If the graph of $y = x^2 + ax + b$ passes through the points $(m, 0)$ and $(n, 0),$ where $m < n,$ what is the

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Fri Jun 19, 2020 2:33 am

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If the graph of $y = x^2 + ax + b$ passes through the points $(m, 0)$ and $(n, 0),$ where $m < n,$ what is the value of $n – m?$

(1) $4b = a^2 – 4$

(2) $b = 0$

[spoiler]OA=A[/spoiler]

Source: Manhattan GMAT

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Source: Beat The GMAT — Data Sufficiency | Fri Jun 19, 2020 2:33 am

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

Re: If the graph of $y = x^2 + ax + b$ passes through the points $(m, 0)$ and $(n, 0),$ where $m < n,$ what is t

by deloitte247 » Sun Jun 21, 2020 9:15 am

The root of the equation x^2 + ax + b is given to be m and n, where m < n

Target question => What is the value of n - m?
i.e find the difference of the 2 roots

$$For\ f\left(x\right)=Ax^2-Bx+c$$
$$difference\ of\ roots\ =>x_1-x_2=\sqrt{\frac{b^2-4ac}{a^2}}$$
$$where\ x_1>x_2\ or\ x_2<x_1$$
Expressing the given quadratic equation in terms of the above expression
$$n-m=\sqrt{\frac{a^2-4\left(1\right)\left(b\right)}{1^2}}$$
$$n-m=\sqrt{a^2-4b}$$

$$Statement1=>\ 4b=a^2-4$$
$$a^2-4=4b$$
$$a^2-4-4b=0$$
$$a^2-4b=4............eqn\ 1$$
Substituting value of a^2 - 4b into the expression from the question stem
$$n-m=\sqrt{a^2-4b}\ where\ a^2-4b=4$$
$$n-m=\sqrt{4}$$
$$n-m=2$$
$$statement\ 1\ is\ SUFFICIENT$$

$$Statement\ 2\ =>b=0$$
$$n-m=\sqrt{a^2-4b}where\ b=0$$
$$n-m=\sqrt{a^2-0}$$
The value of a is unknown, so n - m cannot be evaluated. Statement 2 is NOT SUFFICIENT

Since only statement 1 is SUFFICIENT
Answer = A

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