Solve the following equations: 3x2 - 5 = 3 and 22x+2 - 2^x+1 + 1 = 2^x

Here are the solutions for the equations in your image. For the first equation: 3x^2 - 5 = 3 Step 1: Isolate the x^2 term. Add 5 to both sides of the equation: 3x^2 = 3 + 5 3x^2 = 8 Step 2: Divide by 3. x^2 = (8)/(3) Step 3: Take the square root of both sides. Remember to include both positive and negative roots. x = ±sqrt((8)/(3)) x = ±sqrt(8)sqrt(3) x = ±2sqrt(2)sqrt(3) Step 4: Rationalize the denominator. Multiply the numerator and denominator by sqrt(3): x = ±2sqrt(2) · sqrt(3)sqrt(3) · sqrt(3) x = ±2sqrt(6)3 The solutions for the first equation are: x=±2sqrt(6)3 For the second equation: 2^2x+2 - 2^x+1 + 1 = 2^x Step 1: Rewrite the terms using exponent rules a^m+n = a^m · a^n and a^mn = (a^m)^n. 2^2x+2 = 2^2x · 2^2 = (2^x)^2 · 4 2^x+1 = 2^x · 2^1 Substitute these into the equation: 4 · (2^x)^2 - 2 · (2^x) + 1 = 2^x Step 2: Let y = 2^x. Substitute y into the equation. Note that for real x, y must be a positive real number (y > 0). 4y^2 - 2y + 1 = y Step 3: Rearrange the equation into the standard quadratic form ay^2+by+c=0. 4y^2 - 2y - y + 1 = 0 4y^2 - 3y + 1 = 0 Step 4: Solve the quadratic equation for y using the quadratic formula y = -b ± sqrt(b^2-4ac)2a. Here, a=4, b=-3, c=1. First, calculate the discriminant = b^2 - 4ac: = (-3)^2 - 4(4)(1) = 9 - 16 = -7 Since the discriminant is negative ( < 0), there are no real solutions for y. Because y = 2^x must be a positive real number, and there are no real values of y that satisfy the quadratic equation, there are no real solutions for x. The second equation has: No real solutions for x Last free one today — make it count tomorrow, or type /upgrade for unlimited.