Question

Determine which equation has exactly one real solution. PLS HELPP

Answer 1

Answer:

b)  $4x^2-12x+9=0$

Step-by-step explanation:

We can use the discriminant $b^2-4ac$ to determine how many real solutions a quadratic equation in the form $ax^2+bx+c=0$ has.

• discriminant > 0 ⇒ 2 real solutions
• discriminant = 0 ⇒ 1 real solution
• discriminant < 0 ⇒ no real solutions

$5x^2+9x-4=0$

$b^2-4ac=9^2-4(5)(-4)=161 > 0\implies \textsf{2 solutions}$

$4x^2-12x+9=0$

$b^2-4ac=(-12)^2-4(4)(9)=0 \implies \textsf{1 solution}$

$-9x^2+6x+1=0$

$b^2-4ac=6^2-4(-9)(1)=72 > 0 \implies \textsf{2 solutions}$

$-2x^2-7=0$

$b^2-4ac=0^2-4(-2)(-7)=-56 < 0 \implies \textsf{no solutions}$