SAT test 8 section question 10 and 12

# The lengths of the sides of a rectangle are a and b…

The lengths of the sides of a rectangle are a and b, where a > b The sum of the lengths of the two shorter sides and one of the longer sides of the rectangle is 36. What value of a maximizes the area of the rectangle?

A .9

B. 12

C. 18

D. 24

The answer is C. I suspect there is an easier way to solve than completing the square and finding the vertex of the resulting quadratic function. What is your most direct, easy to understand solution to this calculator-allowed question?

# Test 10 Question 23

Test 10 Question 23

# Can you explain a more direct way to solve College Board Official Practice Test 9, Math Section 4 #19, than the College Board’s explanation?

Can you explain a more direct way to solve College Board Official Practice Test 9, Math Section 4 #19, than the College Board’s explanation? I seem to remember something about making a chart to solve mixture problems. Would that work here?

# In question #10 of the backsolving chapter…

In question #10 of the backsolving chapter : in the xy plane, a line containing the points (a, a^3) and (10,40) passes through the origin. Which of the following could be the value of a?

I found the explanation in the answer key to be too time-consuming if I were to solve the equation with backsolving. Can you explain how to solve this question algebraically instead?

# Test 7 Section 4 Question 6

Hi Mike…SAT 7, Section 4, Q6: I now see the shortcut here (that both sides of the equation are perfect squares,) but if I did expand and FOIL the left side, wouldn’t I still get the correct “a” values even though it takes longer? I can’t get it to work !! Can you please show the alternate path math steps? Or is recognizing the perfect squares the ONLY way to solve this one ? Thanks!

# SAT 8, Section 3, Number 7

Hi Mike, I’m asking about SAT 8, Section 3, Number 7. Is it always best to immediately plug in answer choices on questions like this? For the algebraic practice, I rewrote the equation as a quadratic and solved for (x=5) and (x= -1). Then sub’d each value back into the given equation to find that only (x=5) worked. OK…but what a time-killer at #7 out of 20. Any other solution path to consider? Thanks!

# College Board Test 4 Section 3 #9

College Board Test 4 Section 3 #9:

____

√x-a = x-4

If a=2 what is the solution set of the preceding equation?

A. {3, 6}

B. {2}

C. {3}

D. {6}

Is there another way to solve this quickly besides plugging in the answer choices?

# Test 8 Section 4 #8

Test 8 Section 4 #8

# Test 5 Section 4 Number 23

Test 5 Section 4 Number 23

# Test 3 Section 4 #24

Test 3 Section 4 #24

# Test 7 Section 3 #15

Test 7 Section 3 #15

# Could you help me with question 9 on page 38 please?

Could you help me with question 9 on page 38 please? I don’t understand how to solve it even using back-solving

# How would you use backsolving to solve practise test 2, section 4 question 29?

Hello, I know you’ve already solved practise test 2, section 4 question 29 (by either using your graphic calculator or by looking at the equation of a parabola) but how would you use backsolving? Lets say I try in option C and im getting y as 3 (which means my equations do NOT have 2 real solutions), how do I know whether to try out option B or D next?

Thank you so much!

# For practise test 2, section 3 Q6, how exactly could I use backsolving to solve this?

Hi Mike! For practise test 2, section 3 Q6, how exactly could I use backsolving to solve this? Lets say I start with C and I plug in 8. My gradient of line l is 2/5. If I plug in p as 8, I’m getting gradient of line k as 4/8. Do I now compare the fractions? How do I know if I should try plugging in a bigger or smaller number to get closer towards 2/5 (initial gradient)?

Thanks!