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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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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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" DS Forum Moderator Joined: 19 Oct 2018 Posts: 1997 Location: India Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two [#permalink] ### Show Tags 01 Jun 2020, 18:38 1 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. $$9$$ PS Forum Moderator Joined: 18 Jan 2020 Posts: 1136 Location: India GPA: 4 Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two [#permalink] ### Show Tags 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? Posted from my mobile device DS Forum Moderator Joined: 19 Oct 2018 Posts: 1997 Location: India Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two [#permalink] ### Show Tags 01 Jun 2020, 19:42 1 1 yashikaaggarwal Attachment: Untitled.png [ 5.85 KiB | Viewed 449 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. PS Forum Moderator Joined: 18 Jan 2020 Posts: 1136 Location: India GPA: 4 Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two [#permalink] ### Show Tags 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? Posted from my mobile device DS Forum Moderator Joined: 19 Oct 2018 Posts: 1997 Location: India Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two [#permalink] ### Show Tags 01 Jun 2020, 20:21 1 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 PS Forum Moderator Joined: 18 Jan 2020 Posts: 1136 Location: India GPA: 4 Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two [#permalink] ### Show Tags 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. CEO Joined: 03 Jun 2019 Posts: 3228 Location: India GMAT 1: 690 Q50 V34 WE: Engineering (Transportation) Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two [#permalink] ### Show Tags 02 Jun 2020, 08:01 1 1 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 _________________ Kinshook Chaturvedi Email: kinshook.chaturvedi@gmail.com Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 9164 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The intersections of the graph y = x2 + ax + 4 and y = 3x + b are two [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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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