Question

drag each equation to the correct location on the table. for each equation, determine the number of solutions and place on the appropriate field in the table. 3x² + 24x = -48 4x² - 16x = 0 5x² + 2 = 4x 3(x + 5)² = -2 no real solutions exactly one real solution two real solutions

Explanation:

Step1: Rewrite to standard quadratic form

For a quadratic equation $ax^2+bx+c=0$, use discriminant $\Delta=b^2-4ac$:

• If $\Delta>0$: 2 real solutions; $\Delta=0$: 1 real solution; $\Delta<0$: no real solutions.

Equation 1: $3x^2+24x=-48$

Step1: Rearrange to standard form

$3x^2+24x+48=0$

Step2: Simplify and compute $\Delta$

Divide by 3: $x^2+8x+16=0$
$\Delta=8^2-4(1)(16)=64-64=0$

Equation 2: $4x^2-16x=0$

Step1: Identify $a,b,c$

$a=4, b=-16, c=0$

Step2: Compute $\Delta$

$\Delta=(-16)^2-4(4)(0)=256-0=256>0$

Equation 3: $5x^2+2=4x$

Step1: Rearrange to standard form

$5x^2-4x+2=0$

Step2: Compute $\Delta$

$\Delta=(-4)^2-4(5)(2)=16-40=-24<0$

Equation 4: $3(x+5)^2=-2$

Step1: Analyze left-hand side

$(x+5)^2\geq0$, so $3(x+5)^2\geq0$

Step2: Compare to right-hand side

Right-hand side is $-2<0$, no real $x$ satisfies this, so $\Delta<0$

Answer:

No Real Solutions	Exactly One Real Solution	Two Real Solutions