Question

1. solve for x and check the possible solution/s: √(4 - 3x) + 2 = 7
2. solve for x and check the possible solution/s: √(5x - 6) = x

Explanation:

Problem 3: Solve \(\boldsymbol{\sqrt{4 - 3x}+2 = 7}\) and check the solution

Step 1: Isolate the square root term

Subtract 2 from both sides of the equation to get the square root by itself.
\(\sqrt{4 - 3x}+2 - 2=7 - 2\)
Simplifies to: \(\sqrt{4 - 3x}=5\)

Step 2: Eliminate the square root

Square both sides of the equation to eliminate the square root.
\((\sqrt{4 - 3x})^2 = 5^2\)
Simplifies to: \(4 - 3x = 25\)

Step 3: Solve for \(x\)

Subtract 4 from both sides:
\(4 - 3x - 4=25 - 4\)
\(-3x = 21\)
Divide both sides by \(-3\):
\(x=\frac{21}{-3}=-7\)

Step 4: Check the solution

Substitute \(x = -7\) back into the original equation:
Left - hand side (LHS): \(\sqrt{4 - 3(-7)}+2=\sqrt{4 + 21}+2=\sqrt{25}+2 = 5 + 2=7\)
Right - hand side (RHS): \(7\)
Since \(LHS = RHS\) when \(x=-7\), the solution is valid.

Step 1: Eliminate the square root

Square both sides of the equation to eliminate the square root.
\((\sqrt{5x - 6})^2=x^2\)
Simplifies to: \(5x - 6=x^2\)

Step 2: Rearrange into standard quadratic form

Rearrange the equation to \(ax^2+bx + c = 0\) form.
\(x^2-5x + 6 = 0\)

Step 3: Solve the quadratic equation

Factor the quadratic equation. We need two numbers that multiply to 6 and add to - 5. The numbers are - 2 and - 3.
\(x^2-5x + 6=(x - 2)(x - 3)=0\)
Set each factor equal to zero:
\(x - 2 = 0\) or \(x - 3 = 0\)
So, \(x = 2\) or \(x = 3\)

Step 4: Check the solutions

• Check \(x = 2\):

Substitute \(x = 2\) into the original equation.
LHS: \(\sqrt{5(2)-6}=\sqrt{10 - 6}=\sqrt{4}=2\)
RHS: \(2\)
Since \(LHS = RHS\), \(x = 2\) is a valid solution.

• Check \(x = 3\):

Substitute \(x = 3\) into the original equation.
LHS: \(\sqrt{5(3)-6}=\sqrt{15 - 6}=\sqrt{9}=3\)
RHS: \(3\)
Since \(LHS = RHS\), \(x = 3\) is a valid solution.

Answer:

(Problem 3):
\(x=-7\)

Problem 4: Solve \(\boldsymbol{\sqrt{5x - 6}=x}\) and check the solution