Question

which equation has exactly one solution?
select one:
a. $3x^2 - 2 = -8$
b. $2x^2 - 14 = 4x - 1$
c. $5x^2 - 4x - 6 = x^2 - 7$
d. $6x^2 = 18 - 23x$
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what is the solution set to the following equation?
$4m^2 - 8m + 9 = 0$
select one:
a. $m = 1 \pm 4i\sqrt{5}$
b. $m = \frac{2\pm i\sqrt{5}}{2}$
c. $m = -1 \pm 4i\sqrt{5}$
d. $m = \frac{-2\pm i\sqrt{5}}{2}$

Explanation:

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First Question: Which equation has exactly one solution?

For quadratic equations in the form $ax^2+bx+c=0$, we use the discriminant $\Delta = b^2-4ac$:

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

Step1: Simplify option a

$3x^2 - 2 = -8 \implies 3x^2 +6=0$
$\Delta = 0^2 -4(3)(6)=-72<0$

Step2: Simplify option b

$2x^2 -14=4x-1 \implies 2x^2-4x-13=0$
$\Delta = (-4)^2 -4(2)(-13)=16+104=120>0$

Step3: Simplify option c

$5x^2-4x-6=x^2-7 \implies 4x^2-4x+1=0$
$\Delta = (-4)^2 -4(4)(1)=16-16=0$

Step4: Simplify option d

$6x^2=18-23x \implies 6x^2+23x-18=0$
$\Delta = 23^2 -4(6)(-18)=529+432=961>0$

Use quadratic formula $m=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for $ax^2+bx+c=0$

Step1: Identify coefficients

$a=4,\ b=-8,\ c=9$

Step2: Calculate discriminant

$\Delta = (-8)^2 -4(4)(9)=64-144=-80$
$\sqrt{\Delta}=\sqrt{-80}=4i\sqrt{5}$

Step3: Apply quadratic formula

$m=\frac{8\pm4i\sqrt{5}}{2(4)}=\frac{8\pm4i\sqrt{5}}{8}=\frac{2\pm i\sqrt{5}}{2}$

Answer:

c. $5x^2 - 4x - 6 = x^2 - 7$

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Second Question: Solution set for $4m^2 -8m +9=0$