For what values of x does the following hold true: a. (5x+3)^2=5(x+3) b. (3x+10)^2=3(x+10) c. (3x−8)^2=3x^2−8x

a. x ∈ {-1.2, 0.2}

b. x ∈ {-4 2/3, -1 2/3}

c. x ∈ {2 2/3, 4}

For this sort of exercise, I find a graphing calculator to be very handy. While the one shown in the attachment (Desmos) can solve the equation as written, I find it convenient to recast the equation to the form f(x) = 0. The calculator finds the values of zero crossings very nicely. Some, like my TI-84, will show the value to 8 or 10 significant digits. Here, 4 digits is sufficient to determine the exact solution.

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If you want to work these by hand, you rewrite them to standard form, then use factoring, completing the square, or the quadratic formula to solve them.

a. 25x² +30x +9 -5x -15 = 0 . . . . subtract the right side

... 25x² +25x -6 = 0 . . . . . . . . . . . collect terms

... (5x+6)(5x-1) = 0 . . . . . . . . . . . . . factor (the graphing calc helps here)

... x = -6/5, 1/5

b. 9x² +60x +100 -3x -30 = 0 . . . . subtract the right side

... 9x² +57x +70 = 0 . . . . . . . . . . . . collect terms

... (3x +14)(3x +5) = 0 . . . . . . . . . . . factor

... x = -14/3, -5/3

c. 9x² -48x +64 -3x² +8x = 0 . . . . . subtract the right side

... 6x² -40x +64 = 0 . . . . . . . . . . . . collect terms

... 3x² -20x +32 = 0 . . . . . . . . . . . . factor out 2

... (3x -8)(x -4) = 0 . . . . . . . . . . . . . . factor

... x = 8/3, 4

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: