Question

1. using the quadratic formula solve the following quadratic: $2x^2 - 3x + 5 = 0$

a) $\frac{3pm7i}{4}$
b) $\frac{3pm isqrt{31}}{4}$
c) $\frac{-3pmsqrt{-31}}{4}$
d) $\frac{5i}{2}, -i$
options: c, b, a, d

Explanation:

Step1: Recall Quadratic Formula

For \( ax^2 + bx + c = 0 \), \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b=-3 \), \( c = 5 \).

Step2: Calculate Discriminant (\( D \))

\( D=b^2 - 4ac=(-3)^2-4(2)(5)=9 - 40=-31 \).

Step3: Apply Quadratic Formula

\( x=\frac{-(-3)\pm\sqrt{-31}}{2(2)}=\frac{3\pm i\sqrt{31}}{4} \)? Wait, no, wait: Wait, \( \sqrt{-31}=\sqrt{31}\times\sqrt{-1}=i\sqrt{31} \)? Wait, no, wait original calculation: Wait, \( b=-3 \), so \( -b = 3 \). \( 4ac=4*2*5 = 40 \). So \( b^2 - 4ac=9 - 40=-31 \). Then \( \sqrt{-31}=i\sqrt{31} \)? Wait, but option a is \( \frac{3\pm7i}{4} \), option b is \( \frac{3\pm i\sqrt{31}}{4} \), option c is \( \frac{-3\pm\sqrt{-31}}{4} \), option d is \( \frac{5i}{2}, -i \). Wait, maybe I made a mistake. Wait, let's recalculate discriminant: \( (-3)^2=9 \), \( 4*2*5 = 40 \), so \( 9 - 40=-31 \). Then \( \sqrt{-31}=i\sqrt{31} \), but option a: \( 3\pm7i \), 7 squared is 49, 3 squared is 9, 49 - 9=40? No. Wait, maybe the equation is \( 2x^2 - 3x - 5 = 0 \)? No, the equation is \( 2x^2 - 3x + 5 = 0 \). Wait, maybe I miscalculated. Wait, no, let's check option a: numerator \( 3\pm7i \), denominator 4. Then discriminant would be \( (7i)^2= -49 \), so \( b^2 - 4ac=-49 \), then \( 9 - 40=-31 \), not -49. Wait, maybe the equation is \( 2x^2 - 3x - 5 = 0 \), then discriminant is \( 9 + 40=49 \), square root 7, then \( x=\frac{3\pm7}{4} \), which is \( \frac{10}{4}=\frac{5}{2} \) or \( \frac{-4}{4}=-1 \), which is option d. But the equation given is \( 2x^2 - 3x + 5 = 0 \). Wait, maybe there's a typo, but assuming the equation is \( 2x^2 - 3x - 5 = 0 \), then discriminant is 49, roots \( \frac{3\pm7}{4} \), which is \( \frac{10}{4}=\frac{5}{2} \) and \( \frac{-4}{4}=-1 \), which is option d. But the original equation is \( +5 \). Wait, maybe the user made a typo. But according to the options, option a: \( \frac{3\pm7i}{4} \), let's check discriminant for option a: \( (7i)^2=-49 \), so \( b^2 - 4ac=-49 \), so \( 9 - 40=-31
eq -49 \). Wait, maybe the equation is \( 2x^2 - 3x + 2 = 0 \)? No. Wait, maybe I messed up the sign of b. Wait, \( b=-3 \), so \( -b = 3 \). So quadratic formula is \( \frac{3\pm\sqrt{9 - 40}}{4}=\frac{3\pm\sqrt{-31}}{4}=\frac{3\pm i\sqrt{31}}{4} \), which is option b. Wait, but option b is \( \frac{3\pm i\sqrt{31}}{4} \), which matches. Wait, earlier I thought option a, but no. Wait, let's re-express: \( \sqrt{-31}=i\sqrt{31} \), so \( x=\frac{3\pm i\sqrt{31}}{4} \), which is option b. Wait, but the user's options: option b is \( \frac{3\pm i\sqrt{31}}{4} \), yes. So the correct answer is option b? Wait, no, wait the original calculation: discriminant is -31, so \( \sqrt{-31}=i\sqrt{31} \), so \( x=\frac{3\pm i\sqrt{31}}{4} \), which is option b. Wait, but earlier I thought option a, but no. Let's confirm:

\( a = 2 \), \( b = -3 \), \( c = 5 \)

Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{9 - 40}}{4} = \frac{3 \pm \sqrt{-31}}{4} = \frac{3 \pm i\sqrt{31}}{4} \), which is option b. Wait, but option b is written as \( \frac{3\pm i\sqrt{31}}{4} \), yes. So the correct option is b? Wait, but let's check option a: \( \frac{3\pm7i}{4} \), 7 squared is 49, so discriminant would be 49, so \( b^2 - 4ac = 49 \), so \( 9 - 40 = -31
eq 49 \). So option b is correct.

Wait, but maybe I made a mistake. Wait, the equation is \( 2x^2 - 3x + 5 = 0 \), so a=2, b=-3, c=5. Then discriminant is 9 - 40 = -31. So square root of -31 is i√31. So x = [3 ± i√31]/4, which is option b. So the correct answer is option b.

Answer:

B. \( \frac{3\pm i\sqrt{31}}{4} \)