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. 4x² - 16x = 0, 3(x + 5)² = -2, 3x² + 24x = -48, 5x² + 2 = 4x. table with columns: no real solutions, exactly one real solution, two real solutions.

Explanation:

Step1: Recall discriminant rule

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

• $\Delta < 0$: No real solutions
• $\Delta = 0$: One real solution
• $\Delta > 0$: Two real solutions

Step2: Analyze $4x^2 - 16x = 0$

Rewrite: $4x^2-16x+0=0$, $a=4, b=-16, c=0$
$\Delta = (-16)^2 - 4(4)(0) = 256 > 0$

Step3: Analyze $3(x+5)^2 = -2$

Expand: $3(x^2+10x+25)=-2 \to 3x^2+30x+77=0$
$a=3, b=30, c=77$
$\Delta = 30^2 - 4(3)(77) = 900 - 924 = -24 < 0$

Step4: Analyze $3x^2 + 24x = -48$

Rewrite: $3x^2+24x+48=0$, $a=3, b=24, c=48$
$\Delta = 24^2 - 4(3)(48) = 576 - 576 = 0$

Step5: Analyze $5x^2 + 2 = 4x$

Rewrite: $5x^2-4x+2=0$, $a=5, b=-4, c=2$
$\Delta = (-4)^2 - 4(5)(2) = 16 - 40 = -24 < 0$

Answer:

No Real Solutions	Exactly One Real Solution	Two Real Solutions