determine the number of real solutions each quadratic equation has:
$y = 12x^2 - 9x + 4$
real solution(s)
$10x + y = x^2 + 1$
real solution(s)
$4y - 3 = 5x^2 - x + 12 + 3y$
real solution(s)
$y = 6x + 4x^2$
real solution(s)

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To determine the number of real solutions for each quadratic equation, we use the discriminant of a quadratic equation in the form $ax^2 + bx + c = 0$, which is given by $D = b^2 - 4ac$. The number of real solutions is determined as follows: - If $D>0$, there are 2 distinct real solutions. - If $D = 0$, there is 1 real solution (a repeated root). - If $D<0$, there are no real solutions (two complex conjugate solutions). ### 1. For the equation $y = 12x^2-8x + 4$ (rewrite as $12x^2-8x + 4 - y=0$, but since we are interested in the quadratic in $x$, we consider $12x^2-8x + 4 = 0$) ## Step 1: Identify $a$, $b$, $c$ For the quadratic equation $12x^2-8x + 4=0$, we have $a = 12$, $b=- 8$, $c = 4$ ## Step 2: Calculate the discriminant $D=b^2-4ac$ $$ \begin{align*} D&=(-8)^2-4\times12\times4\ &=64 - 192\ &=- 128 \end{align*} $$ Since $D=- 128<0$, the quadratic equation $12x^2-8x + 4 = 0$ has 0 real solutions. ### 2. For the equation $4y-3=5x^2 - x+2 + 3y$ (rewrite in standard quadratic form) ## Step 1: Simplify the equation $$ \begin{align*} 4y-3&=5x^2 - x+2 + 3y\ 4y-3y-3 - 2&=5x^2 - x\ y - 5&=5x^2 - x\ 5x^2 - x-y + 5&=0 \end{align*} $$ We are interested in the quadratic in $x$, so we consider $5x^2 - x+(5 - y)=0$. For the quadratic in $x$, $a = 5$, $b=-1$, $c=5 - y$ ## Step 2: Calculate the discriminant $D=b^2-4ac$ $$ \begin{align*} D&=(-1)^2-4\times5\times(5 - y)\ &=1-100 + 20y\ &=20y-99 \end{align*} $$ But if we consider the equation as a quadratic in $x$ (treating $y$ as a constant), we can also think of it as $5x^2 - x+(5 - y)=0$. However, if we want to find the number of real solutions for $x$ (regardless of $y$), we can analyze the discriminant in terms of $x$. But if we assume we are looking for real solutions in $x$ (for any $y$), the discriminant $D = (-1)^2-4\times5\times(5 - y)=1-100 + 20y=20y - 99$. But if we consider the equation as a quadratic in $x$ and we want to find if there are real solutions for $x$ (for some $y$), we can also note that for a quadratic $ax^2+bx + c = 0$ ($a eq0$), since $a = 5>0$ and the discriminant $D = 20y-99$ can take both positive and negative values depending on $y$. But if we made a mistake and we want to consider the equation as a quadratic in $y$: Rewrite the equation $4y-3=5x^2 - x+2 + 3y$ as $y-(5x^2 - x + 5)=0$. This is a linear equation in $y$, not a quadratic. So we should consider it as a quadratic in $x$. For a quadratic in $x$ $5x^2 - x+(5 - y)=0$, the discriminant $D=(-1)^2-4\times5\times(5 - y)=1 - 100 + 20y=20y-99$. If we assume we are looking for real solutions for $x$ (for a given $y$), when $D\geq0$ (i.e., $20y-99\geq0$ or $y\geq\frac{99}{20} = 4.95$), there are real solutions for $x$. But if we consider the equation as a quadratic in $x$ and we want to find the number of real solutions for $x$ (regardless of $y$), the quadratic $5x^2 - x+(5 - y)=0$ will have real solutions for $x$ when $20y-99\geq0$ (i.e., $y\geq4.95$) and no real solutions when $y < 4.95$. But this seems a bit off. Maybe we made a mistake in the approach. Let's re - express the original equation: $4y-3=5x^2 - x+2 + 3y$ $4y-3y=5x^2 - x + 2+3$ $y=5x^2 - x + 5$ This is a quadratic function in $x$. The graph of $y = 5x^2 - x + 5$ is a parabola opening upwards ($a = 5>0$) with vertex at $x=-\frac{b}{2a}=-\frac{-1}{2\times5}=\frac{1}{10}$ The $y$ - coordinate of the vertex is $y=5\times(\frac{1}{10})^2-\frac{1}{10}+5=5\times\frac{1}{100}-\frac{1}{10}+5=\frac{1}{20}-\frac{2}{20}+5=\frac{- 1}{20}+5=\frac{99}{20}=4.95$ Since the parabola opens upwards, for all real values of $x$, $y\geq4.95$. But if we are looking for the number of real solutions of the equation (treating it as a relation between $x$ and $y$), the equation $y = 5x^2 - x + 5$ represents a parabola, and for each real $x$, there is a unique real $y$. But if we consider the equation as a quadratic in $x$ (solving for $x$ in terms of $y$): $5x^2 - x+(5 - y)=0$ The discriminant $D=(-1)^2-4\times5\times(5 - y)=1-100 + 20y=20y - 99$ When $D\geq0$ (i.e., $y\geq\frac{99}{20}$), we have two real solutions for $x$ (since $a = 5>0$ and $D>0$ gives two distinct roots, $D = 0$ gives one root). When $D<0$ (i.e., $y<\frac{99}{20}$), we have no real solutions for $x$. But this is a bit ambiguous. Maybe there was a typo and the equation is supposed to be a quadratic in one variable. If we assume the equation is $4y-3=5x^2 - x+2$ (ignoring the $+ 3y$ term by mistake), then: Rewrite as $5x^2 - x+2-(4y - 3)=0$ or $5x^2 - x+(5 - 4y)=0$ $a = 5$, $b=-1$, $c=5 - 4y$ $D=(-1)^2-4\times5\times(5 - 4y)=1-100 + 80y=80y - 99$ Still, it's a quadratic in $x$ with discriminant depending on $y$. ### 3. For the equation $18x + y=x^2+1$ (rewrite as $x^2-18x - y + 1=0$, quadratic in $x$) ## Step 1: Identify $a$, $b$, $c$ For the quadratic in $x$: $x^2-18x+(1 - y)=0$, we have $a = 1$, $b=-18$, $c=1 - y$ ## Step 2: Calculate the discriminant $D=b^2-4ac$ $$ \begin{align*} D&=(-18)^2-4\times1\times(1 - y)\ &=324-4 + 4y\ &=320 + 4y\ &=4(y + 80) \end{align*} $$ If we consider the equation as a quadratic in $x$ (treating $y$ as a constant), when $D\geq0$ (i.e., $y\geq - 80$), there are real solutions for $x$ (two distinct solutions when $D>0$, one solution when $D = 0$) and when $D<0$ (i.e., $y<-80$), there are no real solutions for $x$. ### 4. For the equation $y=6x + 4x^2$ (rewrite as $4x^2+6x - y=0$, quadratic in $x$) ## Step 1: Identify $a$, $b$, $c$ For the quadratic in $x$: $4x^2+6x - y=0$, we have $a = 4$, $b = 6$, $c=-y$ ## Step 2: Calculate the discriminant $D=b^2-4ac$ $$ \begin{align*} D&=6^2-4\times4\times(-y)\ &=36 + 16y \end{align*} $$ If we consider the equation as a quadratic in $x$ (treating $y$ as a constant), when $D\geq0$ (i.e., $16y\geq - 36$ or $y\geq-\frac{9}{4}=-2.25$), there are real solutions for $x$ (two distinct solutions when $D>0$, one solution when $D = 0$) and when $D<0$ (i.e., $y<-\frac{9}{4}$), there are no real solutions for $x$. If we assume that the first equation $y = 12x^2-8x + 4$ (quadratic in $x$) has $a = 12$, $b=-8$, $c = 4$ and discriminant $D=-128<0$, so 0 real solutions. For the second equation, if we made a mistake and the equation is $4y-3=5x^2 - x+2$ (quadratic in $x$) with $a = 5$, $b=-1$, $c = 5$ (assuming $y$ is a typo), then $D=(-1)^2-4\times5\times5=1 - 100=-99<0$, 0 real solutions. But this is just an assumption. For the third equation $18x + y=x^2+1$ (quadratic in $x$) with $a = 1$, $b=-18$, $c = 1 - y$, the discriminant $D = 320+4y$. If $y$ is such that $D\geq0$, we have real solutions. For the fourth equation $y=6x + 4x^2$ (quadratic in $x$) with $a = 4$, $b = 6$, $c=-y$, the discriminant $D=36 + 16y$. If $y$ is such that $D\geq0$, we have real solutions. Since the first equation $y = 12x^2-8x + 4$ (quadratic in $x$) has a clear discriminant calculation: # Answer for $y = 12x^2-8x + 4$: 0 real solutions (Note: The other equations have ambiguous interpretations due to the presence of two variables. If you can provide more context or correct the equations, we can give more accurate answers.)

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Explanation:

1. For the equation \(y = 12x^2-8x + 4\) (rewrite as \(12x^2-8x + 4 - y=0\), but since we are interested in the quadratic in \(x\), we consider \(12x^2-8x + 4 = 0\))

Step 1: Identify \(a\), \(b\), \(c\)

Step 2: Calculate the discriminant \(D=b^2-4ac\)

2. For the equation \(4y-3=5x^2 - x+2 + 3y\) (rewrite in standard quadratic form)

Step 1: Simplify the equation

Step 2: Calculate the discriminant \(D=b^2-4ac\)

Answer: