrow\: \: \: \:\: \: \: one \: \: \: \: real \: \:root$

$\displaystyle D< 0 \: \: \: \: \rightarrow\: \: \: \:\:two \: \:complex \: \:roots$

a) Two Real Roots 

When the discriminate is greater than zero the function has 2 real-valued roots. 

$\displaystyle D>0\: \: \: \: \rightarrow \: \: \: \:49A-8B>0 \: \: \: \: \rightarrow\: \: \: \:A>\frac{8}{49}B$

b) One Real Root 

When the discriminate is zero, the function has 1 real valued root: 

$\displaystyle D = 0$

Apply to the discriminate and solve for $\displaystyle A$. 

$\displaystyle 49A=8B\: \: \: \: \rightarrow \: \: \: \:A=\frac{8}{49}B$

c) Two Complex Roots 

Similarly, for two complex roots solve $\displaystyle D< 0$ , we have: 

$\displaystyle A< \frac{8}{49}B$

 Summary 

a)  $\displaystyle A>\frac{8}{49}B$

b)  $\displaystyle A=\frac{8}{49}B$

c) $\displaystyle A< \frac{8}{49}B$

$\displaystyle 1-Ax^2=Bx+C$                                         (1)

 $\displaystyle B^2 = 4(1-C)A$                                            (2)

The first step is conceptualize. What kind of equation is equation (1)? If we rearrange it's clearly a quadratic equation. 

$\displaystyle 1-Ax^2=Bx+C$

 $\displaystyle Ax^2 -Bx+(1-C)=0$                            (3)

To determine the number of solutions use the discriminate: 

______________________________________________________________

Reminder

Recall that for a quadratic equation  $\displaystyle ax^2+bx+c=0$ the general formula for the solution in terms of the constant coefficients is given by:  

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ca}}{2a}$

The quantity under the radical is known as the discriminate. If the discriminate is less than zero, there are two complex valued solution. In cases where the discriminate is zero, there is just one real solution $\displaystyle -\frac{b}{2a}$. If the discriminate is positive, then there are two real solutions. 

 $\displaystyle D = b^2 - 4ca$                                                        

 _____________________________________________________________

 The discriminate for equation (3) is written: 

$\displaystyle D = B^2 -4(1-C)A$

We are given that  $\displaystyle B^2 = 4(1-C)A$.  Therefore we have: 

$\displaystyle D = B^2 -4(1-C)A =0$ 

The discriminate is therefore zero, meaning there is only one real solution. 

The correct answer is $\displaystyle 21$. The discriminant is equal to $\displaystyle b^2-4ac$ portion of the quadratic formula. In this case, "$\displaystyle a$" corresponds to the coefficient of $\displaystyle 3x^2$, "$\displaystyle b$" corresponds to the coefficient of $\displaystyle -9x$, and "$\displaystyle c$" corresponds to $\displaystyle +5$. So, the answer is $\displaystyle (-9)^2-4(3)(5)$, which is equal to $\displaystyle 21$. 

Given the quadratic equation of $\displaystyle ax^{2}+bx+c=0$

$\displaystyle a=2, b=5, c=-4$

The formula for the discriminant is $\displaystyle b^{2}-4ac$ (remember this as a part of the quadratic formula?)

Plugging in values to the discriminant equation:

$\displaystyle 5^{2}-4(2)(-4)$

$\displaystyle 25+32=57$

So the discriminant is 57. What does that mean for our solutions? Since it is a positive number, we know that we will have 2 real solutions. So the answer is:

57, 2 real solutions

Using the quadratic formula, with $\displaystyle a = 1, b = 4, c = -17$:

$\displaystyle x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$\displaystyle x = \frac{-(4) \pm \sqrt{(4)^{2}-4 (1)(-17)}}{2 (1)}$

$\displaystyle x = \frac{-4 \pm \sqrt{16-(-68)}}{2 }$

$\displaystyle x = \frac{-4 \pm \sqrt{84}}{2 }$

$\displaystyle x = \frac{-4 \pm \sqrt{4} \cdot \sqrt{21}}{2 }$

$\displaystyle x = \frac{-4 \pm 2 \sqrt{21}}{2 }$

$\displaystyle x = -2 \pm \sqrt{21}$

Using the quadratic formula, with $\displaystyle a = 1, b = 4, c = 13$:

$\displaystyle x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$\displaystyle x = \frac{-(4) \pm \sqrt{(4)^{2}-4 (1)(13)}}{2 (1)}$

$\displaystyle x = \frac{-4 \pm \sqrt{16-52}}{2 }$

$\displaystyle x = \frac{-4 \pm \sqrt{-36}}{2 }$

$\displaystyle x = \frac{-4 \pm i \sqrt{36}}{2 }$

$\displaystyle x = \frac{-4 \pm 6 i }{2 }$

$\displaystyle x = -2 \pm 3 i$

Write in the form $\displaystyle (x-p)(x-q)=0$ where p and q are the roots.

Substitute in the roots:

$\displaystyle (x-2)(x-(-10))=0$

Simplify:

$\displaystyle (x-2)(x+10)=0$

Use FOIL and simplify to get

$\displaystyle x^2+8x-20=0$.

To find the roots of this equation, we need to find which values of $\displaystyle x$ make the polynomial equal zero; we do this by factoring. Factoring is a lot of "guess and check" work, but we can figure some things out. If our binomials are in the form $\displaystyle (ax+b)(cx+d)$, we know $\displaystyle a$ times $\displaystyle c$ will be $\displaystyle 6$ and $\displaystyle b$ times $\displaystyle d$ will be $\displaystyle -28$. With that in mind, we can factor our polynomial to 

$\displaystyle (2x-4)(3x+7)$

To find the roots, we need to find the $\displaystyle x$-values that make each of our binomials equal zero. For the first one it is $\displaystyle 2$, and for the second it is $\displaystyle -\frac{7}{3}$, so our roots are $\displaystyle 2, -\frac{7}{3}$.

1. Write the equation in the form $\displaystyle (x-a)(x-b)=0$ where $\displaystyle a$ and $\displaystyle b$ are the given roots.

$\displaystyle (x-4)(x-(-8))=0$

$\displaystyle (x-4)(x+8)=0$

2. Simplify using FOIL method.

$\displaystyle x^{2}+4x-32=0$