Quadratic Equation with Variables: x-intercepts at x = d, e

[Vertciel]

Hello everyone,

I am having some trouble with the following problem. I have arrived at an answer but it differs from the given one, and so, I would appreciate any help or hints.

My work is shown below.

Thanks.

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1. The parabola of equation \(\displaystyle y = ax^2 + bx + c\) has x-intercepts at A(d,0) and at B(e,0). Express a in terms of c, d, and e.

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My Work:

Since A and B are points on the parabola, substitute the coordinates into the parabola to get the following 2 equations:

\(\displaystyle 0 = ad^2 + bd + c\)

\(\displaystyle 0 = ae^2 + be + c\)

a does not need to be expressed in terms of b, so rearrange equations in terms of b and set both equations equal:

\(\displaystyle b = \frac{-ad^2 - c}{d}\)

\(\displaystyle b = \frac{-ae^2 -c}{e}\)

\(\displaystyle \frac{-ad^2 - c}{d} = \frac{-ae^2 -c}{e}\)

\(\displaystyle e(-ad^2 - c) = d(-ae^2 - c)\)

\(\displaystyle -aed^2 + ade^2 = -dc + ec\)

\(\displaystyle a(-ed^2 + de^2) = -dc + ec\)

\(\displaystyle a = \frac{-dc + ec}{-ed^2 + de^2}\)

The provided answer is: \(\displaystyle a = \frac{c}{de}\)

 

[galactus]

Re: Quadratic Equation with Variables

Try it this way. The algebra may be fooling you. A lot of times these things simplify down to an unrecognizable form.

\(\displaystyle ad^{2}+bd+e=0\)....[1]

\(\displaystyle ae^{2}+be+c=0\)...[2]

Solve [1] for \(\displaystyle b=\frac{-c-ad^{2}}{d}\)

Sub into [2] and we get, after some simplification:

\(\displaystyle ae^{2}-\frac{ce}{d}-ade=-c\)

\(\displaystyle ae^{2}-ade=-c+\frac{ce}{d}\)

\(\displaystyle a(e^{2}-de)=-c+\frac{ce}{d}\)

\(\displaystyle a=\frac{-c+\frac{ce}{d}}{e^{2}+ed}=\frac{c}{de}\)

 

[Mrspi]

Re: Quadratic Equation with Variables

Hello everyone,

I am having some trouble with the following problem. I have arrived at an answer but it differs from the given one, and so, I would appreciate any help or hints.

My work is shown below.

Thanks.

---

1. The parabola of equation \(\displaystyle y = ax^2 + bx + c\) has x-intercepts at A(d,0) and at B(e,0). Express a in terms of c, d, and e.

---

My Work:

Since A and B are points on the parabola, substitute the coordinates into the parabola to get the following 2 equations:

\(\displaystyle 0 = ad^2 + bd + c\)

\(\displaystyle 0 = ae^2 + be + c\)

a does not need to be expressed in terms of b, so rearrange equations in terms of b and set both equations equal:

\(\displaystyle b = \frac{-ad^2 - c}{d}\)

\(\displaystyle b = \frac{-ae^2 -c}{e}\)

\(\displaystyle \frac{-ad^2 - c}{d} = \frac{-ae^2 -c}{e}\)

\(\displaystyle e(-ad^2 - c) = d(-ae^2 - c)\)

\(\displaystyle -aed^2 + ade^2 = -dc + ec\)

\(\displaystyle a(-ed^2 + de^2) = -dc + ec\)

\(\displaystyle a = \frac{-dc + ec}{-ed^2 + de^2}\)

The provided answer is: \(\displaystyle a = \frac{c}{de}\)

Factor the numerator and denominator of the fraction....remove a common factor of c from both terms of the numerator, and a common factor of de from both terms of the denominator:

\(\displaystyle a = \frac{c (-d + e)}{de(-d + e}\)

And reduce the fraction....

 

[galactus]

Re: Quadratic Equation with Variables

See there, you had it all along. I reckon I should've paid closer attention.

 

[Vertciel]

Re: Quadratic Equation with Variables

Thank you for both of your replies!

I should have just gone one step further...