Question

(k) \sqrt{2x^2 + 17x} = 3

Explanation:

Step1: Eliminate the square root

To solve the equation \(\sqrt{2x^{2}+17x}=3\), we square both sides of the equation to eliminate the square root. Squaring both sides gives us \((\sqrt{2x^{2}+17x})^{2}=3^{2}\), which simplifies to \(2x^{2}+17x = 9\).

Step2: Rearrange into standard quadratic form

Subtract 9 from both sides of the equation \(2x^{2}+17x = 9\) to get the quadratic equation in standard form \(ax^{2}+bx + c = 0\). So we have \(2x^{2}+17x-9=0\).

Step3: Solve the quadratic equation

We can solve the quadratic equation \(2x^{2}+17x - 9=0\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 2\), \(b = 17\), and \(c=- 9\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(17)^{2}-4\times2\times(-9)=289 + 72=361\). Then, find the roots: \(x=\frac{-17\pm\sqrt{361}}{2\times2}=\frac{-17\pm19}{4}\).
We have two cases:

• Case 1: \(x=\frac{-17 + 19}{4}=\frac{2}{4}=\frac{1}{2}\)
• Case 2: \(x=\frac{-17-19}{4}=\frac{-36}{4}=-9\)

Step4: Check for extraneous solutions

We need to check if these solutions satisfy the original equation because we squared both sides (which can introduce extraneous solutions).

• For \(x=\frac{1}{2}\): Left - hand side of the original equation \(\sqrt{2\times(\frac{1}{2})^{2}+17\times\frac{1}{2}}=\sqrt{2\times\frac{1}{4}+\frac{17}{2}}=\sqrt{\frac{1}{2}+\frac{17}{2}}=\sqrt{\frac{18}{2}}=\sqrt{9}=3\), which is equal to the right - hand side.
• For \(x = - 9\): Left - hand side of the original equation \(\sqrt{2\times(-9)^{2}+17\times(-9)}=\sqrt{2\times81-153}=\sqrt{162 - 153}=\sqrt{9}=3\), which is equal to the right - hand side.

Answer:

\(x=\frac{1}{2}\) or \(x=-9\)