Solve for x square root of 5x+4-1=2x

Problem

√(,5*x+4)−1=2*x

Solution

1. Isolate the radical term by adding 1 to both sides of the equation.

√(,5*x+4)=2*x+1

2. Square both sides of the equation to eliminate the square root.

(√(,5*x+4))^2=(2*x+1)^2

3. Expand the right side using the binomial expansion formula (a+b)^2=a^2+2*a*b+b^2

5*x+4=4*x^2+4*x+1

4. Rearrange the equation into standard quadratic form a*x^2+b*x+c=0 by subtracting 5*x and 4 from both sides.

0=4*x^2−x−3

5. Factor the quadratic equation.

0=(4*x+3)*(x−1)

6. Solve for x by setting each factor equal to zero.

4*x+3=0⇒x=−3/4

x−1=0⇒x=1

7. Check for extraneous solutions by substituting the values back into the original equation.
   For x=1 √(,5*(1)+4)−1=√(,9)−1=3−1=2 Since $2(1) = 2,=1i*s(a)*s(o)*l*u*t*i*o*n.F*o*r = -\frac{3}{4}:sqrt{5(-\frac{3}{4})+4}-1 = \sqrt{-\frac{15}{4}+\frac{16}{4}}-1 = \sqrt{\frac{1}{4}}-1 = \frac{1}{2}-1 = -\frac{1}{2}.S*i*n*c*e(-\frac{3}{4}) = -\frac{3}{2},a*n*d\frac{1}{2} \neq -\frac{3}{2}, = -\frac{3}{4}$ is extraneous.

Final Answer

√(,5*x+4)−1=2*x⇒x=1

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