Difference between revisions of "1963 AHSME Problems/Problem 4"
Rockmanex3 (talk | contribs) (Solution to Problem 4) |
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| − | == Problem | + | == Problem == |
For what value(s) of <math>k</math> does the pair of equations <math>y=x^2</math> and <math>y=3x+k</math> have two identical solutions? | For what value(s) of <math>k</math> does the pair of equations <math>y=x^2</math> and <math>y=3x+k</math> have two identical solutions? | ||
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\textbf{(C)}\ \frac{9}{4}\qquad | \textbf{(C)}\ \frac{9}{4}\qquad | ||
\textbf{(D)}\ -\frac{9}{4}\qquad | \textbf{(D)}\ -\frac{9}{4}\qquad | ||
| − | \textbf{(E)}\ \pm\frac{9}{4} </math> | + | \textbf{(E)}\ \pm\frac{9}{4} </math> |
==Solution== | ==Solution== | ||
Latest revision as of 16:35, 2 June 2018
Problem
For what value(s) of 
does the pair of equations 
and 
have two identical solutions?
Solution
If the system of equations has two identical solutions, then only one 
pair will satisfy both equations.
Substitute 
in one equation into another equation.
![\[x^2 = 3x + k\]](http://latex.artofproblemsolving.com/d/0/f/d0f3170cb91d66fc730555e1809840ce353d9a34.png)
![\[x^2 - 3x = k\]](http://latex.artofproblemsolving.com/f/c/7/fc75276a8639cfc52774637ac90d74805a13dc3a.png)
Complete the square to get
![\[x^2 - 3x + \frac{9}{4} = k + \frac{9}{4}\]](http://latex.artofproblemsolving.com/e/2/4/e24c979807a27bd5c7bc0b0419d6ecc5992fb391.png)
![\[(x - \frac{3}{2})^2 = k + \frac{9}{4}\]](http://latex.artofproblemsolving.com/7/f/f/7ffe12232121f6ed1ef57fa24112f49cd16d24f6.png)
In order for the equation to have one solution, the right side must be 
, so 
, which is answer choice 
.
See Also
| 1963 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
