Question

solve for x, where x is a real number.\\(\sqrt{-19 + 7x} = x - 1\\)if there is more than one solution, separate them with commas. if there is no solution, click on
o solution\.\\(x = \square\\)no solution\\(\sqrt{\square}\\)\\(\sqrt\square{\square}\\)\\(\frac{\square}{\square}\\)\\(\square,\square,...\\)

Explanation:

Step1: Square both sides to eliminate square root

To solve the equation \(\sqrt{-19 + 7x}=x - 1\), we first square both sides. Squaring the left side gives \(-19 + 7x\) and squaring the right side gives \((x - 1)^2=x^{2}-2x + 1\). So we get the equation:

$$ -19 + 7x=x^{2}-2x + 1 $$

Step2: Rearrange into standard quadratic form

Rearrange the equation to \(ax^{2}+bx + c = 0\) form. Move all terms to one side:

$$ x^{2}-2x + 1+19 - 7x = 0 $$

Simplify the like terms:

$$ x^{2}-9x + 20=0 $$

Step3: Factor the quadratic equation

Factor the quadratic \(x^{2}-9x + 20\). We need two numbers that multiply to \(20\) and add up to \(-9\). The numbers are \(-4\) and \(-5\). So,

$$ (x - 4)(x - 5)=0 $$

Step4: Solve for x

Set each factor equal to zero:

• For \(x - 4=0\), we get \(x = 4\)
• For \(x - 5=0\), we get \(x = 5\)

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

• Check \(x = 4\):

Left side: \(\sqrt{-19+7\times4}=\sqrt{-19 + 28}=\sqrt{9} = 3\)
Right side: \(4 - 1=3\). So \(x = 4\) is a valid solution.

• Check \(x = 5\):

Left side: \(\sqrt{-19+7\times5}=\sqrt{-19 + 35}=\sqrt{16}=4\)
Right side: \(5 - 1 = 4\). So \(x = 5\) is also a valid solution.

Answer:

4, 5