What are the solutions to the equation 1+3x-(3/4 x+5/4)^2=0 ?

The solutions to the equation 1+3x-(3/4 x+5/4)^2=0 are x=1

Find the zeros of 1+3x-(3/4 x+5/4)^2=0

The zeros of 1+3x-(3/4 x+5/4)^2=0 are x=1

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1 + 3 x - (\frac{3}{4} x + \frac{5}{4})^{2} = 0

x = 1

Solution steps

1 + 3 x - (\frac{3}{4} x + \frac{5}{4})^{2} = 0

Expand

1 + 3 x - (\frac{3}{4} x + \frac{5}{4})^{2} : - \frac{9}{16} - \frac{9}{16} x^{2} + \frac{9}{8} x

- \frac{9}{16} - \frac{9}{16} x^{2} + \frac{9}{8} x = 0

Write in the standard form

a x^{2} + b x + c = 0

- \frac{9 x ^{2}}{16} + \frac{9 x}{8} - \frac{9}{16} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- \frac{9}{8} \pm ( \frac{9}{8} ) ^{2} - 4 ( - \frac{9}{16} ) ( - \frac{9}{16} )}{2 ( - \frac{9}{16} )}

(\frac{9}{8})^{2} - 4 (- \frac{9}{16}) (- \frac{9}{16}) = 0

x_{1, 2} = \frac{- \frac{9}{8} \pm 0}{2 ( - \frac{9}{16} )}

x = \frac{- \frac{9}{8}}{2 ( - \frac{9}{16} )}

\frac{- \frac{9}{8}}{2 ( - \frac{9}{16} )} = 1

x = 1

The solution to the quadratic equation is:

x = 1

(\frac{5 a ^{- 5}}{3 b ^{- \frac{1}{6}}})^{- 1}

\frac{6}{2 π}

f = \frac{( 26 - 2 )}{3}

∣ 3 x + 7 ∣ \leq 0

81 x^{2} + 180 x + 100

What are the solutions to the equation 1+3x-(3/4 x+5/4)^2=0 ?
The solutions to the equation 1+3x-(3/4 x+5/4)^2=0 are x=1
Find the zeros of 1+3x-(3/4 x+5/4)^2=0
The zeros of 1+3x-(3/4 x+5/4)^2=0 are x=1