-2x^2-7x-2=0?

Question

Compute the value of the discriminant and give the number of real solutions to the quadratic equation
-2x^2 -7x -2 = 0

I have used the quadratic equation and get
-(2x^2+7x+2)=0
which is the same thing
where from here?

Answer 1

Remember the quadratic equation:
-b +/- (plus or minus) the square root of {b^2 - 4(a)(c)} all divided by 2a

Using 2x^2 + 7x + 2 = 0, you find the terms a, b, and c with this equation:
ax^2 + bx + c = 0
So, a = 2, b = 7, and c = 2

Plug the numbers into the quadratic equation and solve for the answer.

It should look something like
-7 +/- the square root of {7^2 - 4(2)(2)} all divided by 2(2)

Using the rule of PEMDAS, you work with the parantheses first, so 4(2)(2) = 4 * 4 = 16 in the square root
For the denominator, 2(2), you get 4.

Next is the exponent, so 7^2 = 49.

Now, you must simplify for the square root, and in it, you get 49 - 16 = 33
However, only 1, 3 and 11 make up 33, so there is no perfect square. Leave it in radical form (square root of 33).

Now you're left with:
7 +/- the square root of 33 all divided by 4

You can't add or subtract 7 and the square root of 33 while it's in its radical form, so you keep it that way.
And, since 7 and the square root of 33 are not divisable by 4 (in radical form), 7 +/- the square root of 33 all divided by 4 is your answer.

Answer 2

The 'standard form' ax²+bx+c = 0 quadratic has roots at
x = (-b ± √(b²-4ac))/(2a) so do the substitutions and the arithmetic.


Doug

Answer 3

ax^2+bx+c=0
a=-2
b=-7
c=-2

D=b^2-4ac --->
D=(-7)(-7)-4(-2)(-2)=49-16=33

Solutions :( -(-7)+- root(33) )/ ((2) (-2) = -7/4 +- root(33)/4