BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

If x^2+9+x^2-3=6, what is

Expert replies
by Max@Math Revolution » Tue Feb 25, 2020 1:21 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

[GMAT math practice question]
If \(\sqrt{x^2+9}+\sqrt{x^2-3}=6\) , what is \(\sqrt{x^2+9}-\sqrt{x^2-3}\) ?

A. 1
B. 2
C. 3
D. 4
E. 5
Join the discussion
Source: — Problem Solving |

Re: If x^2+9+x^2-3=6, what is

by Max@Math Revolution » Thu Feb 27, 2020 1:04 am
=>
\(\left(\sqrt{x^2+9}+\sqrt{x^2-3}\right)\left(\sqrt{x^2+9}-\sqrt{x^2-3}\right)\)
\(=\left(\sqrt{x^2+9}\right)\left(\sqrt{x^2+9}\right)-\left(\sqrt{x^2+9}\right)\left(\sqrt{x^2-3}\right)+\left(\sqrt{x^2-3}\right)\left(\sqrt{x^2+9}\right)-\left(\sqrt{x^2-3}\right)\left(\sqrt{x^2-3}\right)\) (by multiplying the 2 binomials)
= \(\left(x^2+9\right)-\left(x^2-3\right)\) (by simplifying)
= \(x^2+9-x^2-3\) (by multiplying -1 through the second bracket)
= 12

Then we have \(6\left(\sqrt{x^2+9}-\sqrt{x^2-3}\right)=12\) or \(\sqrt{x^2+9}-\sqrt{x^2-3}=2\) .

Therefore, B is the answer.
Answer: B
Join the discussion