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The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two

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The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 01 Jun 2020, 03:08
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[GMAT math practice question]

The intersections of the graph \(y = x^2 + ax + 4\) and \(y = 3x + b\) are two points \(P\) and \(Q\). The length of \(PQ\) is \(4\sqrt{10}.\) What is the maximum of \(b\)?

A. \(5\)

B. \(6\)

C. \(7\)

D. \(8 \)

E. \(9\)

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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 01 Jun 2020, 18:38
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Suppose the difference in the x coordinates of P and that of Q = c

Now since the slope of the line \(y = 3x + b\) is 3, the difference in the y coordinates of P and that of Q = 3c

\(c^2 +(3c)^2 \)= \([4\sqrt{10}]^2\)

\(c^2= 16\)

c= +4 or -4

Hence the roots of the equation \(x^2 + ax + 4 = 3x+b\) are t and t+4

\(x^2+(a-3)x+4-b = 0\)

\(t+t+4 = 3-a\)

\(a= -2t-1\)

\(t^2+(-2t-1-3)t+4-b = 0\)

\(t^2+4t+b-4 = 0\)

Since t is a real number, D≥0

\(16-4(b-4)≥0\)

\(32≥ 4b\)

\(8≥ b\)


MathRevolution wrote:
[GMAT math practice question]

The intersections of the graph \(y = x^2 + ax + 4\) and \(y = 3x + b\) are two points \(P\) and \(Q\). The length of \(PQ\) is \(4\sqrt{10}.\) What is the maximum of \(b\)?

A. \(5\)

B. \(6\)

C. \(7\)

D. \(8 \)

E. \(9\)
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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 01 Jun 2020, 18:57
nick1816 wrote:
Hence the roots of the equation \(x^2 + ax + 4 = 3x+b\) are t and t+4

\(x^2+(a-3)x+4-b = 0\)

\(t+t+4 = 3-a\)

\(a= -2t-1\)

\(t^2+(-2t-1-3)t+4-b = 0\)

\(t^2+4t+b-4 = 0\)

Since t is a real number, D≥0

\(16-4(b-4)≥0\)

\(32≥ 4b\)

\(8≥ b\)



Sir can you elaborate this? If C value is±4 why we took roots as t and t+4.
Then what did you substitute in highlighted part. And why?

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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 01 Jun 2020, 19:42
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yashikaaggarwal
Attachment:
Untitled.png
Untitled.png [ 5.85 KiB | Viewed 438 times ]


\(PO=x_2 - x_1 = c\)
\(QO = y_2-y_1 = 3c\)

\(PO^2+QO^2 = c^2+9c^2 = PQ^2\)

c= +-4

Since \(x_2>x_1\)

\(x_2 = x_1+4 \)

I just assumed x_1=t and x_2 = t+4 in my previous solution.

As, Line and parabola intersect at t and t+4, both values must satisfies \(x^2 + ax + 4 = 3x+b\) or are the roots of this equation.


Sum of the roots of \(x^2+(a-3)x+4-b = 0\) is -(a-3)
t+t+4 = 3-a

If you still have any doubt, you can ask.
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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 01 Jun 2020, 19:52
nick1816 wrote:
yashikaaggarwal
Attachment:
Untitled.png


\(PO=x_2 - x_1 = c\)
\(QO = y_2-y_1 = 3c\)

\(PO^2+QO^2 = c^2+9c^2 = PQ^2\)

c= +-4

Since \(x_2>x_1\)

\(x_2 = x_1+4 \)

I just assumed x_1=t and x_2 = t+4 in my previous solution.

As, Line and parabola intersect at t and t+4, both values must satisfies \(x^2 + ax + 4 = 3x+b\) or are the roots of this equation.


Sum of the roots of \(x^2+(a-3)x+4-b = 0\) is -(a-3)
t+t+4 = 3-a

If you still have any doubt, you can ask.

Just last question why did you took slope 3 as the difference. Won't constant play any role?

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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 01 Jun 2020, 20:21
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constant term won't affect the slope of the line. Change in constant term just shifts the line upwards or downwards on coordinate plane.

y=3x+b; slope is 3 and y-intercept of the line = b


Slope of the line\(=3 = \frac{y_2-y_1}{x_2-x_1} \)

\(3 = \frac{y_2-y_1}{c}\)

\(y_2-y_1 = 3c \)


yashikaaggarwal wrote:
nick1816 wrote:
yashikaaggarwal
Attachment:
Untitled.png


\(PO=x_2 - x_1 = c\)
\(QO = y_2-y_1 = 3c\)

\(PO^2+QO^2 = c^2+9c^2 = PQ^2\)

c= +-4

Since \(x_2>x_1\)

\(x_2 = x_1+4 \)

I just assumed x_1=t and x_2 = t+4 in my previous solution.

As, Line and parabola intersect at t and t+4, both values must satisfies \(x^2 + ax + 4 = 3x+b\) or are the roots of this equation.


Sum of the roots of \(x^2+(a-3)x+4-b = 0\) is -(a-3)
t+t+4 = 3-a

If you still have any doubt, you can ask.

Just last question why did you took slope 3 as the difference. Won't constant play any role?

Posted from my mobile device
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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 01 Jun 2020, 20:28
nick1816 wrote:
constant term won't affect the slope of the line. Change in constant term just shifts the line upwards or downwards on coordinate plane.

y=3x+b; slope is 3 and y-intercept of the line = b


Slope of the line\(=3 = \frac{y_2-y_1}{x_2-x_1} \)

\(3 = \frac{y_2-y_1}{c}\)

\(y_2-y_1 = 3c \)


yashikaaggarwal wrote:
nick1816 wrote:
yashikaaggarwal
Attachment:
Untitled.png


\(PO=x_2 - x_1 = c\)
\(QO = y_2-y_1 = 3c\)

\(PO^2+QO^2 = c^2+9c^2 = PQ^2\)

c= +-4

Since \(x_2>x_1\)

\(x_2 = x_1+4 \)

I just assumed x_1=t and x_2 = t+4 in my previous solution.

As, Line and parabola intersect at t and t+4, both values must satisfies \(x^2 + ax + 4 = 3x+b\) or are the roots of this equation.


Sum of the roots of \(x^2+(a-3)x+4-b = 0\) is -(a-3)
t+t+4 = 3-a

If you still have any doubt, you can ask.

Just last question why did you took slope 3 as the difference. Won't constant play any role?

Posted from my mobile device

Got it. Thank you for being patient.
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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 02 Jun 2020, 08:01
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MathRevolution wrote:
[GMAT math practice question]

The intersections of the graph \(y = x^2 + ax + 4\) and \(y = 3x + b\) are two points \(P\) and \(Q\). The length of \(PQ\) is \(4\sqrt{10}.\) What is the maximum of \(b\)?

A. \(5\)

B. \(6\)

C. \(7\)

D. \(8 \)

E. \(9\)


Given:
1. The intersections of the graph \(y = x^2 + ax + 4\) and \(y = 3x + b\) are two points \(P\) and \(Q\).
2. The length of \(PQ\) is \(4\sqrt{10}.\)

Asked: What is the maximum of \(b\)?

\(y = x^2 + ax + 4 = 3x + b\)
\(x^2 + (a-3)x + (4-b) = 0\)

\(x_1 + x_2 = (3-a)\)
\(x_1*x_2 = (4-b)\)

\((x_1 - x_2)^2 = (3-a)^2 - 4(4-b) = 9 + a^2 - 6a - 16 + 4b = a^2 - 6a + 4b - 7\)

\((y_1 - y_2)^2 = (3(x_1-x_2))^2 = 9(a^2 - 6a - 7 + 4b)\)

\(PQ^2 = (x_1 - x_2)^2 + (y_1- y_2)^2 = 10(a^2 - 6a - 7 + 4b)\)

\(PQ = \sqrt{10(9 + a^2 - 6a - 15 + 4b)} = 4\sqrt{10}\)

\(a^2 - 6a - 7 + 4b = 16\)
\((a - 3)^2 = 32 - 4b >=0\)
b <= 8

IMO D
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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two  [#permalink]

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New post 03 Jun 2020, 03:18
=>


Assume the intersections are \(P(p, 3p+b)\) and \(Q(q, 3q+b).\)

Then the length of \(PQ\) is

\(\sqrt{(p-q)^2+(3p+b-3q-b)^2}\)

\(= \sqrt{(p-q)^2+(3p-3q)^2}\)

\(= \sqrt{(p-q)^2+9(p-q)^2}\)

\(= \sqrt{10(p-q)^2}\)

\(= \sqrt{10[(p+q)^2-4pq]}\)

\(= 4\sqrt{10} = \sqrt{160} \)

Then we have \(10[(p+q)^2-4pq]=160\) or \((p + q)^2 - 4pq = 16.\)

Since \(p\) and \(q\) are roots of the equation \(x^2 + ax + 4 = 3x + b\) or \(x^2 + (a - 3)x + 4 - b = 0,\) we have \((x - p)(x - q) = x^2 - (p + q)x + pq = x^2 + (a - 3)x + 4 - b, -(p + q) = a – 3\), or \(p + q = -a + 3\) and \(pq = 4 - b.\)

Then \((-a + 3)^2 - 4(4 - b) = 16, (-a + 3)^2 - 16 + 4b = 16\), or \(4b = -(-a + 3)^2 + 32.\)

We have \(b = -(\frac{1}{4})(-a + 3)^2 + 8\) and the maximum value of \(b = 8\) at \(a = 3.\)

Therefore, D is the answer.
Answer: D
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Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two   [#permalink] 03 Jun 2020, 03:18

The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two

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