Answer :
To determine whether an equation can be used to show that Vladas' hypothesis is incorrect, we need to check if any of the given equations represents a function of [tex]\( x \)[/tex]. A function of [tex]\( x \)[/tex] means for every value of [tex]\( x \)[/tex], there should be exactly one value of [tex]\( y \)[/tex].
Let's go through each equation step by step:
1. Equation: [tex]\( x + y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = 25 - x \][/tex]
[tex]\[ y = \pm \sqrt{25 - x} \][/tex]
For a given [tex]\( x \)[/tex], [tex]\( y \)[/tex] can be either [tex]\( \sqrt{25 - x} \)[/tex] or [tex]\( -\sqrt{25 - x} \)[/tex]. This implies that for each [tex]\( x \)[/tex] there are two possible values of [tex]\( y \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it provides more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
2. Equation: [tex]\( x^2 - y = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y = x^2 - 25 \][/tex]
For any given [tex]\( x \)[/tex], there is exactly one corresponding value of [tex]\( y \)[/tex], which is [tex]\( x^2 - 25 \)[/tex]. Therefore, this equation is indeed a function because it satisfies the condition of providing exactly one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
3. Equation: [tex]\( x^2 + y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = 25 - x^2 \][/tex]
[tex]\[ y = \pm \sqrt{25 - x^2} \][/tex]
For any given [tex]\( x \)[/tex] within [tex]\( -5 \leq x \leq 5 \)[/tex], there are two possible values of [tex]\( y \)[/tex]: [tex]\( \sqrt{25 - x^2} \)[/tex] and [tex]\( -\sqrt{25 - x^2} \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it can provide more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
4. Equation: [tex]\( x^2 - y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = x^2 - 25 \][/tex]
[tex]\[ y = \pm \sqrt{x^2 - 25} \][/tex]
For [tex]\( |x| \geq 5 \)[/tex], there are two possible values of [tex]\( y \)[/tex]: [tex]\( \sqrt{x^2 - 25} \)[/tex] and [tex]\( -\sqrt{x^2 - 25} \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it provides more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
From the above examination, we observe that only the equation [tex]\( x^2 - y = 25 \)[/tex] (i.e., the second equation) is a function of [tex]\( x \)[/tex]. This disproves Vladas' hypothesis that an equation with a squared term can never be a function of [tex]\( x \)[/tex].
Therefore, the equation [tex]\( x^2 - y = 25 \)[/tex] can be used to show Vladas that his hypothesis is incorrect.
Let's go through each equation step by step:
1. Equation: [tex]\( x + y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = 25 - x \][/tex]
[tex]\[ y = \pm \sqrt{25 - x} \][/tex]
For a given [tex]\( x \)[/tex], [tex]\( y \)[/tex] can be either [tex]\( \sqrt{25 - x} \)[/tex] or [tex]\( -\sqrt{25 - x} \)[/tex]. This implies that for each [tex]\( x \)[/tex] there are two possible values of [tex]\( y \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it provides more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
2. Equation: [tex]\( x^2 - y = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y = x^2 - 25 \][/tex]
For any given [tex]\( x \)[/tex], there is exactly one corresponding value of [tex]\( y \)[/tex], which is [tex]\( x^2 - 25 \)[/tex]. Therefore, this equation is indeed a function because it satisfies the condition of providing exactly one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
3. Equation: [tex]\( x^2 + y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = 25 - x^2 \][/tex]
[tex]\[ y = \pm \sqrt{25 - x^2} \][/tex]
For any given [tex]\( x \)[/tex] within [tex]\( -5 \leq x \leq 5 \)[/tex], there are two possible values of [tex]\( y \)[/tex]: [tex]\( \sqrt{25 - x^2} \)[/tex] and [tex]\( -\sqrt{25 - x^2} \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it can provide more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
4. Equation: [tex]\( x^2 - y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = x^2 - 25 \][/tex]
[tex]\[ y = \pm \sqrt{x^2 - 25} \][/tex]
For [tex]\( |x| \geq 5 \)[/tex], there are two possible values of [tex]\( y \)[/tex]: [tex]\( \sqrt{x^2 - 25} \)[/tex] and [tex]\( -\sqrt{x^2 - 25} \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it provides more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
From the above examination, we observe that only the equation [tex]\( x^2 - y = 25 \)[/tex] (i.e., the second equation) is a function of [tex]\( x \)[/tex]. This disproves Vladas' hypothesis that an equation with a squared term can never be a function of [tex]\( x \)[/tex].
Therefore, the equation [tex]\( x^2 - y = 25 \)[/tex] can be used to show Vladas that his hypothesis is incorrect.
