Quadratics (1 Viewer)

Lukybear · Jul 26, 2009

Find the value of K that will make equation perfect square.

(3k-2)x^2 - 8x + (9k+7)

Solving discriminate = 0 will give
k=1 or -10/9

I just want to know, why -10/9 is discounted (as stated in the answer) and which step discount it?

lychnobity · Jul 26, 2009

Lukybear said:
Find the value of K that will make equation perfect square.

(3k-2)x^2 - 8x + (9k+7)

Solving discriminate = 0 will give
k=1 or -10/9

I just want to know, why -10/9 is discounted (as stated in the answer) and which step discount it?

To be a perfect square, the constant term must be positive.

eg (x±2)² = x² ± 4x + 4

For the equation (3k-2)x² - 8x + (9k+7), 9k +7 must be positive, therefore more than zero (as if 9k +7 equalled zero, it would be impossible for the original eqn to be a perfect square)

ie 9k+7 > 0 ...1

But if you sub in k = -10/9 to eqn 1, then LHS = -3, which is <0 . Therefore k=-10/9 must not be counted as an answer.

Lukybear · Jul 26, 2009

Thanks lychnobity! I see

What about this question; show that the roots of the simultaneous equation x^2 + y^2 = 1 , 2ax + 2by + c =0 are real, provided that 4(a^2 + b^2 >= c^2.

And also: If (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 ; prove that a=b=c. Then, show that (x-b)(x-c)+(x-c)(x-a)+ (x-a)(x-b) = 0 cannot have equal roots unless a= b =c

kurt.physics · Jul 27, 2009

Lukybear said:
If $(a-b)^2 + (b-c)^2 + (c-a)^2 = 0$ ; prove that $a=b=c$ . Then, show that $(x-b)(x-c)+(x-c)(x-a)+ (x-a)(x-b) = 0$ cannot have equal roots unless $a= b =c$

Consider the first equation:

$(a-b)^2 + (b-c)^2 + (c-a)^2 = 0$

Then,

$(a-b)^2 \geq 0$

$(b-c)^2 \geq 0$

$(c-a)^2 \geq 0$

(This is because they are all squares)

So it can only equal zero when

$(a-b)^2 = 0$

$(b-c)^2 = 0$

$(c-a)^2 = 0$

ie when

$a= b$

$b= c$

$c= a$

Combining these three statements,

$\therefore a = b = c$

Consider the second equation

$(x-b)(x-c)+(x-c)(x-a)+ (x-a)(x-b) = 0$

$\left(x^2 - (b + c)x + bc\right) + \left(x^2 - (c + a)x + ca\right) + \left(x^2 - (a + b) + ab\right) = 0$

$x^2 - (b + c)x + bc + x^2 - (c + a)x + ca + x^2 - (a + b) + ab = 0$

$3x^2 -2x(a + b + c) + (ab + bc + ca) = 0$

If a quadratic has equal roots, then $\triangle = 0$

$\therefore 4(a + b + c)^2 - 12(ab + bc + ca) = 0$

$(a + b + c)^2 - 3(ab + bc + ca) = 0$

$a^2 + b^2 + c^2 + 2(ab + bc + ca) - 3(ab + bc + ca) = 0$

$a^2 + b^2 + c^2 - (ab + bc + ca) = 0$

$2a^2 + 2b^2 + 2c^2 - 2(ab + bc + ca) = 0 \times 2$

$a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 = 0$

$(a-b)^2 + (b-c)^2 + (c-a)^2 = 0$

As we have already proven this only is true when

$a = b = c$

as required

kurt.physics · Jul 27, 2009

Lukybear said:
Thanks lychnobity! I see

What about this question; show that the roots of the simultaneous equation x^2 + y^2 = 1 , 2ax + 2by + c =0 are real, provided that 4(a^2 + b^2 >= c^2.

To solve this problem use the second equation (the linear one) and make x the subject of the equation (or whatever), you will find it is

$y = \frac{-(2ax + c)}{2b}$

Squaring gives

$y^ = \frac{\left(2ax + c\right)^2}{4b^2}$

plug this into the first equation

$x^2 + \frac{\left(2ax + c\right)^2}{4b^2} = 1$

$4b^2 \cdot x^2 + \left(2ax + c\right)^2 = 4b^2$

$4b^2 \cdot x^2 + 4a^2 \cdot x^2 + 4ac \cdot x + c^2 = 4b^2$

$\left(4b^2 + 4a^2\right)\cdot x^2 + 4ac \cdot x + (c^2 - 4b^2) = 0$

For real values of x, $\triangle \geq 0$

$16a^2c^2 - 4(4a^2 + 4b^2)(c^2 - 4b^2) \geq 0$

$16a^2c^2 - 16(a^2 + b^2)(c^2 - 4b^2) \geq 0$

$\therefore a^2c^2 - (a^2 + b^2)(c^2 - 4b^2) \geq 0$

$a^2c^2 - (a^2c^2 - 4a^2b^2 + b^2c^2 - 4b^4) \geq 0$

$4a^2b^2 - b^2c^2 + 4b^4 \geq 0$

$4a^2b^2 + 4b^4 \geq b^2c^2$

Now consider the condition they said was true, ie

$4a^2 + 4b^2 \geq c^2$

If we multiply both sides by $b^2$ then equality still remains true, so

$b^2(4a^2 + 4b^2) \geq b^2c^2$

$\therefore 4a^2b^2 + 4b^4 \geq b^2c^2$

as required.

Lukybear · Jul 27, 2009

If x and y are real and x+y=1, find the minimum value of 7x^2 - 4xy + 9y^2 and the values that x and y then assume

Lukybear · Jul 27, 2009

kurt?

kurt.physics · Jul 27, 2009

Lukybear said:
If x and y are real and x+y=1, find the minimum value of 7x^2 - 4xy + 9y^2 and the values that x and y then assume

What makes this problem hard is that it is in terms of 2 variables. But we are given that x + y = 1, so y = 1 - x. Substitute this into the equation

$7x^2 - 4x(1-x) + 9(1-x)^2 = 7x^2 - 4x + 4x^2 + 9 - 18x + 9x^2$

$= 7x^2 + 4x^2 + 9x^2 - 4x - 18x + 9$

$=20x^2 - 22x + 9$

The minimum value that this expression can have occurs at axis of symmetry (as it is a parabola). So the minimum x value is

$x = \frac{-b}{2a} = \frac{22}{40} = \frac{11}{20}$

And so the y value is

$y = 1 - \frac{11}{20} = \frac{9}{20}$

And then you just sub that value into the equation (or plug the x value back into the equation). I would do it but i dont have a calculator on me =(

Hope my solution is correct +)

Lukybear · Jul 27, 2009

Yes it is correct.

Thank you.

kurt.physics · Jul 27, 2009

Lukybear said:
Yes it is correct.

Thank you.

Your welcome. Good luck

Lukybear · Jul 27, 2009

Can I just inquire kurt, when you approached this questions, what sort of method did you use. Did you approach this graphically etc... because in all trufully, this dosent make any sense to me whatsoever in why equn 1 was subbed into equn 2...

kurt.physics · Jul 27, 2009

Lukybear said:
Can I just inquire kurt, when you approached this questions, what sort of method did you use. Did you approach this graphically etc... because in all trufully, this dosent make any sense to me whatsoever in why equn 1 was subbed into equn 2...

The last question you asked involved a bit more problem solving. I had no idea what to do. So when you start a problem, instead of jumping in and jumbling up equations to try some how push a solution out, think about ways to go about solving the problem.

Doing it graphically doesn't look to promising because its such an ugly expression, if it is even possible to graph it then it would take a lot of time and all you could get is an approximation. So we can just forget about graphing for this problem. This is a problem, and with mathematics problems we must always assume that there is a nice solution, that there is a logical way to arrive at a solution.

One problem solving technique is "Identify what makes the problem hard".

So look at the problem, what makes it difficult. Well we have this ugly expression. But the hardest part is that it is expressed in 2 variables - x and y. So what do we want to do? We want to get rid of the hard part. So we want to find a way to have this expression in just one variable, say x. If we get this then we can easily reduce this problem to finding the minimum of a simple function.

How do we do this? Another important part of problem solving is identifying important facts in the question, these are clues. Usually you will have to eventually use all the information given in a mathematics problem. So what is given

Well the thing that stands out at us is

x + y = 1

Why would they just drop this into the question. From this we get the idea to make x the subject ie y = 1 - x. So here we have it, a way to take out the uglyness =)

Lukybear · Jul 28, 2009

Yea ok, it perhaps will take a long time for me to be adept at these things. Thanks again kurt.

Ive also a questionabout discriminants.

Generally speaking, when proving that the discrimant is >0 , is it possible to just say using example find that discrim = 16(m^2+16m+1). And then state that discrim>0 for all values of m.

Timothy.Siu · Jul 28, 2009

Lukybear said:
Yea ok, it perhaps will take a long time for me to be adept at these things. Thanks again kurt.

Ive also a questionabout discriminants.

Generally speaking, when proving that the discrimant is >0 , is it possible to just say using example find that discrim = 16(m^2+16m+1). And then state that discrim>0 for all values of m.

nup, u'd have to prove that m^2+16m+1 is greater than 0 for all m.
and btw its not.

Lukybear · Jul 28, 2009

o mybad typo. it was actually 16(m^2+m+1). Yea and i assume you complete the square etc...

But say, the typically speaking, the discrim = 4(x^2+c^2+y^2) .... one would then say that; for all real values of x , c, y the discrim is >=0 thus more than 1 root exist etc... right?

kurt.physics · Jul 28, 2009

Lukybear said:
o mybad typo. it was actually 16(m^2+m+1). Yea and i assume you complete the square etc...

But say, the typically speaking, the discrim = 4(x^2+c^2+y^2) .... one would then say that; for all real values of x , c, y the discrim is >=0 thus more than 1 root exist etc... right?

$\triangle = 4\left(x^2 + c^2 + y^2\right)$

if $x = y = c = 0$ then

$\triangle = 4 \times 0 = 0$

$\therefore$ there is just one root.

if any of x, y or c is >0 for all values then your statement "thus more than 1 root exists" would be correct. But if $\triangle \ge 0$ then there is either 1 or 2 roots.

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