Mathematics

Mathematics

Solve for “y” If there is more than one solution, separate them with commas

 

Equation to solve is:

y²−5y−6=0

The solutions are:

y1= −1

y2= 6

Explanation:

STEP1: Find coefficients a , b and c.

In this example a=1,b=−5,c=−6

STEP2: Plug in the values for a, b, and c into the quadratic formula

y1,2 = −b± √(b2−4ac)/2a

 

y1,2=−(−5)± √ (−5)²−4*1*(−6)/2*1

STEP3: Simplify expression under the square root.

Y 1,2=5± √49/2

STEP4: Solve for y

y1=5 + √49/2          =−1

y2=5 − √49/2        =6

 

 

 

 

 

2. Solve for “y” If there is more than one solution, separate them with commas.
   3y2 +13y +4 =0

 

The solutions are:

y1= −4

y2=−13

Explanation:

STEP1: Find coefficients a , b and c.

In this example a=3,b=13,c=4

STEP2: Plug in the values for a, b, and c into the quadratic formula

y1,2 = −b± √(b2−4ac)/2a

y1,2= −13±√(13² −4*3*4)/2*3

STEP3: Simplify expression under the square root.

y1,2=−13±√121/6

STEP4: Solve for y

y1=−13 + √121     = −4

y2=−13 − √121/6 = −1/3

 

3. Fill in the blank to make the expression a perfect square y^2+8y=16

 

The solutions are:

y1 =−4−4√2

y2=−4+4√2

Explanation:

STEP 1: Keep all terms containing x on one side. Move the constant to the right.

y2+8y=16

STEP 2: Take half of the x-term coefficient and square it. Add this value to both sides. In this example we have:

The y-term coefficient = 8

The half of the y-term coefficient = 4

After squaring we have 42=16

When we add 16 to both sides we have:

y2+8y+16=16+16

STEP 3: Simplify right side

y²+8y+16=32

STEP 4: Write the perfect square on the left.

(y+4)2=32

STEP 5: Take the square root of both sides.

y+4= ±√32

STEP 6: Solve for y.

Y = −4± √32

that is,

y1= −4−4√2

y2= −4+4√2

4. Use the quadratic formula to solve for x

 

The solutions are:

 

X1= −5/4−1/4 √33

x2=−5/4+1/4 √33

Explanation:

STEP1: Find coefficients a , b and c.

In this example a=2,b=5,c=−1

STEP2: Plug in the values for a, b, and c into the quadratic formula

x1,2= −b± √(b²−4ac)/2a

x1,2=−5± √(5²−4*2*(−1)/ 2*2

STEP3: Simplify expression under the square root.

x1, 2= −5 √±33/4

STEP4: Solve for x

x1= −5 + √33/4     =−5/4−1/4 √33

x2=−5 − √33/4 =−5/4+1/4   √33

 

 

5. Write the quadratic equation whose roots are and , and whose leading coefficient is

(x-5) (x+2) = 0

4(x-5) (x+2) =0

(4x-20) (4x+8) = 0

Notice that you could divide through by 4

(x-5) (x+2) = 0

 

6. Compute the value of the discriminant and give the number of real solutions of the quadratic equation

-5x^2+9x-3=0

 

 

y = 5x² +9x -3

 

Discriminant        =b²−4 (a) (c)

=9²−4 (5) (3) =21

(2 real solutions)
The Work

 

−b±√{b2−4(a)(c)}/2 (a)

 

−9±√ {9²−4 (5) (3)}/2(5)

−9±√21/10

The Actual Solutions

x=−0.44174243050441603

x=−1.3582575694955838

 

 

7. Graph the parabola.plot the vertex and four additional points, two on each side of the vertex
8. Graph the parabola.

 

9. Graph the solution to the following inequality on the number line.

X² -6x-7 = 0

 

 

 

 

1. Answer the questions below about the quadratic function.

 

Does the function have a minimum or maximum value ?

 

Yes

Where does the minimum or maximum value occur

X-min= -10

X- Max = 10

What is the functions minimum or maximum value ?

Y-Min  = -7.28

Y-Max = 7.28

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