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Difference between revisions of "1955 AHSME Problems/Problem 19"
 
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Let the roots of the equation be <math>x</math> and <math>y</math>. Therefore, we can set up a system of equation:
 
Let the roots of the equation be <math>x</math> and <math>y</math>. Therefore, we can set up a system of equation:
 
<cmath>x+y=6</cmath><cmath>|x-y|=8</cmath>Therefore, we get <math>x=7</math> and <math>y=-1</math>. So, <math>(x-7)(x+1)=\boxed{x^2-6x-7}</math>
 
<cmath>x+y=6</cmath><cmath>|x-y|=8</cmath>Therefore, we get <math>x=7</math> and <math>y=-1</math>. So, <math>(x-7)(x+1)=\boxed{x^2-6x-7}</math>
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- kante314 -
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:11, 12 July 2021

Problem 19

Two numbers whose sum is $6$ and the absolute value of whose difference is $8$ are roots of the equation:

$\textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0$

Solution

The first two hints can be expressed as the following system of equations: \[\begin{cases} (1) & a + b = 6 \\ (2) & a - b = 8 \end{cases}\] From this, we can clearly see that $a = 7$, and that $b = -1$.

Since quadratic equations can generally be expressed in the form of $(x - a)(x - b) = 0$, where a and b are roots, the correct quadratic, once factored, would look like $(x - 7)(x + 1) = 0$

Expanding the above equation gets us $\textbf{(B)} x^2 - 6x - 7 = 0$

Solution 2

Let the roots of the equation be $x$ and $y$. Therefore, we can set up a system of equation: \[x+y=6\]\[|x-y|=8\]Therefore, we get $x=7$ and $y=-1$. So, $(x-7)(x+1)=\boxed{x^2-6x-7}$

- kante314 -

See Also

1955 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AHSME Problems and Solutions


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