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9 (a) Find the range of values of pp for which 2x2+5x=px−82 x^{2}+5 x=p x-8 is always p...
Jun 26, 2024
Generated Graph
Solution by Steps
step 1
Given the inequality 2 x^{2}+5 x=p x-8>0, we need to solve for pp
step 2
Rearrange the inequality: 2 x^{2} + 5 x - p x + 8 > 0
step 3
Combine like terms: 2 x^{2} + (5 - p) x + 8 > 0
step 4
For the quadratic expression to be always positive, its discriminant must be less than zero: (5 - p)^2 - 4 \cdot 2 \cdot 8 < 0
step 5
Simplify the discriminant inequality: (5 - p)^2 - 64 < 0
step 6
Solve for pp: (5 - p)^2 < 64
step 7
Take the square root of both sides: |5 - p| < 8
step 8
Solve the absolute value inequality: -8 < 5 - p < 8
step 9
Isolate pp: -3 < p < 13
Answer
-3 < p < 13
Part (b)
step 1
Given the curve y=−x2+3x−1y = -x^{2} + 3x - 1, we need to find the equation of the tangent to the curve that is parallel to the xx-axis
step 2
The slope of the tangent line parallel to the xx-axis is 00
step 3
Find the derivative of the curve: ddx(−x2+3x−1)=−2x+3\frac{d}{dx}(-x^{2} + 3x - 1) = -2x + 3
step 4
Set the derivative equal to 00 to find the xx-coordinate where the slope is 00: −2x+3=0-2x + 3 = 0
step 5
Solve for xx: x=32x = \frac{3}{2}
step 6
Substitute x=32x = \frac{3}{2} back into the original equation to find the yy-coordinate: y=−(32)2+3(32)−1y = -\left(\frac{3}{2}\right)^{2} + 3\left(\frac{3}{2}\right) - 1
step 7
Simplify: y=−94+92−1=−94+184−44=54y = -\frac{9}{4} + \frac{9}{2} - 1 = -\frac{9}{4} + \frac{18}{4} - \frac{4}{4} = \frac{5}{4}
step 8
The point of tangency is (32,54)\left(\frac{3}{2}, \frac{5}{4}\right)
step 9
The equation of the tangent line parallel to the xx-axis is y=54y = \frac{5}{4}
Answer
y=54y = \frac{5}{4}
Key Concept
Quadratic Inequality and Tangent Line
Explanation
For part (a), the quadratic expression must have a negative discriminant to be always positive. For part (b), the tangent line parallel to the xx-axis has a slope of zero, and we find the point of tangency by setting the derivative to zero.
Generated Graph
Solution by Steps
step 1
Given the roots of the equation 3x2−5x+9=0 3x^2 - 5x + 9 = 0 are α+1 \alpha + 1 and β+1 \beta + 1 , we need to find the quadratic equation whose roots are α2 \alpha^2 and β2 \beta^2
step 2
Let α+1 \alpha + 1 and β+1 \beta + 1 be the roots of the given equation. Then, the sum of the roots is (α+1)+(β+1)=α+β+2 (\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 and the product of the roots is (α+1)(β+1)=αβ+α+β+1 (\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1
step 3
For the quadratic equation 3x2−5x+9=0 3x^2 - 5x + 9 = 0 , the sum of the roots is −(−5)3=53 \frac{-(-5)}{3} = \frac{5}{3} and the product of the roots is 93=3 \frac{9}{3} = 3
step 4
Therefore, α+β+2=53 \alpha + \beta + 2 = \frac{5}{3} and αβ+α+β+1=3 \alpha\beta + \alpha + \beta + 1 = 3 . Solving these, we get α+β=−13 \alpha + \beta = \frac{-1}{3} and αβ=103 \alpha\beta = \frac{10}{3}
step 5
The quadratic equation with roots α2 \alpha^2 and β2 \beta^2 is x2−(α2+β2)x+α2β2=0 x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0
step 6
Using the identity (α+β)2=α2+β2+2αβ (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta , we get α2+β2=(α+β)2−2αβ=(−13)2−2(103)=19−203=1−609=−599 \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{-1}{3}\right)^2 - 2\left(\frac{10}{3}\right) = \frac{1}{9} - \frac{20}{3} = \frac{1 - 60}{9} = \frac{-59}{9}
step 7
Also, α2β2=(αβ)2=(103)2=1009 \alpha^2\beta^2 = (\alpha\beta)^2 = \left(\frac{10}{3}\right)^2 = \frac{100}{9}
step 8
Therefore, the quadratic equation is x2−(−599)x+1009=0 x^2 - \left(\frac{-59}{9}\right)x + \frac{100}{9} = 0 . Multiplying through by 9 to clear the fractions, we get 9x2+59x+100=0 9x^2 + 59x + 100 = 0
Answer
The quadratic equation is 9x2+59x+100=0 9x^2 + 59x + 100 = 0 .
Key Concept
Finding a new quadratic equation based on transformed roots.
Explanation
We used the relationships between the sums and products of the roots of the original and new equations to derive the new quadratic equation.
Question 2: Simplify 20(42x)(2x−3)(8x+2) \frac{20(4^{2x})}{(2^{x-3})(8^{x+2})} .
step 1
Rewrite the expression using the same base: 42x=(22)2x=24x 4^{2x} = (2^2)^{2x} = 2^{4x} and 8x+2=(23)x+2=23(x+2)=23x+6 8^{x+2} = (2^3)^{x+2} = 2^{3(x+2)} = 2^{3x+6}
step 2
Substitute these into the original expression: 20⋅24x2x−3⋅23x+6 \frac{20 \cdot 2^{4x}}{2^{x-3} \cdot 2^{3x+6}}
step 3
Combine the exponents in the denominator: 2x−3⋅23x+6=2(x−3)+(3x+6)=24x+3 2^{x-3} \cdot 2^{3x+6} = 2^{(x-3) + (3x+6)} = 2^{4x+3}
step 4
Simplify the expression: 20⋅24x24x+3=20⋅24x−(4x+3)=20⋅2−3=20⋅18=208=52 \frac{20 \cdot 2^{4x}}{2^{4x+3}} = 20 \cdot 2^{4x - (4x+3)} = 20 \cdot 2^{-3} = 20 \cdot \frac{1}{8} = \frac{20}{8} = \frac{5}{2}
Answer
The simplified expression is 52 \frac{5}{2} .
Key Concept
Simplifying expressions with exponents.
Explanation
We used properties of exponents to rewrite and simplify the given expression.
Question 3: Find the height of the triangle in the form a−b3 a - b\sqrt{3} given the area and base length.
step 1
Given the area of the triangle is (63−8) cm2 (6\sqrt{3} - 8) \, \text{cm}^2 and the base length is (1+23) cm (1 + 2\sqrt{3}) \, \text{cm} , we use the formula for the area of a triangle: Area=12×base×height \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
step 2
Let the height be h h . Then, (63−8)=12×(1+23)×h (6\sqrt{3} - 8) = \frac{1}{2} \times (1 + 2\sqrt{3}) \times h
step 3
Multiply both sides by 2 to clear the fraction: 2(63−8)=(1+23)×h 2(6\sqrt{3} - 8) = (1 + 2\sqrt{3}) \times h
step 4
Simplify: 123−16=(1+23)×h 12\sqrt{3} - 16 = (1 + 2\sqrt{3}) \times h
step 5
Divide both sides by (1+23) (1 + 2\sqrt{3}) : h=123−161+23 h = \frac{12\sqrt{3} - 16}{1 + 2\sqrt{3}}
step 6
Rationalize the denominator: h=(123−16)(1−23)(1+23)(1−23)=(123−16)(1−23)1−(23)2=(123−16)(1−23)1−12=(123−16)(1−23)−11 h = \frac{(12\sqrt{3} - 16)(1 - 2\sqrt{3})}{(1 + 2\sqrt{3})(1 - 2\sqrt{3})} = \frac{(12\sqrt{3} - 16)(1 - 2\sqrt{3})}{1 - (2\sqrt{3})^2} = \frac{(12\sqrt{3} - 16)(1 - 2\sqrt{3})}{1 - 12} = \frac{(12\sqrt{3} - 16)(1 - 2\sqrt{3})}{-11}
step 7
Expand the numerator: (123−16)(1−23)=123−24⋅3−16+323=123−72−16+323=443−88 (12\sqrt{3} - 16)(1 - 2\sqrt{3}) = 12\sqrt{3} - 24\cdot3 - 16 + 32\sqrt{3} = 12\sqrt{3} - 72 - 16 + 32\sqrt{3} = 44\sqrt{3} - 88
step 8
Simplify: h=443−88−11=−43+8 h = \frac{44\sqrt{3} - 88}{-11} = -4\sqrt{3} + 8
Answer
The height of the triangle is 8−43 cm 8 - 4\sqrt{3} \, \text{cm} .
Key Concept
Using the area formula to find the height of a triangle.
Explanation
We used the given area and base length to solve for the height by rationalizing the denominator.
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