Let c be a real number, and consider the system of quadratic equations

y = 6x^{2 }- 9x + c,

y = 5x^{2} - 3x.

For which values of does this system have:

(a) Exactly one real solution (x,y)Solutions to the quadratics are pairs.

(b) More than one real solution?

(c) No real solutions?

Solutions to the quadratics are (x,y) pairs.

ZBRS7311 Jul 9, 2023

#1**0 **

\(\text{Hello ZBRS7311,} \\ \text{This question is about system of two quadratic equations, and it asks about which values of c does this system have: } \\ \text{1-Exactly one real solution} \\ \text{2-More than one real solution} \\ \text{3- No real solutions} \\ \text{The strategy is to make use of the discriminant of a quadratic equation. Namely the quantity: } \\ \Delta = b^2-4ac \space\space\space \text{Where } \Delta \space\text{is called the discriminant}\\ \text{But the problem is, which quadratic equation shall we apply this on? } \\ \text{Well, the quadratic equation that satisifies both "constraints" I.e. the intersection of both equations, I.e. equate both equations} \\ \text{Hence, setting both equations equal to each other, (eliminating y), yields:} \\ \implies 6x^2-9x+c=5x^2-3x \\ \iff x^2-6x+c=0 \\ \text{Notice: a=1,b=-6,c=c} \\ \text{Now for part (a) of the question, to have one real solution, the discriminant must be equal to zero} \\ \iff (-6)^2-4(1)(c)=0 \implies c=9 \\\text{Thus, this is the value of c such that the system has exactly one pair of (x,y) I.e. one solution}\\ \text{Next, for part (b), we want the system to have multiple solutions, this will be satisifed if and only if the discriminant is greater than 0} \\ \iff (-6)^2-4(1)(c)>0 \implies 36-4c> 0 \implies c<9 \\ \text{Thus, whenever c<9, the system will have more than one real solution} \\ \text{For the last part, (c), for the system to have no solutions, the discriminant must be less than zero} \\ (-6)^2-4(1)(c)<0 \implies 36-4c<0 \implies c>9 \text{Therefore, to have no real solutions, c>9} \)I hope this helps, and do not hesitate to ask further questions.

Guest Jul 10, 2023

#3**0 **

Hi,

No, a,b, and c are just letters for the "coefficients"

I.e.:

Any quadratic equation can be written as: \(ax^2+bx+c\)

And, in general, the discriminant is \(b^2-4ac\).

So for example (Try to solve these)

\(2x^2-4x+3\)

Here, a=2, b=-4, c=3

\(7x^2-\dfrac{13}{2}x-13=0\)

Here, a=7, b=-13/2, c=-13

Guest Jul 10, 2023