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# If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what

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If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what  [#permalink]

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Updated on: 29 Jul 2018, 20:42
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If the sum of the squares of x and y is 3, and $$x^4 = y^4 + 25$$, what is the value of $$x^2$$?

A) $$\frac{-8}{3}$$

B) $$\frac{14}{3}$$

C) $$\frac{17}{3}$$

D) $$\frac{28}{3}$$

E) $$\frac{34}{3}$$

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Originally posted by CAMANISHPARMAR on 29 Jul 2018, 13:05.
Last edited by Bunuel on 29 Jul 2018, 20:42, edited 2 times in total.
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Re: If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what  [#permalink]

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29 Jul 2018, 13:09
1
$$x^4-y^4=25$$
It is given that $$(x^2+y^2)=3$$
$$(x^2-y^2)=25/3$$
$$2x^2=25/3+3=34/3$$
$$x^2=17/3$$
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Re: If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what  [#permalink]

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29 Jul 2018, 18:12
CAMANISHPARMAR wrote:
If the sum of the squares of x and y is 3, and $$x^4 = y^4 + 25$$, what is the value of $$x^2$$?

A) $$\frac{-8}{3}$$
B) $$\frac{14}{3}$$
C) $$\frac{17}{3}$$
D) $$\frac{28}{3}$$
E) $$\frac{34}{3}$$

if x^2+y^2=3, then
y^2=3-x^2 and
y^4=x^4-6x^2+9 (eq1)

if x^4=y^4+25, then
y^4=x^4-25 (eq2)

subtracting eq2 from eq1,
6x^2=34→
x^2=34/6=17/3
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Re: If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what  [#permalink]

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29 Jul 2018, 19:31
2
CAMANISHPARMAR wrote:
If the sum of the squares of x and y is 3, and $$x^4 = y^4 + 25$$, what is the value of $$x^2$$?

A) $$\frac{-8}{3}$$
B) $$\frac{14}{3}$$
C) $$\frac{17}{3}$$
D) $$\frac{28}{3}$$
E) $$\frac{34}{3}$$

The sum of the squares of x and y is 3..
$$x^2+y^2=3........y^2=3-x^2$$..
$$x^4=y^4+25........x^4-y^4=25........(x^2-y^2)(x^2+y^2)=25.......(x^2-y^2)*3=25$$..
Substitute y^2 in it..
$$x^2-(3-x^2)=\frac{25}{3}...........x^2-3+x^2=\frac{25}{3}.....2x^2=\frac{25}{3}+3=\frac{34}{3}.........x^2=\frac{17}{3}$$

C
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Re: If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what  [#permalink]

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29 Jul 2018, 20:08
CAMANISHPARMAR wrote:
If the sum of the squares of x and y is 3, and $$x^4 = y^4 + 25$$, what is the value of $$x^2$$?

A) $$\frac{-8}{3}$$
B) $$\frac{14}{3}$$
C) $$\frac{17}{3}$$
D) $$\frac{28}{3}$$
E) $$\frac{34}{3}$$

$$x^2+ y^2 = 3$$ ----- (1)
$$x^4 = y^4 + 25$$

This can be written as -

$$x^4 - y^4 = 25$$ ----- (2)

Now using formula $$a^2 - b^2 = (a-b)*(a+b)$$ on --- (2)

$$(x^2 - y^2) (x^2 + y^2) = 25$$ ----- (3)

Substituting (1) in (3), we get

$$(x^2 - y^2) * (3) = 25$$

$$(x^2 - y^2) = \frac{25}{3}$$ ----- (4)

Using (1) we can get,

$$y^2 = 3 - x^2$$ ------ (5)

Using (5) in (4)

$$(x^2 - (3 - x^2)) = \frac{25}{3}$$

Solving further,

$$(x^2 - 3 + x^2) = \frac{25}{3}$$

$$(2 * x^2) - 3 = \frac{25}{3}$$

$$(2 * x^2) = \frac{25}{3} + 3$$

$$(2 * x^2) = \frac{25}{3} + \frac{9}{3}$$

$$(2 * x^2) = \frac{34}{3}$$

$$x^2 = \frac{34}{6}$$

$$x^2 = \frac{17}{3}$$

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Re: If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what  [#permalink]

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03 Aug 2018, 15:05
CAMANISHPARMAR wrote:
If the sum of the squares of x and y is 3, and $$x^4 = y^4 + 25$$, what is the value of $$x^2$$?

A) $$\frac{-8}{3}$$

B) $$\frac{14}{3}$$

C) $$\frac{17}{3}$$

D) $$\frac{28}{3}$$

E) $$\frac{34}{3}$$

We are given that x^2 + y^2 = 3 and x^4 = y^4 + 25.

Let’s simplify the second equation:

x^4 = y^4 + 25

x^4 - y^4 = 25

(x^2 + y^2)(x^2 - y^2) = 25

Since x^2 + y^2 = 3, we have:

3(x^2 - y^2) = 25

x^2 - y^2 = 25/3

Now, adding x^2 - y^2 = 25/3 and x^2 + y^2 = 3, we have:

2x^2 = 34/3

x^2 = 17/3

Alternate Solution:

Let’s raise each side of x^2 + y^2 = 3 to the second power:

(x^2 + y^2)^2 = 3^2
x^4 + 2x^2y^2 + y^4 = 9

Let’s substitute y^4 = x^4 - 25:

x^4 + 2x^2y^2 + x^4 - 25 = 9

2x^4 + 2x^2y^2 = 34

2x^2(x^2 + y^2) = 34

Recall that x^2 + y^2 was given to be 3, thus:

2x^2 * 3 = 34

x^2 = 34/6 = 17/3

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Director
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Re: If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what  [#permalink]

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09 Aug 2018, 04:26
1. let x^2 be x and y^2 be y, similarly x^4 will be x^2 and y^4 will be y^2

then substitute the value of x from first equation to the second equation.

And we get the solution.

Simple.
Re: If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what   [#permalink] 09 Aug 2018, 04:26
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