Question

stion
ve for all possible values of x.
\\(\sqrt{11 - 10x} = x - 2\\)
answer attempt 1 out of 2
additional solution no solution
x =

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{11 - 10x}=x - 2\), we first square both sides. Squaring the left side gives us \(11-10x\) (since \((\sqrt{a})^2=a\) for \(a\geq0\)), and squaring the right side gives us \((x - 2)^2\). So we have the equation:

$$11-10x=(x - 2)^2$$

Step2: Expand the right - hand side

We know that \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\) and \(b = 2\). So \((x - 2)^2=x^{2}-4x + 4\). Our equation becomes:

$$11-10x=x^{2}-4x + 4$$

Step3: Rearrange the equation to form a quadratic equation

Move all terms to one side of the equation. Subtract \(11\) and add \(10x\) to both sides:

$$0=x^{2}-4x + 4-11 + 10x$$

Simplify the right - hand side:

$$x^{2}+6x - 7 = 0$$

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(-7\) and add up to \(6\). The numbers are \(7\) and \(-1\). So we can factor the quadratic as:

$$(x + 7)(x - 1)=0$$

Step5: Solve for \(x\) using the zero - product property

If \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we have two cases:

• Case 1: \(x+7 = 0\), then \(x=-7\)
• Case 2: \(x - 1=0\), then \(x = 1\)

Step6: Check the solutions in the original equation

• For \(x=-7\):

The left - hand side of the original equation is \(\sqrt{11-10\times(-7)}=\sqrt{11 + 70}=\sqrt{81}=9\)
The right - hand side is \(-7-2=-9\)
Since \(9
eq - 9\), \(x=-7\) is an extraneous solution.

• For \(x = 1\):

The left - hand side of the original equation is \(\sqrt{11-10\times1}=\sqrt{11 - 10}=1\)
The right - hand side is \(1-2=-1\)
Since \(1
eq - 1\), wait, there is a mistake above. Wait, when \(x = 1\), \(\sqrt{11-10x}=\sqrt{11 - 10}=1\), and \(x - 2=1-2=-1\), \(1
eq - 1\). Wait, we made a mistake in factoring? Wait, let's go back to step 3.
Wait, when we moved the terms:

$$11-10x=x^{2}-4x + 4$$

Subtract \(11\) and add \(10x\) to both sides:

$$0=x^{2}-4x + 4-11 + 10x$$

$$x^{2}+6x-7 = 0$$

is correct. Factoring: \(x^{2}+6x - 7=(x + 7)(x - 1)\) is correct. But let's check the domain of the square root. The expression inside the square root must be non - negative, so \(11-10x\geq0\), which implies \(x\leq\frac{11}{10}=1.1\). Also, the right - hand side \(x - 2\) must be non - negative (because the square root of a number is non - negative) since \(\sqrt{11 - 10x}\geq0\), so \(x-2\geq0\), which implies \(x\geq2\). But our solutions \(x=-7\) and \(x = 1\) do not satisfy \(x\geq2\). So there is no solution? Wait, let's re - do the squaring step.
Wait, the square root \(\sqrt{11-10x}\) is non - negative, so the right - hand side \(x - 2\) must also be non - negative, so \(x-2\geq0\Rightarrow x\geq2\). But if \(x\geq2\), then \(11-10x\leq11-20=-9\), and the square root of a negative number is not a real number. So there is a contradiction. So the equation \(\sqrt{11 - 10x}=x - 2\) has no solution.

Wait, let's check the squaring step again. When we square both sides of an equation \(A = B\), we get \(A^{2}=B^{2}\), but the solutions of \(A^{2}=B^{2}\) are the solutions of \(A = B\) or \(A=-B\). In our case, \(A=\sqrt{11 - 10x}\) and \(B=x - 2\). So \(A^{2}=B^{2}\) is \((\sqrt{11 - 10x})^{2}=(x - 2)^{2}\), which is \(11-10x=x^{2}-4x + 4\), but we also have the condition that \(A\geq0\) and \(B\) must satisfy \(A = B\), so \(B\geq0\) (since \(A\geq0\)). But as we saw, if \(B=x - 2\geq0\), then \(x\geq2\), and if \(x\geq2\), then \(11-10x\leq11 - 20=-9<0\), so \(\sqrt{11 - 10x}\) is not a real number. So there is no solution.

Answer:

No Solution