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How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have?

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rheam25
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How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   15 Mar 2020, 10:47
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How many solutions does the equation \(0.25x^4\) – \(0.75x^2\) -1 = 0 have?

A) 0
B) 1
C) 2
D) 3
E) 4
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C
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sujoykrdatta
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   15 Mar 2020, 20:37
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rheam25 wrote:
How many solutions does the equation \(0.25x^4\) – \(0.75x^2\) - 1 = 0 have?

A) 0
B) 1
C) 2
D) 3
E) 4



Multiply throughout by 4: \(x^4\) – \(3x^2\) - 4 = 0

Assume \(x^2 = k\): \(k^2 - 3k - 4 = 0\)

=> \((k - 4)(k + 1) = 0\)

=> k = 4 or -1

=> \(x^2 = 4\) (since \(x^2\) cannot be -1 for any real x)

=> x = 2 or -2


Answer C
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exc4libur
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   23 Sep 2020, 07:45
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rheam25 wrote:
How many solutions does the equation \(0.25x^4\) – \(0.75x^2\) -1 = 0 have?

A) 0
B) 1
C) 2
D) 3
E) 4


x^4/4-3x^2/4-1=0
x^4-3x^2-4=0
x^2x^2-3x^2-4=0
(x^2)^2-3(x^2)-4=0
a^2-3a-4=0
(a-4)(a+1)=0
a=4 or a=-1
x^2=-1, cannot square root negative
x^2=4, |x|=2 or -2

(C)
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planetearth9
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   23 Sep 2020, 08:04
The question does not specify if x is real. In that case, would it be wrong to assume that x could be imaginary as well?
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   23 Sep 2020, 08:20
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planetearth9 wrote:
The question does not specify if x is real. In that case, would it be wrong to assume that x could be imaginary as well?


All numbers on the GMAT are real numbers by default.
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   23 Sep 2020, 15:42
There are 2 tricks with this question:
- First is to convert the co-efficients in the equation to whole numbers. Converts the equation to x^4 - 3x^2 - 4 = 0
- Second to substitute x^2 as another variable. This turns it into a quadratic equation which is much simpler to solve => (x^2 = a) => a^2 - 3a -4 = 0

We would be spending much more time to get to the actual answer without applying the above tricks.
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   25 Sep 2020, 07:14
can you guys please help me understand how you got the -4?
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dhichawla
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   25 Sep 2020, 07:27
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Hi omavsp,

If you look at the original equation, it is 0.25x4 – 0.75x2 -1 = 0.
There are 2 ways to go about it:
1) Multiply the entire equation by 4. This being done to change the co-efficient of x^4 from 0.25 to 1.
So, 4 * (0.25x4 – 0.75x2 -1 = 0) ==> 4*0.25x4 - 4*0.75x2 - 4*1 = 0 ==> x4 - 3x2 - 4 = 0.

2) If you convert the decimals to fractions in the original equation, you will end up with below:
0.25x4 – 0.75x2 -1 = 0 ==> 25/100* x4 - 75/100* x2 - 1 = 0
==> x4/4 - 3x2/4 - 1 = 0
Now, multiply by 4 again, you get ==> x4 - 3x2 - 4 = 0.

Hope this clarifies you doubt.
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   12 Oct 2020, 14:51
Hi!

I would like to know if my procedure is right! could someone check it (please)

1)Used sqrt(b^2 - 4ac) (also called discriminant) to identify how many solutions for X are
a) Discriminant >0; then there are 2 different solutions
b) Discriminant =0; then there are 2 equal solutions
c) Discriminant <0; then there are 0 solutions in the Real numbers world

The solution gave me a) meaning that there are 2 solutions
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Re: How many solutions does the equation 0.25x4 – 0.75x2 -1 = 0 have? [#permalink] New post   09 Dec 2020, 02:29
Asked: How many solutions does the equation \(0.25x^4\) – \(0.75x^2\) -1 = 0 have?

Let us multiple the whole equation by 4

xˆ4 - 3xˆ2 - 4 = 0
Let t = xˆ2

tˆ2 - 3t - 4= 0
(t-4)(t+1) = 0

Case 1: t = xˆ2 = 4
x = {-2,2}: 2 real solutions

Case 2: t= xˆ2 = - 1
x = {i, - i} : 2 imaginary solutions but No real solutions

Combining x = {-2,2 }: 2 real solutions

IMO C
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Re: How many solutions does the equation 0.25x4 0.75x2 -1 = 0 have? [#permalink] New post   01 Jun 2022, 07:50
jcerdae wrote:
Hi!

I would like to know if my procedure is right! could someone check it (please)

1)Used sqrt(b^2 - 4ac) (also called discriminant) to identify how many solutions for X are
a) Discriminant >0; then there are 2 different solutions
b) Discriminant =0; then there are 2 equal solutions
c) Discriminant <0; then there are 0 solutions in the Real numbers world

The solution gave me a) meaning that there are 2 solutions

jcerdae

The right term is not "discriminant" but "determinant". Further, we don't require to take sqrt of \(b^2 - 4ac\) because determinant is itself \(b^2 - 4ac\).

IMO, we can use this approach for the ques in hand.
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Re: How many solutions does the equation 0.25x4 0.75x2 -1 = 0 have? [#permalink] New post   05 Feb 2024, 06:00
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Re: How many solutions does the equation 0.25x4 0.75x2 -1 = 0 have? [#permalink]
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