Question

1. $(x + 5)^2-40 = 9$

• add 40 to both sides: $sqrt{(x + 5)^2}=sqrt{49}$
• case 1: $x + 5=7$, then $x=2$ (real - solution)
• case 2: $x + 5=-7$, then $x=-12$ (real - solution)

1. $2x + 3=sqrt{5x + 6}$

• options: a) $x = 1$ b) $-\frac{3}{4}$ c) $\frac{3}{4}$

Explanation:

Step1: Isolate the square - term

Given the equation \((x + 5)^2-40 = 9\). Add 40 to both sides of the equation: \((x + 5)^2=9 + 40=49\).

Step2: Take the square - root of both sides

\(\sqrt{(x + 5)^2}=\pm\sqrt{49}\), which simplifies to \(x + 5=\pm7\).

Step3: Solve for \(x\) when \(x + 5 = 7\)

Subtract 5 from both sides: \(x=7 - 5=2\).

Step4: Solve for \(x\) when \(x + 5=-7\)

Subtract 5 from both sides: \(x=-7 - 5=-12\).

For the second equation \(2x + 3=\sqrt{5x + 6}\):

Step1: Square both sides

\((2x + 3)^2=5x + 6\). Expand the left - hand side using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), so \((2x)^2+2\times(2x)\times3+3^{2}=5x + 6\), which simplifies to \(4x^{2}+12x + 9 = 5x+6\).

Step2: Rearrange to form a quadratic equation

\(4x^{2}+12x-5x + 9 - 6 = 0\), so \(4x^{2}+7x + 3 = 0\).

Step3: Factor the quadratic equation

\(4x^{2}+4x+3x + 3 = 0\), \(4x(x + 1)+3(x + 1)=0\), \((4x + 3)(x + 1)=0\).

Step4: Solve for \(x\)

\(4x+3 = 0\) gives \(x=-\frac{3}{4}\), and \(x + 1 = 0\) gives \(x=-1\). But we need to check for extraneous solutions by substituting back into the original equation \(2x + 3=\sqrt{5x + 6}\).
When \(x = 1\):
Left - hand side: \(2\times1+3=5\).
Right - hand side: \(\sqrt{5\times1+6}=\sqrt{11}\), \(5
eq\sqrt{11}\).
When \(x=-\frac{3}{4}\):
Left - hand side: \(2\times(-\frac{3}{4})+3=-\frac{3}{2}+3=\frac{3}{2}\).
Right - hand side: \(\sqrt{5\times(-\frac{3}{4})+6}=\sqrt{-\frac{15}{4}+6}=\sqrt{\frac{-15 + 24}{4}}=\sqrt{\frac{9}{4}}=\frac{3}{2}\).
When \(x=\frac{3}{4}\):
Left - hand side: \(2\times\frac{3}{4}+3=\frac{3}{2}+3=\frac{9}{2}\).
Right - hand side: \(\sqrt{5\times\frac{3}{4}+6}=\sqrt{\frac{15}{4}+6}=\sqrt{\frac{15 + 24}{4}}=\sqrt{\frac{39}{4}}
eq\frac{9}{2}\).

Answer:

The solutions for \((x + 5)^2-40 = 9\) are \(x = 2\) and \(x=-12\). For \(2x + 3=\sqrt{5x + 6}\), the solution is \(x=-\frac{3}{4}\), so the answer for the multiple - choice part is B. \(-\frac{3}{4}\)