Question

find the approximate solution of this system of equations.
y = x² + 5x + 3
y = √(2x + 5)
a. (2.1, 0.2)
b. (-0.2, 2.1)
c. (-2.1, 0.2)
d. (0.2, 2.1)

Explanation:

Step1: Analyze the domain of the square root function

For \( y = \sqrt{2x + 5} \), the expression inside the square root must be non - negative, so \( 2x+5\geq0\), which gives \( x\geq-\frac{5}{2}=-2.5 \). Also, since the square root of a non - negative number is non - negative, \( y=\sqrt{2x + 5}\geq0 \).

Step2: Test option A \((2.1,0.2)\)

For the function \( y = \sqrt{2x + 5}\), when \( x = 2.1\), \( y=\sqrt{2\times2.1 + 5}=\sqrt{4.2 + 5}=\sqrt{9.2}\approx3.03
eq0.2\). So option A is incorrect.

Step3: Test option B \((- 0.2,2.1)\)

For \( y=x^{2}+5x + 3\), when \( x=-0.2\), \(y=(-0.2)^{2}+5\times(-0.2)+3=0.04-1 + 3=2.04\approx2.1\).
For \( y=\sqrt{2x + 5}\), when \( x = - 0.2\), \(y=\sqrt{2\times(-0.2)+5}=\sqrt{-0.4 + 5}=\sqrt{4.6}\approx2.14\approx2.1\). So this point satisfies both equations approximately.

Step4: Test option C \((-2.1,0.2)\)

For \( y=x^{2}+5x + 3\), when \( x=-2.1\), \(y=(-2.1)^{2}+5\times(-2.1)+3=4.41-10.5 + 3=-3.09
eq0.2\). So option C is incorrect.

Step5: Test option D \((0.2,2.1)\)

For \( y=\sqrt{2x + 5}\), when \( x = 0.2\), \(y=\sqrt{2\times0.2+5}=\sqrt{0.4 + 5}=\sqrt{5.4}\approx2.32
eq2.1\). So option D is incorrect.

Answer:

B. \((-0.2, 2.1)\)