Question

determine the number of real solutions each quadratic equation has.
y = 12x² - 9x + 4
real solution(s)

y = -x² + 2
real solution(s)
no
one
two

4y + 7 = 5x² - x + 2 + 3y
real solution(s)

y = |x + 4|²
real solution(s)
done

Explanation:

First, we need to rewrite each equation in standard quadratic form $ax^2+bx+c=0$, then use the discriminant $\Delta = b^2-4ac$:

• If $\Delta>0$: 2 real solutions
• If $\Delta=0$: 1 real solution
• If $\Delta<0$: 0 real solutions

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Equation 1: $y=12x^2-5x+4$

Step1: Identify $a,b,c$

$a=12$, $b=-5$, $c=4$

Step2: Calculate discriminant

$\Delta = (-5)^2 - 4(12)(4) = 25 - 192 = -167$

Step3: Interpret discriminant

$\Delta<0$, so 0 real solutions

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Equation 2: $4y-1=5x^2-x+2+3y$

Step1: Simplify to standard form

$y = 5x^2 -x +3$, so $a=5$, $b=-1$, $c=3$

Step2: Calculate discriminant

$\Delta = (-1)^2 - 4(5)(3) = 1 - 60 = -59$

Step3: Interpret discriminant

$\Delta<0$, so 0 real solutions

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Equation 3: $y=x^2+2$

Step1: Identify $a,b,c$

$a=1$, $b=0$, $c=2$

Step2: Calculate discriminant

$\Delta = 0^2 - 4(1)(2) = 0 - 8 = -8$

Step3: Interpret discriminant

$\Delta<0$, so 0 real solutions

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Equation 4: $y=(x+4)^2$

Step1: Expand to standard form

$y=x^2+8x+16$, so $a=1$, $b=8$, $c=16$

Step2: Calculate discriminant

$\Delta = 8^2 - 4(1)(16) = 64 - 64 = 0$

Step3: Interpret discriminant

$\Delta=0$, so 1 real solution

Answer:

1. $y=12x^2-5x+4$: no real solution(s)
2. $4y-1=5x^2-x+2+3y$: no real solution(s)
3. $y=x^2+2$: no real solution(s)
4. $y=(x+4)^2$: one real solution(s)