EBUS PC11 Chapter 3 Quadratic Functions Test-2

Step 1: Set the two equations equal to find the intersection points: \(x^2-4x+5=-x^2+6x-3\). Step 2: Move all terms to one side: \(2x^2-10x+8=0\). Step 3: Divide by \(2\): \(x^2-5x+4=0\). Step 4: Factor: \((x-1)(x-4)=0\). Step 5: The solutions are \(x=1\) and \(x=4\). Step 6: Substitute into either equation: when \(x=1\), \(y=2\); when \(x=4\), \(y=5\). Step 7: The correct graph shows intersections at \((1,2)\) and \((4,5)\). Answer: Graph A

Step 1: Factor \(4x^2+5x-9\). Step 2: Find two terms whose product is \(4x^2\cdot(-9)=-36x^2\) and whose sum is \(5x\). Step 3: Use \(9x\) and \(-4x\): \(4x^2+9x-4x-9=0\). Step 4: Factor by grouping: \(x(4x+9)-1(4x+9)=0\). Step 5: Factor: \((4x+9)(x-1)=0\). Step 6: Solve: \(4x+9=0\Rightarrow x=-\frac{9}{4}\), and \(x-1=0\Rightarrow x=1\). Answer: \(1, -\frac{9}{4}\)

Step 1: Move all terms to one side: \(x^2+4x-16+2x=0\). Step 2: Combine like terms: \(x^2+6x-16=0\). Step 3: Factor the quadratic: \((x+8)(x-2)=0\). Step 4: Solve each factor: \(x+8=0\Rightarrow x=-8\), and \(x-2=0\Rightarrow x=2\). Answer: \(x=2, -8\)

Step 1: Start with \(x^2+20x+15=0\). Step 2: Move the constant: \(x^2+20x=-15\). Step 3: Take half of \(20\), which is \(10\), and square it: \(100\). Step 4: Add \(100\) to both sides: \(x^2+20x+100=85\). Step 5: Write the left side as a square: \((x+10)^2=85\). Step 6: Take the square root: \(x+10=\pm\sqrt{85}\). Step 7: Solve: \(x=-10\pm\sqrt{85}\). Answer: \(x=-10\pm\sqrt{85}\)

Step 1: Start with \(4x^2+8x+1=0\). Step 2: Divide by \(4\): \(x^2+2x+\frac{1}{4}=0\). Step 3: Move the constant: \(x^2+2x=-\frac{1}{4}\). Step 4: Complete the square by adding \(1\) to both sides: \(x^2+2x+1=\frac{3}{4}\). Step 5: Write the left side as \((x+1)^2=\frac{3}{4}\). Step 6: Take the square root: \(x+1=\pm\frac{\sqrt{3}}{2}\). Step 7: Solve: \(x=-1\pm\frac{\sqrt{3}}{2}=\frac{-2\pm\sqrt{3}}{2}\). Answer: \(x=\frac{-2\pm\sqrt{3}}{2}\)

Step 1: Identify \(a=4\), \(b=-3\), and \(c=-2\). Step 2: Use the quadratic formula: \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Step 3: Substitute: \(x=\frac{-(-3)\pm\sqrt{(-3)^2-4(4)(-2)}}{2(4)}\). Step 4: Simplify inside the radical: \(9+32=41\). Step 5: Therefore \(x=\frac{3\pm\sqrt{41}}{8}\). Answer: \(\frac{3\pm\sqrt{41}}{8}\)

Step 1: Clear the fraction by multiplying the equation by \(2\): \(2x^2-4=-7x\). Step 2: Move all terms to one side: \(2x^2+7x-4=0\). Step 3: Identify \(a=2\), \(b=7\), and \(c=-4\). Step 4: Use the quadratic formula: \(x=\frac{-7\pm\sqrt{7^2-4(2)(-4)}}{2(2)}\). Step 5: Simplify the discriminant: \(49+32=81\). Step 6: \(\sqrt{81}=9\). Step 7: Therefore \(x=\frac{-7\pm9}{4}\). Answer: \(\frac{-7\pm9}{4}\)

Step 1: A quadratic has two equal real roots when its discriminant equals zero. Step 2: For \(x^2+kx+4=0\), we have \(a=1\), \(b=k\), and \(c=4\). Step 3: Set the discriminant equal to zero: \(b^2-4ac=0\). Step 4: Substitute: \(k^2-4(1)(4)=0\). Step 5: Simplify: \(k^2-16=0\). Step 6: Solve: \(k^2=16\), so \(k=\pm4\). Answer: \(k=\pm4\)

Step 1: Write the rental cost as \(C=50+12.50d\). Step 2: The cost must be less than \(200\), so \(50+12.50d<200\). Step 3: Subtract \(50\): \(12.50d<150\). Step 4: Divide by \(12.50\): \(d<12\). Answer: \(d<12\) days

Step 1: Start with \(6(8-x)\leq12\). Step 2: Divide both sides by \(6\): \(8-x\leq2\). Step 3: Subtract \(8\): \(-x\leq-6\). Step 4: Divide by \(-1\), remembering to reverse the inequality sign: \(x\geq6\). Answer: \(x\geq6\)

Step 1: Move all terms to one side: \(x^2-5x+4\leq0\). Step 2: Factor: \((x-1)(x-4)\leq0\). Step 3: Find the critical values: \(x=1\) and \(x=4\). Step 4: Since the parabola opens upward, the expression is less than or equal to zero between the roots. Step 5: Include the endpoints because the inequality is \(\leq\). Answer: \(1\leq x\leq4\)

Step 1: Move all terms to one side: \(x^2+12x+39\geq0\). Step 2: Calculate the discriminant: \(\Delta=b^2-4ac=12^2-4(1)(39)=144-156=-12\). Step 3: Since \(\Delta0\), the parabola opens upward and is always above the \(x\)-axis. Step 5: Therefore the inequality is true for every real number. Answer: All real numbers