- #1

Avalance789

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Find all possible a when an equation has only one possible solution.

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- MHB
- Thread starter Avalance789
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In summary: Why not?4x^2 - a^2 = 2x + a$$4x^2-a^2=(2x+a)(2x-a)$$$$4x^2-a^2=(2x+a)(2x-a)$$$$4x^2-a^2=(2x+a)(2x-a)$$$$4x^2-a^2=(2x+a)(2x-a)$$$$4x^2-a^2=(2x+a)(2x-a)$$$$4x^2-a^2=(2x+a)(2x-a)$$$$4x^2-a^2=(2x+a)(2x-a)$$

- #1

Avalance789

- 13

- 0

Find all possible a when an equation has only one possible solution.

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- #2

HOI

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- #3

Avalance789

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I am sorry, but I am unable to determine it. But if I won't solve it, I am dead man(

- #4

Greg

Gold Member

MHB

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$$ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The desired quadratic has only one real root so you are looking for $b^2-4ac=0$ (with $a$ as in the equation in this post). Apply that to the quadratic Country Boy gave.

When summarising the given problem I'd pay more attention to the trivial solution mentioned by Country Boy.

- #5

HOI

- 921

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By the quadratic formula, which greg1313 wrote out for you, the solutions to the quadratic equation \(\displaystyle 2y^2- 4y+ a- \sqrt{a}= 0\) are given by

\(\displaystyle \frac{4\pm\sqrt{16- 8(a-\sqrt{a})}}{4}\).

Clearly taking the "+" sign gives a positive root. In order that there be only one positive root we must have \(\displaystyle 4-\sqrt{16- 8(a- \sqrt{a})}< 0\). From that, \(\displaystyle \sqrt{16- 8(a- \sqrt{a})}> 4\) and, squaring both sides, \(\displaystyle 16- 8(a- \sqrt{a})> 16\) so \(\displaystyle 8(a- \sqrt{a})< 0\), \(\displaystyle a< \sqrt{a}\).

Now, what must \(a\) be?

\(\displaystyle \frac{4\pm\sqrt{16- 8(a-\sqrt{a})}}{4}\).

Clearly taking the "+" sign gives a positive root. In order that there be only one positive root we must have \(\displaystyle 4-\sqrt{16- 8(a- \sqrt{a})}< 0\). From that, \(\displaystyle \sqrt{16- 8(a- \sqrt{a})}> 4\) and, squaring both sides, \(\displaystyle 16- 8(a- \sqrt{a})> 16\) so \(\displaystyle 8(a- \sqrt{a})< 0\), \(\displaystyle a< \sqrt{a}\).

Now, what must \(a\) be?

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- #6

Avalance789

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So it doesn't have solution? My teacher states on that it has

- #7

Avalance789

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Forces me to solve it ultimatively

- #8

Avalance789

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Teacher gives me result

(-5/3; -1/2] [2/3; 5/3)

Is it correct?

(-5/3; -1/2] [2/3; 5/3)

Is it correct?

- #9

Avalance789

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sqrt (5-6x)*ln(4x^2-a^2)=sqrt(5-6x)*ln(2x+a)So sorry

- #10

Avalance789

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4x^2-a^2 - 2x - a = 0

So we need to find a values when D=0?

And if a=1, then it's quadrant equation?

So we need to find a values when D=0?

And if a=1, then it's quadrant equation?

- #11

MarkFL

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Avalance789 said:

sqrt (5-6x)*ln(4x^2-a^2)=sqrt(5-6x)*ln(2x+a)So sorry

Okay, so the equation is:

\(\displaystyle \sqrt{5-6x}\ln\left(4x^2-a^2\right)=\sqrt{5-6x}\ln\left(2x+a\right)\)

Let's arrange as:

\(\displaystyle \sqrt{5-6x}\ln\left(4x^2-a^2\right)-\sqrt{5-6x}\ln\left(2x+a\right)=0\)

Factor:

\(\displaystyle \sqrt{5-6x}\left(\ln\left(4x^2-a^2\right)-\ln\left(2x+a\right)\right)=0\)

Apply log rule for subtraction:

\(\displaystyle \sqrt{5-6x}\ln\left(\frac{4x^2-a^2}{2x+a}\right)=0\)

Factor log argument:

\(\displaystyle \sqrt{5-6x}\ln\left(\frac{(2x+a)(2x-a)}{2x+a}\right)=0\)

Divide out common factors:

\(\displaystyle \sqrt{5-6x}\ln(2x-a)=0\)

Okay, now I would first look at the square root, and observe that in order for the equation to have only 1 solution, we need:

\(\displaystyle 5-6x>0\implies x<\frac{5}{6}\)

Now, in order for the equation to be true with respect to the factor involving \(a\), we require:

\(\displaystyle 2x-a=1\implies x=\frac{a+1}{2}\)

And so:

\(\displaystyle \frac{a+1}{2}<\frac{5}{6}\)

\(\displaystyle a+1<\frac{5}{3}\)

\(\displaystyle a<\frac{2}{3}\)

- #12

Wilmer

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Mark, seems to me that above easily/instantly simplifies to:MarkFL said:Okay, so the equation is:

\(\displaystyle \sqrt{5-6x}\ln\left(4x^2-a^2\right)=\sqrt{5-6x}\ln\left(2x+a\right)\)

4x^2 - a^2 = 2x + a

No?

- #13

Avalance789

- 13

- 0

And after that simplification?

- #14

MarkFL

Gold Member

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Wilmer said:Mark, seems to me that above easily/instantly simplifies to:

4x^2 - a^2 = 2x + a

No?

Yes, and I saw that after going through the laborious log shuffle, but we will get to the same place. But, we still need to be mindful of the domain.

- #15

Avalance789

- 13

- 0

Just tell me if a<2/3 is correct answer. I will be happiest person on Earth, guys

- #16

MarkFL

Gold Member

MHB

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Avalance789 said:Just tell me if a<2/3 is correct answer. I will be happiest person on Earth, guys

Something I didn't consider when I simplified is that:

\(\displaystyle 4x^2-a^2>0\)

\(\displaystyle 4\left(\frac{a+1}{2}\right)^2-a^2>0\)

\(\displaystyle (a+1)^2-a^2>0\)

\(\displaystyle (a+1+a)(a+1-a)>0\)

\(\displaystyle 2a+1>0\implies a>-\frac{1}{2}\)

Now, suppose \(\displaystyle x=\frac{5}{6}\)

Then we have:

\(\displaystyle 4\left(\frac{5}{6}\right)^2-a^2>0\)

\(\displaystyle \left(\frac{5}{3}\right)^2-a^2>0\)

\(\displaystyle |a|<\frac{5}{3}\)

So, what does all this imply?

- #17

Greg

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Wilmer said:4x^2 - a^2 = 2x + a

$$4x^2-a^2=(2x+a)(2x-a)$$

- #18

Wilmer

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Yes...but this was the OP's:greg1313 said:$$4x^2-a^2=(2x+a)(2x-a)$$

\(\displaystyle \sqrt{5-6x}\ln\left(4x^2-a^2\right)=\sqrt{5-6x}\ln\left(2x+a\right)\)

- #19

Greg

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Ah, pardon me.

- #20

Wilmer

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Only if you recite the Hooooooly Rosary twice !greg1313 said:Ah, pardon me.

- #21

Avalance789

- 13

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Ок, but if a=1, then x=1.

Cannot be that -5/3<a<5/3

Cannot be that -5/3<a<5/3

The equation represents a relationship between the square root of the difference between 5 and 6 times x, multiplied by the natural logarithm of the square root of 4 times the square root of x minus the square root of a, and the square root of the difference between 5 and 6 times x, multiplied by the natural logarithm of 2 times x plus a.

The variables in this equation are x and a. x represents a number or value and a represents another number or value. They can be any real numbers that satisfy the equation.

This equation is a nonlinear equation and can be solved using various methods such as substitution, elimination, or graphical methods. It can also be solved using numerical methods such as Newton's method or the bisection method.

The natural logarithm, represented by ln, is a mathematical function that shows the relationship between a number and the exponential value of that number. In this equation, the natural logarithm is used to represent the relationship between the two sides of the equation and to find the values of x and a that satisfy the equation.

Yes, there are some restrictions on the values of x and a in this equation. For example, the values inside the square root cannot be negative, so x must be greater than or equal to 5/6 and a must be greater than or equal to 0. Additionally, the values inside the natural logarithm cannot be negative, so 2x+a must be greater than 0.

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