example 1 solve an equation with two real solutions solve x 2 + 3x = 2. x 2 + 3x = 2 write original...
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EXAMPLE 1 Solve an equation with two real solutions
Solve x2 + 3x = 2.x2 + 3x = 2 Write original equation.
x2 + 3x – 2 = 0 Write in standard form.
x =– b + b2 – 4ac2a
Quadratic formula
x =– 3 + 32 – 4(1)(–2) 2(1)
a = 1, b = 3, c = –2
Simplify.x = – 3 + 172
The solutions are x = – 3 + 172
0.56 andx = – 3 – 17
2– 3.56.
ANSWER
EXAMPLE 1 Solve an equation with two real solutions
CHECK
Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about – 3.56.
EXAMPLE 2 Solve an equation with one real solutions
Solve 25x2 – 18x = 12x – 9.25x2 – 18x = 12x – 9. Write original equation.
Write in standard form.
x =30 + (–30)2– 4(25)(9)2(25)
a = 25, b = –30, c = 9
Simplify.
25x2 – 30x + 9 = 0.
x =30 + 0
50
x = 35 Simplify.
35The solution is
ANSWER
EXAMPLE 2 Solve an equation with one real solutions
CHECK
Graph y = –5x2 – 30x + 9 and note that the only x-intercept is 0.6 = .3
5
EXAMPLE 3 Solve an equation with imaginary solutions
Solve –x2 + 4x = 5.–x2 + 4x = 5 Write original equation.
Write in standard form.
x =– 4+ 42– 4(– 1)(– 5)2(– 1)
a = –1, b = 4, c = –5
Simplify.
–x2 + 4x – 5 = 0.
x =– 4+ – 4
– 2– 4+ 2i
x = – 2
Simplify.
Rewrite using the imaginary unit i.
x = 2 + i
The solution is 2 + i and 2 – i.
ANSWER
EXAMPLE 3 Solve an equation with imaginary solutions
CHECK
Graph y = 2x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + i is shown.
–(2 + i)2 + 4(2 + i) = 5?
–3 – 4i + 8 + 4i = 5?
5 = 5
EXAMPLE 4 Use the discriminant
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
a. x2 – 8x + 17 = 0 b. x2 – 8x + 16 = 0 c. x2 – 8x + 15 = 0
SOLUTION
Equation Discriminant Solution(s)
ax2 + bx + c = 0 b2 – 4ac x =– b+ b2– 4ac2ac
a. x2 – 8x + 17 = 0 (–8)2 – 4(1)(17) = – 4 Two imaginary: 4 + i
b. x2 – 8x + 16 = 0 (–8)2 – 4(1)(16) = 0 One real: 4
b. x2 – 8x + 15 = 0 (–8)2 – 4(1)(15) = 0 Two real: 3,5
EXAMPLE 1 Graph a quadratic inequality
Graph y > x2 + 3x – 4.
SOLUTION
STEP 1
Graph y = x2 + 3x – 4. Because the inequality symbol is >, make the parabola dashed.
Test a point inside the parabola, such as (0, 0).
STEP 2
y > x2 + 3x – 4
0 > 02 + 3(0) – 4?
0 > – 4
EXAMPLE 1 Graph a quadratic inequality
So, (0, 0) is a solution of the inequality.
STEP 3
Shade the region inside the parabola.
EXAMPLE 2 Use a quadratic inequality in real life
A manila rope used for rappelling down a cliff can safely support a weight W (in pounds) provided
Rappelling
W ≤ 1480d2
where d is the rope’s diameter (in inches). Graph the inequality.
SOLUTION
Graph W = 1480d2 for nonnegative values of d. Because the inequality symbol is ≤, make the parabola solid. Test a point inside the parabola, such as (1, 2000).
EXAMPLE 2 Use a quadratic inequality in real life
W ≤ 1480d2
2000 ≤ 1480(1)2
2000 ≤ 1480
Because (1, 2000) is not a solution, shade the region below the parabola.
EXAMPLE 3 Graph a system of quadratic inequalities
Graph the system of quadratic inequalities.
y < – x2 + 4 Inequality 1
y > x2 – 2x – 3 Inequality 2
SOLUTION
STEP 1
Graph y ≤ – x2 + 4. The graph is the red region inside and including the parabola y = – x2 + 4.
EXAMPLE 3 Graph a system of quadratic inequalities
STEP 2
Graph y > x2– 2x – 3. The graph is the blue region inside (but not including) the parabola y = x2 –2x – 3.
Identify the purple region where the two graphs overlap. This region is the graph of the system.
STEP 3
EXAMPLE 4 Solve a quadratic inequality using a table
Solve x2 + x ≤ 6 using a table.
SOLUTION
Rewrite the inequality as x2 + x – 6 ≤ 0. Then make a table of values.
Notice that x2 + x –6 ≤ 0 when the values of x are between –3 and 2, inclusive.
The solution of the inequality is –3 ≤ x ≤ 2.
ANSWER
EXAMPLE 5 Solve a quadratic inequality by graphing
Solve 2x2 + x – 4 ≥ 0 by graphing.
SOLUTION
The solution consists of the x-values for which the graph of y = 2x2 + x – 4 lies on or above the x-axis. Find the graph’s x-intercepts by letting y = 0 and using the quadratic formula to solve for x.
0 = 2x2 + x – 4
x =– 1+ 12– 4(2)(– 4)2(2)
x =– 1+ 33
4x 1.19 or x –1.69
EXAMPLE 5 Solve a quadratic inequality by graphing
Sketch a parabola that opens up and has 1.19 and –1.69 as x-intercepts. The graph lies on or above the x-axis to the left of (and including) x = – 1.69 and to the right of (and including) x = 1.19.
The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19.
ANSWER
EXAMPLE 1 Write a quadratic function in vertex form
Write a quadratic function for the parabola shown.
SOLUTION
Use vertex form because the vertex is given.
y = a(x – h)2 + k Vertex form
y = a(x – 1)2 – 2 Substitute 1 for h and –2 for k.
Use the other given point, (3, 2), to find a.2 = a(3 – 1)2 – 2 Substitute 3 for x and 2 for y.
2 = 4a – 2 Simplify coefficient of a.
1 = a Solve for a.
EXAMPLE 1 Write a quadratic function in vertex form
A quadratic function for the parabola is y = (x – 1)2 – 2.
ANSWER
EXAMPLE 2 Write a quadratic function in intercept form
Write a quadratic function for the parabola shown.
SOLUTION
Use intercept form because the x-intercepts are given.
y = a(x – p)(x – q) Intercept form
y = a(x + 1)(x – 4) Substitute –1 for p and 4 for q.
EXAMPLE 2 Write a quadratic function in intercept form
Use the other given point, (3, 2), to find a.
2 = a(3 + 1)(3 – 4) Substitute 3 for x and 2 for y.
2 = – 4a Simplify coefficient of a.
Solve for a.12
– = a
A quadratic function for the parabola is
12
– (x + 1)(x – 4) .y =
ANSWER
EXAMPLE 3 Write a quadratic function in standard form
Write a quadratic function in standard form for the parabola that passes through the points (–1, –3), (0, – 4), and (2, 6).
SOLUTION
STEP 1
Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.
EXAMPLE 3 Write a quadratic function in standard form
–3 = a(–1)2 + b(–1) + c Substitute –1 for x and 23 for y.
–3 = a – b + c Equation 1
–3 = a(0)2 + b(0) + c Substitute 0 for x and – 4 for y.
– 4 = c Equation 2
6 = a(2)2 + b(2) + c Substitute 2 for x and 6 for y.
6 = 4a + 2b + c Equation 3
Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 4 for c in Equations 1 and 3.
STEP 2
EXAMPLE 3 Write a quadratic function in standard form
a – b + c = – 3 Equation 1
a – b – 4 = – 3 Substitute – 4 for c.
a – b = 1 Revised Equation 1
4a + 2b + c = 6 Equation 3
4a + 2b + 4 = 6 Substitute – 4 for c.
4a + 2b = 10 Revised Equation 3
STEP 3Solve the system consisting of revised Equations 1 and 3. Use the elimination method.
EXAMPLE 3 Write a quadratic function in standard form
a – b = 1 2a – 2b = 2
4a + 2b = 10 4a + 2b = 10
6a = 12
a = 2
So 2 – b = 1, which means b = 1.
The solution is a = 2, b = 1, and c = – 4.
A quadratic function for the parabola is y = 2x2 + x – 4.
ANSWER