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solve the square root equation \\(\\sqrt{2x + 9} = x + 5\\). (1 point) …

Question

solve the square root equation \\(\sqrt{2x + 9} = x + 5\\). (1 point)
the solution to the square root equation is \\(x = \square\\).

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{2x + 9}=x + 5\), we first square both sides of the equation. Squaring the left side \((\sqrt{2x + 9})^2\) gives \(2x+9\), and squaring the right side \((x + 5)^2\) expands to \(x^2+10x + 25\) using the formula \((a + b)^2=a^2+2ab + b^2\) where \(a=x\) and \(b = 5\). So we get the equation:

$$2x+9=x^2+10x + 25$$

Step2: Rearrange into standard quadratic form

Subtract \(2x\) and \(9\) from both sides to set the equation to zero:

$$x^2+10x+25-(2x + 9)=0$$
$$x^2+10x+25 - 2x-9=0$$

Combine like terms:

$$x^2+(10x-2x)+(25 - 9)=0$$
$$x^2+8x + 16=0$$

Step3: Solve the quadratic equation

Notice that \(x^2+8x + 16\) is a perfect square trinomial, which factors as \((x + 4)^2=0\) (since \((a + b)^2=a^2+2ab + b^2\) with \(a=x\) and \(b = 4\), \(2ab=8x\) and \(b^2 = 16\)). Setting \((x + 4)^2=0\), we take the square root of both sides:

$$x+4=0$$

Subtract 4 from both sides:

$$x=-4$$

Step4: Check the solution

We need to check if \(x=-4\) satisfies the original equation because squaring both sides can introduce extraneous solutions. Substitute \(x=-4\) into the left side: \(\sqrt{2(-4)+9}=\sqrt{-8 + 9}=\sqrt{1}=1\). Substitute into the right side: \(-4 + 5=1\). Since both sides are equal, \(x=-4\) is a valid solution.

Answer:

\(-4\)