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# Help ASAP

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

y = 6x- 9x + c,

y = 5x2 - 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.

Jul 9, 2023

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

Jul 10, 2023
#2
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Thanks for your help. I do have one question. would a and b = x,y?

ZBRS7311  Jul 10, 2023
#3
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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