Question

1. 2x = \sqrt{-13x - 10}

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(2x = \sqrt{-13x - 10}\), we first square both sides. Squaring the left side gives \((2x)^2 = 4x^2\), and squaring the right side gives \((\sqrt{-13x - 10})^2=-13x - 10\). So we get the quadratic equation:
\(4x^2=-13x - 10\)

Step2: Rearrange into standard quadratic form

We rearrange the equation \(4x^2=-13x - 10\) to the standard form \(ax^2+bx + c = 0\) by adding \(13x\) and \(10\) to both sides:
\(4x^2 + 13x+ 10=0\)

Step3: Factor the quadratic equation

We factor the quadratic \(4x^2 + 13x + 10\). We need two numbers that multiply to \(4\times10 = 40\) and add up to \(13\). The numbers are \(5\) and \(8\). So we can rewrite the middle term:
\(4x^2+5x + 8x + 10 = 0\)
Group the terms:
\(x(4x + 5)+2(4x + 5)=0\)
Factor out \((4x + 5)\):
\((4x + 5)(x + 2)=0\)

Step4: Solve for x

Set each factor equal to zero:
- For \(4x+5 = 0\), we have \(4x=-5\), so \(x =-\frac{5}{4}\)
- For \(x + 2=0\), we have \(x=-2\)

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because we squared both sides (which can introduce extraneous solutions).
- For \(x =-\frac{5}{4}\):
Left side: \(2\times(-\frac{5}{4})=-\frac{5}{2}\)
Right side: \(\sqrt{-13\times(-\frac{5}{4})-10}=\sqrt{\frac{65}{4}-10}=\sqrt{\frac{65 - 40}{4}}=\sqrt{\frac{25}{4}}=\frac{5}{2}\)
Since \(-\frac{5}{2} eq\frac{5}{2}\), \(x =-\frac{5}{4}\) is extraneous.
- For \(x=-2\):
Left side: \(2\times(-2)=-4\)
Right side: \(\sqrt{-13\times(-2)-10}=\sqrt{26 - 10}=\sqrt{16}=4\)
Since \(-4 eq4\), \(x = - 2\) is extraneous? Wait, that can't be right. Wait, maybe I made a mistake in calculation. Wait, let's re - check the original equation. The right - hand side \(\sqrt{-13x - 10}\) requires that \(-13x-10\geq0\), i.e., \(13x\leq - 10\), \(x\leq-\frac{10}{13}\approx - 0.769\). Also, the left - hand side \(2x\), if \(x\) is a solution, since the square root is non - negative, \(2x\geq0\), so \(x\geq0\). But our solutions \(x =-\frac{5}{4}\) and \(x=-2\) are both negative, which contradicts \(x\geq0\). Wait, maybe I made a mistake in factoring. Let's use the quadratic formula for \(4x^2+13x + 10 = 0\), where \(a = 4\), \(b = 13\), \(c = 10\). The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\)
\(x=\frac{-13\pm\sqrt{13^2-4\times4\times10}}{2\times4}=\frac{-13\pm\sqrt{169 - 160}}{8}=\frac{-13\pm3}{8}\)
So \(x=\frac{-13 + 3}{8}=\frac{-10}{8}=-\frac{5}{4}\) and \(x=\frac{-13 - 3}{8}=\frac{-16}{8}=-2\)
Wait, but the domain of the original function: \(\sqrt{-13x - 10}\) requires \(-13x-10\geq0\Rightarrow x\leq-\frac{10}{13}\approx - 0.769\), and \(2x=\sqrt{\text{non - negative}}\), so \(2x\geq0\Rightarrow x\geq0\). The intersection of \(x\leq-\frac{10}{13}\) and \(x\geq0\) is empty? That means there is no solution? Wait, maybe I made a mistake in the sign when squaring. Wait, the original equation is \(2x=\sqrt{-13x - 10}\). The left - hand side \(2x\) must be non - negative (because the square root is non - negative), so \(2x\geq0\Rightarrow x\geq0\). The right - hand side \(\sqrt{-13x - 10}\) requires \(-13x-10\geq0\Rightarrow x\leq-\frac{10}{13}\approx - 0.769\). There is no number \(x\) that is both \(\geq0\) and \(\leq-\frac{10}{13}\). So the equation has no solution. Wait, but let's check my factoring again. Wait, \(4x^2+13x + 10\): discriminant \(D = 13^2-4\times4\times10=169 - 160 = 9\). So roots are \(\frac{-13\pm3}{8}\), which are \(\frac{-10}{8}=-\frac{5}{4}\) and \(\frac{-16}{8}=-2\). But as we saw, both do not satisfy the original equation because of the domain restrictions. So the equation has no solution.

Answer:

The equation \(2x=\sqrt{-13x - 10}\) has no solution.