<p>Hi,
I need help on this particular question below:</p>
<p>(x-8)(x-k)=x^2 - 5kx + m
in the equation above, k and m are constants. If the equation is true for all values of x, what is the value of m? </p>
<p>(A) 8
(B) 16
(C) 24
(D) 32
(E) 40</p>
<p>the answer is "B", but how do you solve it to find 16?</p>
<p>(x-8)(x-k)= x^2-kx-8x+8k= x^2-(k+8)x+8k=x^2-5kx+m
k+8=5k => 8=4k => k=2
8k=m => m=16</p>
<p>Quickest solution: The left hand side is 0 when x=8 and x=k. So 8k=m and 8+k=5k. From the second equation 8=4k so that k=2. So m=8k=8(2)=16.</p>
<p>Note: In this solution I have used the fact that in the quadratic equation x^2+bx+c=0, the product of the roots is c and the sum of the roots is -b.</p>
<p>^ dr. steve… i didn’t get it ._.’’</p>
<p>The roots of the expression on the LHS are 8 and k. Their sum is 8+k, and their product is 8k.</p>
<p>Now looking at the RHS, the sum of the roots is -b/a = 5k, and the product is c/a = m. </p>
<p>The LHS and RHS must have the same roots (since they’re the same expression) so we equate and get</p>
<p>8+k = 5k
8k = m</p>
<p>Solving the first equation yields k = 2. Substitute that to get m = 8*2 = 16.</p>
<p>Alternatively, you can expand the LHS to get the two equations described above.</p>
<p>Slightly different approach. The given quadratic equation holds true for all values of x, meaning x could be 0, 1, -4, a, k, any number. </p>
<p>If we replace x with 0, the relationship becomes:
(0-8)(0-k) = 0^2 - 5k(0) + m
or 8k = m
or k = m/8</p>
<p>Then set x to be equal to k(Why is that a good choice?):
(8-8)(8-k) = 8^2 - 5k(8) + m
or 0 = 64-40k + m. </p>
<p>Substitute k = m/8, which yields:
0 = 64 - 40(m/8) + m
or 0 = 64-5m + m
or 4m = 64 or m = 16.</p>
<p>You can choose any of the above methods.</p>
<p>@SATQuantum Looks like you set x=8, not k.</p>
<p>@2200 Do you understand rspence’s clarifications or do you need more explanation?</p>
<p>@DrSteve Thanks, yes it should be x = 8 in the second step:</p>
<p>Then set x to be equal to 8(Why is that a good choice?):
(8-8)(8-k) = 8^2 - 5k(8) + m
or 0 = 64-40k + m.</p>
<p>thanks a lot everyone for your contributions, you’ve really helped…</p>
<p>If you all don’t mind, I’ve got two other similar questions and I want to know the fastest way of solving them</p>
<p>Question 1)
Ray and Jane live 150 miles apart. Each drives
toward the other’s house along a straight road
connecting the two, Ray at a constant rate of 30
miles per hour and Jane at a constant rate of 50
miles per hour. If Ray and Jane leave their houses
at the same time, how many miles are they from
Ray’s house when they meet?
(A) 40
(B)51 1/2
(C)56 1/4
(D) 75
(E)93 1/4</p>
<p>the answer is C …</p>
<p>Question 2) student response</p>
<p>Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?</p>
<p>yes dr. steve… I got it thanks…
thanks rspence
But can u guys give me another example of it
another example of this one’s type?
“(x-8)(x-k)=x^2 - 5kx + m
in the equation above, k and m are constants. If the equation is true for all values of x, what is the value of m?” </p>
<p>Would help a lot… :)</p>
<p>As for question 2:
“Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?” </p>
<p>I got the answer 18…
Is it correct?</p>
<p>Couldn’t do question 1 tho… are these from cb book? i think question 2 is… But i would like to know where u got the others from…</p>
<p>(1) Combined they are travelling 80 mph. Using distance = rate * time we see that they meet in t=d/r=150/80=1.875 hours. Finally, d=r*t=(30)(1.875)=56.25, choice (C).</p>
<p>(2) Method 1: We can use the harmonic mean formula (aka Xiggi’s formula): avg rate = 2(rate1)(rate2)/(rate1+rate2) = 2<em>45</em>30(45+30) = 36. So round trip distance = (36)(1) = 36. But we only want half this distance, so the answer is 18.</p>
<p>Method 2: d/45+d/30=1. So 30d+45d=45<em>30, or 75d=45</em>30. So d=45*30/75=18.</p>
<p>In question 1, can we just combine the 2 speeds like that? Seems weird…
hahahahaha… I dunno, thanks… i’ll make sure i remember that afterwards…
U have another example for that very first question that confused me?
thanks. :D</p>
<p>thanks a lot dr steve your methods are faster and easier to use :)…
thanks 2200 for your contributions; I got these from the updated BB…</p>
<p>By the way I need help on this question from student response:</p>
<p>Let the function f(x)=x^2 + 18. If m is a positive number such that f(2m)=2f(m), what is the value of m ?</p>
<p>f(2m) = (2m)^2 + 18 = 4(m^2) + 18
2f(m) = 2(m^2 + 18) = 2(m^2) + 36
Set them equal to each other:
4(m^2) + 18 = 2(m^2) + 36
2(m^2) = 18
m^2 = 9
m= 3</p>
<p>Also for the first question posted with the LHS and RHS, why do you use the sum 8+k? I still don’t understand this one… I’m kinda slow</p>
<p>All I’m doing is finding the sum & product of the roots of the LHS. I do the same for the RHS, then equate them.</p>
<p>@2200andbeyondXD</p>
<p>Another example of a similar question. This is from the PSAT. </p>
<p>x^2 + kx + 4 = (x+c)^2 </p>
<p>In the equation above, k and c are positive constants.
If the equation is true for all values of x, what is the value of k?
A) 2 B) 4 C) 8 D) 16 E) 32</p>
<p>the answer is B ?</p>
<p>Cheerio SATQuantum! Thanks… 
Yeah, I got it B) too… but i find this one waay easier…
The other one is complicated for some reason… :/</p>
<p>@2200andbeyondXD</p>
<p>Here is another one, but there is a caveat, this question is from the GMAT and I am a little bit on the fence as to how much relevant it is for the SAT. However, it is written by ETS(writers of SAT). </p>
<p>Question:
If (t-8) is a factor of t² - kt - 48, then k=</p>
<p>A) -6
B) -2
C) 2
D) 6
E) 14</p>
<p>Let me see if I can find some other official SAT questions on the same concept.</p>