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Points (x,y) in the rectangular coordinate plane such that
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03 Sep 2018, 05:09
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[GMATH course practice question] Points \((x,y)\) in the rectangular coordinate plane such that \(\,y = {x^2} + mx + \left( {8  m} \right)\,\) are presented in the graph shown, where \(m\) is constant. What is the value of \(k+p\) ? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5
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Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our highlevel "quant" preparation starts here: https://gmath.net



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Re: Points (x,y) in the rectangular coordinate plane such that
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03 Sep 2018, 08:19
fskilnik wrote: [GMATH course practice question]
Points \((x,y)\) in the rectangular coordinate plane such that \(\,y = {x^2} + mx + \left( {8  m} \right)\,\) are presented in the graph shown, where \(m\) is constant. What is the value of \(k+p\) ?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 I am sorry no one contributed (yet), but I hope you all enjoy my solution! \(? = k + p\) \(y = {x^2} + mx + \left( {8  m} \right)\,\,\,\,\,\,,\,\,\,m\,\,{\text{cte}}\,\,\,\,\,\left( * \right)\) \(k = {x_{{\text{vertex}}}}\mathop = \limits^{\left( \odot \right)} \,\,  \frac{m}{{2 \cdot 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,m =  2k\,\,\,\left( {**} \right)\) \(\left( {k,0} \right)\,\,\, \in \,\,\,{\text{graph}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,0 = \,\,{k^2} + mk + \left( {8  m} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,\,\,0 =  {k^2} + 2k + 8\) \(0 =  {k^2} + 2k + 8\,\,\,\,\mathop \Rightarrow \limits^{{\text{Sum}}\, = \,2\,\,,\,\,\,{\text{Product}}\, = \,\,  8} \,\,\,\,k =  2\,\,\,{\text{or}}\,\,\,k = 4\,\,\,\mathop \Rightarrow \limits^{k\, < \,\,0\,\,\,\left( {f{\text{igure}}} \right)} \,\,\,\,k =  2\) \(\left( {0,p} \right)\,\,\, \in \,\,\,{\text{graph}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,p = \,\,{0^2} + m \cdot 0 + \left( {8  m} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,\,\,p = 8 + 2k = 8 + 2\left( {  2} \right) = 4\) \(? = k + p =  2 + 4 = \boxed2\) Reminder: \(\left( \odot \right)\,\,y = a{x^2} + bx + c\,\,\,\,\left( {a \ne 0} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\,{x_{{\text{vertex}}}} =  \frac{b}{{2a}}\) The above follows the notations and rationale taught in the GMATH method. Regards, fskilnik.
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Re: Points (x,y) in the rectangular coordinate plane such that
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03 Sep 2018, 10:14
let's say that m = 4
y = sqr(x)+4x+2 =sqr (x+2)
when y = 0 x=2 (k)
when x=0 y =4 (p)
k+p =42 = 2 Anwser B



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Re: Points (x,y) in the rectangular coordinate plane such that
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03 Sep 2018, 10:15
let's say that m = 4
y = sqr(x)+4x+2 =sqr (x+2)
when y = 0 x=2 (k)
when x=0 y =4 (p)
k+p =42 = 2 Anwser B



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Points (x,y) in the rectangular coordinate plane such that
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03 Sep 2018, 10:57
amascarenhas wrote: let's say that m = 4
y = sqr(x)+4x+2 =sqr (x+2)
when y = 0 x=2 (k)
when x=0 y =4 (p)
k+p =42 = 2 Anwser B Hi, amascarenhas! I am sure you have meant x^2 in the place of sqr(x) and (x+2)^2 in the place of sqr(x+2) ... The fact that m = 4 makes the expression y = (x+2)^2 DOES guarantee that the x_vertex of the parabola is 2 (=k), exactly the opposite of half the value of this value of m, what we know (a posteriori) is the correct relationship between these two. More than that, y_vertex will be zero (for x=2), exactly as the figure suggests. (The vertex of the parabola must be on the xaxis.) Thank you for your contribution! Regards, fskilnik. P.S.: let me be more clear. If you had chosen (say) m=2, you would find the x_vertex = 1 (negative, fine) but the y_vertex would be 5, therefore the vertex would be point (1,5) , that violates the question stem (figure). In other words, m=4 is not "free" as we wished, therefore exploring "any" particular case would not be possible here.
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Re: Points (x,y) in the rectangular coordinate plane such that
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03 Sep 2018, 11:29
Let me present a second possible solution for the interested readers: \(? = k + p\) \(y = {x^2} + mx + \left( {8  m} \right)\,\,\,\,\,\,,\,\,\,m\,\,{\text{cte}}\,\,\,\,\,\left( * \right)\) \(0\mathop = \limits^{{\text{single}}\,\,{\text{root}}} \Delta = {m^2}  4\left( {8  m} \right)\,\,\,\,\, \Rightarrow \,\,\,\,{m^2} + 4m  32 = 0\,\,\,\mathop \Rightarrow \limits^{{\text{Sum}}\, = \,  4\,\,,\,\,\,{\text{Product}}\, = \,\,  32} \,\,\,\,m =  8\,\,\,{\text{or}}\,\,\,m = 4\) \(0\mathop > \limits^{{\text{figure}}} k = {x_{{\text{vertex}}}}\mathop = \limits^{\left( \odot \right)} \,\,  \frac{m}{{2 \cdot 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,m > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,m = 4\,\,\,\,\,\mathop \Rightarrow \limits^{k =  m/2} \,\,\,\,\,k =  2\) \(\left. \begin{gathered} \left( {0,p} \right)\,\,\, \in \,\,\,{\text{graph}} \hfill \\ m = 4 \hfill \\ \end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,p = \,\,{0^2} + 4 \cdot 0 + \left( {8  4} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,p = 4\) \(? = k + p =  2 + 4 = \boxed2\) Reminder: \(\left( \odot \right)\,\,y = a{x^2} + bx + c\,\,\,\,\left( {a \ne 0} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\,{x_{{\text{vertex}}}} =  \frac{b}{{2a}}\) The above follows the notations and rationale taught in the GMATH method. Regards, fskilnik.
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Re: Points (x,y) in the rectangular coordinate plane such that
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03 Sep 2018, 17:18
fskilnik wrote: [GMATH course practice question]
Points \((x,y)\) in the rectangular coordinate plane such that \(\,y = {x^2} + mx + \left( {8  m} \right)\,\) are presented in the graph shown, where \(m\) is constant. What is the value of \(k+p\) ?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Since the graph intercepts the xaxis to the left of the yaxis and has only one solution, its graph is as follows: y = (xk)², where k<0. The equation above has an xintercept at (k, 0), as shown in the figure above. The answer choices represent the sum of k (the xintercept) and p (the yintercept). Since the answer choices are very small, test negative values for k that are close to 0. Case 1: k=1 Plugging k=1 into y = (xk)², we get: y = (x(1))² = (x+1)² = x² + 2x + 1. Since the prompt indicates that y = x² + mx + (8m), Case 1 implies that m=2 and that 8m = 1. The equations in red contradict each other, implying that Case 1 is not viable. Case 2: k=2 Plugging k=2 into y = (xk)², we get: y = (x(2))² = (x+2)² = x² + 4x + 4. Since the prompt indicates that y = x² + mx + (8m), Case 2 implies that m=4 and that 8m = 4. The equations in green are both viable for m=4, implying that Case 2 is correct and that the equation of the graph is y = (x+2)². Since y=4 when x=0, the value of p = 4. Thus, k+p = 2 + 4 = 2.
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Points (x,y) in the rectangular coordinate plane such that
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Updated on: 05 Sep 2018, 19:10
Hi,
When we look at the equation y=x^2+mx+(8−m) and at the graph we see two things:
1) The equation has only one root. 2) At x=0 , it has a positive intercept i.e. (8m) > 0
Now When we have only one root ( which are also equal and real ) then according to quadratic equations the determinant shall be 0 (B^24AC=0) so solve for m^2  4(8m) = 0
We shall get two values 8, 4 but these should support the 2nd point above. Hence we conclude that m=4. On placing the value of m in the equation we get x^2 + 4x + 4 which is (x +2 )^2 and hence x=2
k is the value of x when y=0 so in other terms k is the root of the equation hence k=2 p is the y intercept when x=0 i.e. p = ( 8 m ) = 4 Hence 4+(2) = 2 (Answer)
Originally posted by anse on 05 Sep 2018, 18:32.
Last edited by anse on 05 Sep 2018, 19:10, edited 1 time in total.



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Re: Points (x,y) in the rectangular coordinate plane such that
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05 Sep 2018, 18:59
amascarenhas wrote: let's say that m = 4
y = sqr(x)+4x+2 =sqr (x+2)
when y = 0 x=2 (k)
when x=0 y =4 (p)
k+p =42 = 2 Anwser B Hi, Thank you for the explanation. How can we assume a value for m?????




Re: Points (x,y) in the rectangular coordinate plane such that
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05 Sep 2018, 18:59






