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01 Dec 2020, 04:51
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Nishant1795
Bunuel
7. If x^4 = 29x^2 - 100, then which of the following is NOT a product of three possible values of x?
I. -50
II. 25
III. 50
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
Re-arrange and factor for x^2: \((x^2-25)(x^2-4)=0\).
So, we have that \(x=5\), \(x=-5\), \(x=2\), or \(x=-2\).
\(-50=5*(-5)*2\);
\(50=5*(-5)*(-2)\).
Only 25 is NOT a product of three possible values of x
Answer: B.
I. -50
II. 25
III. 50
A. I only
B. II only
C. III only
D. I and II only
E. I and III only
Re-arrange and factor for x^2: \((x^2-25)(x^2-4)=0\).
So, we have that \(x=5\), \(x=-5\), \(x=2\), or \(x=-2\).
\(-50=5*(-5)*2\);
\(50=5*(-5)*(-2)\).
Only 25 is NOT a product of three possible values of x
Answer: B.
Hi Bunuel, Can you please explain how did you rearrange the equation into a quadratic equation? Thanks
x^4 = 29x^2 - 100;
x^4 - 29x^2 + 100 = 0;
Say x^2 = a, so we have a^2 - 29a + 100 = 0.
Now, you can solve for a or factor.
Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm
Hope it helps.
06 Dec 2020, 10:49
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Hi Bunuel,
In first question why x cannot be negative . i know the square root is always positive . but when i put the value of x =-2 it staisfy both the side of equation which is creating a confusion for me.
In first question why x cannot be negative . i know the square root is always positive . but when i put the value of x =-2 it staisfy both the side of equation which is creating a confusion for me.
06 Dec 2020, 10:55
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abhinavsodha800
Hi Bunuel,
In first question why x cannot be negative . i know the square root is always positive . but when i put the value of x =-2 it staisfy both the side of equation which is creating a confusion for me.
In first question why x cannot be negative . i know the square root is always positive . but when i put the value of x =-2 it staisfy both the side of equation which is creating a confusion for me.
If x = -2, then \(\sqrt[4]{x^3+6x^2}=\sqrt[4]{-8+24}=\sqrt[4]{16}=2\).
As you can see \(x≠\sqrt[4]{x^3+6x^2}\).
