which of these equations could have solutions that are non - real? assume m and n are both negative real numbers. select all that apply. $nx^{2}-12 = mx$ $4x^{2}=mnx$ $mx^{2}-nx + 2 = 0$ $nx^{2}-m = 0$

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To determine which quadratic equations could have non - real solutions, we analyze the discriminant $D = b^{2}-4ac$ for each quadratic equation (after rewriting them in standard form $ax^{2}+bx + c = 0$). A quadratic equation has non - real solutions when $D<0$. We know that $m<0$ and $n<0$. ### 1. For the equation $nx^{2}-mx - 12=0$ (rewritten from $nx^{2}-12 = mx$) Here, $a = n$, $b=-m$, and $c = - 12$. The discriminant $D=(-m)^{2}-4\times n\times(-12)=m^{2}+48n$. Since $m^{2}>0$ (because $m$ is a non - zero real number, as $m$ is negative) and $n<0$, but $m^{2}$ is positive and $48n$ is negative. However, we can't be sure if $m^{2}+48n<0$ for all negative $m$ and $n$. Let's check the other equations. ### 2. For the equation $4x^{2}-mnx = 0$ (rewritten from $4x^{2}=mnx$) This is a quadratic equation with $a = 4$, $b=-mn$, and $c = 0$. The discriminant $D=(-mn)^{2}-4\times4\times0=m^{2}n^{2}$. Since $m$ and $n$ are non - zero real numbers, $m^{2}n^{2}>0$. So, the equation $4x^{2}-mnx = 0$ has two real solutions (one of which is $x = 0$ and the other is $x=\frac{mn}{4}$). So this equation does not have non - real solutions. ### 3. For the equation $mx^{2}-nx + 2=0$ Here, $a = m$, $b=-n$, and $c = 2$. The discriminant $D=(-n)^{2}-4\times m\times2=n^{2}-8m$. Since $n^{2}>0$ (because $n$ is a non - zero real number) and $m<0$, then $-8m>0$. So $D=n^{2}-8m>0$ (the sum of two positive numbers). Thus, this equation has two real solutions. ### 4. For the equation $nx^{2}-m = 0$ (rewritten as $nx^{2}+0x - m=0$) Here, $a = n$, $b = 0$, and $c=-m$. The discriminant $D=0^{2}-4\times n\times(-m)=4mn$. Since $m<0$ and $n<0$, the product $mn>0$, so $D = 4mn>0$. Thus, this equation has two real solutions ($x=\pm\sqrt{\frac{m}{n}}$, and since $m$ and $n$ are both negative, $\frac{m}{n}>0$). Wait, we made a mistake in the first equation's analysis. Let's re - analyze the equation $nx^{2}-mx - 12=0$ (from $nx^{2}-12=mx$): The correct standard form is $nx^{2}-mx - 12 = 0$, so $a=n$, $b=-m$, $c = - 12$. $D=b^{2}-4ac=(-m)^{2}-4\times n\times(-12)=m^{2}+48n$ Since $m<0$, $m^{2}>0$, and $n<0$. Let's take an example: let $m=-1$, $n=-1$. Then $D=(-1)^{2}+48\times(-1)=1 - 48=-47<0$. So when $m$ and $n$ are negative, it is possible for $D<0$ in this equation. The equation $4x^{2}-mnx=0$: $a = 4$, $b=-mn$, $c = 0$, $D = b^{2}-4ac=m^{2}n^{2}\geq0$, so it has real solutions. The equation $mx^{2}-nx + 2=0$: $a = m$, $b=-n$, $c = 2$, $D=n^{2}-8m$. Since $n<0$, $n^{2}>0$, and $m<0$, $-8m>0$, so $D=n^{2}-8m>0$, real solutions. The equation $nx^{2}-m = 0$: $a = n$, $b = 0$, $c=-m$, $D=0 - 4\times n\times(-m)=4mn$. Since $m<0$ and $n<0$, $mn>0$, so $D>0$, real solutions. So the only equation that could have non - real solutions is $nx^{2}-mx - 12 = 0$ (i.e., $nx^{2}-12=mx$). But the original selection in the problem is wrong for the other options. But according to the problem's initial check marks, if we have to go with the analysis of which can have non - real solutions: The equation $nx^{2}-12=mx$ (rewritten as $nx^{2}-mx - 12 = 0$) has a discriminant $D=m^{2}+48n$. Since $m$ and $n$ are negative, if $|48n|>m^{2}$, then $D<0$ and the equation has non - real solutions. The correct equation that could have non - real solutions is $nx^{2}-12 = mx$ (the first option). The other options as marked in the original problem are incorrect. If we consider the original problem's marking (even though our analysis shows some mistakes in the marking), but based on the idea of discriminant: For $nx^{2}-12 = mx$: rewrite as $nx^{2}-mx - 12=0$, $a = n$, $b=-m$, $c=-12$, $D=m^{2}+48n$. Since $m<0$, $n<0$, it's possible for $D<0$. For $4x^{2}=mnx$: $4x^{2}-mnx = 0$, $D=m^{2}n^{2}\geq0$, real solutions. For $mx^{2}-nx + 2=0$: $D=n^{2}-8m$, $n^{2}>0$, $m<0$ so $-8m>0$, $D>0$, real solutions. For $nx^{2}-m = 0$: $D = 4mn$, $m<0$, $n<0$ so $mn>0$, $D>0$, real solutions. So only the first equation $nx^{2}-12 = mx$ could have non - real solutions. But the original problem's check marks are wrong for the other three. However, if we follow the problem's requirement to select all that apply (assuming the initial marking is correct in the problem's context, maybe some miscalculations in our part), but based on proper discriminant analysis: The equation that could have non - real solutions is $nx^{2}-12 = mx$ (the first option). If we consider the original problem's marked options, but our analysis shows that only the first one is correct. But perhaps there is a different approach. Wait, maybe we made a mistake in the sign of $mn$ in the second equation. Let's re - check: The equation $4x^{2}=mnx$ can be written as $4x^{2}-mnx = 0$. $a = 4$, $b=-mn$, $c = 0$. $D=b^{2}-4ac=m^{2}n^{2}$. Since $m$ and $n$ are real numbers, $m^{2}n^{2}\geq0$, so the equation has real roots ($x = 0$ and $x=\frac{mn}{4}$). So it can't have non - real roots. The equation $mx^{2}-nx + 2=0$: $a = m$, $b=-n$, $c = 2$. $D=n^{2}-8m$. Since $n<0$, $n^{2}>0$, $m<0$, so $-8m>0$, $D=n^{2}-8m>0$, real roots. The equation $nx^{2}-m = 0$: $a = n$, $b = 0$, $c=-m$. $D=0 - 4\times n\times(-m)=4mn$. Since $m<0$, $n<0$, $mn>0$, $D>0$, real roots. The equation $nx^{2}-mx - 12=0$: $a = n$, $b=-m$, $c=-12$. $D=m^{2}+48n$. Let $m=-2$, $n=-2$, then $D = 4+48\times(-2)=4 - 96=-92<0$, so it has non - real roots. So the only equation that could have non - real solutions is $nx^{2}-12 = mx$ (the first option). So the correct answer is the equation $nx^{2}-12 = mx$ (the first boxed equation in the given options). If we have to select from the given options (with the check marks), but our analysis shows that only the first one is correct. However, if we assume that there is a mistake in our analysis and follow the problem's initial marking (maybe the problem has a different approach), but based on standard quadratic equation discriminant analysis, only $nx^{2}-12 = mx$ could have non - real solutions. # Answer: The equation $nx^{2}-12 = mx$ (the first option among the given equations) could have non - real solutions. If we consider the options as given in the problem (with the check marks, but our analysis shows otherwise), but based on correct discriminant analysis, the answer is the equation $nx^{2}-12 = mx$ (the first equation in the list).

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Explanation:

1. For the equation \(nx^{2}-mx - 12=0\) (rewritten from \(nx^{2}-12 = mx\))

2. For the equation \(4x^{2}-mnx = 0\) (rewritten from \(4x^{2}=mnx\))

3. For the equation \(mx^{2}-nx + 2=0\)

4. For the equation \(nx^{2}-m = 0\) (rewritten as \(nx^{2}+0x - m=0\))

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