Math  /  Algebra

Question11x234x+28=11 x^{2}-34 x+28=

Studdy Solution

STEP 1

1. The given expression is a quadratic equation.
2. The goal is to factor the quadratic expression or solve for xx.
3. The quadratic equation can be factored if it has rational roots.

STEP 2

1. Identify the quadratic equation's coefficients aa, bb, and cc.
2. Use the quadratic formula or factorization to solve for xx.
3. Simplify the solutions for xx.

STEP 3

Identify the coefficients aa, bb, and cc in the quadratic equation 11x234x+2811x^2 - 34x + 28. a=11,b=34,c=28 a = 11, \quad b = -34, \quad c = 28

STEP 4

To factor the quadratic equation, we need to find two numbers that multiply to ac=1128=308a \cdot c = 11 \cdot 28 = 308 and add to b=34b = -34. We look for pairs of numbers that multiply to 308 and sum to -34.

STEP 5

Finding the pair: The numbers 22-22 and 14-14 multiply to 308 and add to -34.

STEP 6

Rewrite the middle term of the quadratic equation using the pair found: 11x222x14x+28 11x^2 - 22x - 14x + 28

STEP 7

Group the terms to factor by grouping: (11x222x)+(14x+28) (11x^2 - 22x) + (-14x + 28)

STEP 8

Factor out the common factors in each group: 11x(x2)14(x2) 11x(x - 2) - 14(x - 2)

STEP 9

Factor out the common binomial factor (x2)(x - 2): (x2)(11x14) (x - 2)(11x - 14)

STEP 10

The factored form of the quadratic equation 11x234x+2811x^2 - 34x + 28 is: (x2)(11x14)=0 (x - 2)(11x - 14) = 0

STEP 11

Set each factor to zero and solve for xx: x2=0or11x14=0 x - 2 = 0 \quad \text{or} \quad 11x - 14 = 0

STEP 12

Solve the first equation: x=2 x = 2

STEP 13

Solve the second equation: 11x=14 11x = 14 x=1411 x = \frac{14}{11}

STEP 14

The solutions to the quadratic equation 11x234x+28=011x^2 - 34x + 28 = 0 are: x=2andx=1411 x = 2 \quad \text{and} \quad x = \frac{14}{11}

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