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Zorluk: OrtaSolving Quadratic Equations by Factoring

For each quadratic equation on the left, solve for xx by factoring and match it to its correct solution set on the right.

  • 2x2+5x=32x^2 + 5x = 3{−3,12}\{-3, \frac{1}{2}\}
  • 3x2−10x=−83x^2 - 10x = -8{43,2}\{\frac{4}{3}, 2\}
  • x(x−4)=12x(x - 4) = 12\{-2, 6\}

Cevap

The equation 2x2+5x=32x^2 + 5x = 3 matches the solution set {−3,12}\{-3, \frac{1}{2}\}; the equation 3x2−10x=−83x^2 - 10x = -8 matches the solution set {43,2}\{\frac{4}{3}, 2\}; and the equation x(x−4)=12x(x - 4) = 12 matches the solution set {−2,6}\{-2, 6\}.
Each of the quadratic equations can be solved by first rearranging the terms to set the equation equal to zero. After rewriting them in the standard form ax2+bx+c=0ax^2 + bx + c = 0, they can be factored into a product of linear binomials. Setting each factor equal to zero and solving for xx yields the solutions. Specifically: 2x2+5x=32x^2 + 5x = 3 simplifies to 2x2+5x−3=02x^2 + 5x - 3 = 0, which factors as (2x−1)(x+3)=0(2x - 1)(x + 3) = 0 and gives the solution set {−3,12}\{-3, \frac{1}{2}\}. 3x2−10x=−83x^2 - 10x = -8 simplifies to 3x2−10x+8=03x^2 - 10x + 8 = 0, which factors as (3x−4)(x−2)=0(3x - 4)(x - 2) = 0 and gives the solution set {43,2}\{\frac{4}{3}, 2\}. x(x−4)=12x(x - 4) = 12 simplifies to x2−4x−12=0x^2 - 4x - 12 = 0, which factors as (x−6)(x+2)=0(x - 6)(x + 2) = 0 and gives the solution set {−2,6}\{-2, 6\}.

Adım Adım Çözüm

1
Rearrange the equation 2x2+5x=32x^2 + 5x = 3 into the standard form ax2+bx+c=0ax^2 + bx + c = 0.
2x2+5x−3=02x^2 + 5x - 3 = 0
To apply factoring and the zero product property, the quadratic expression must equal zero.
2
Factor the trinomial 2x2+5x−32x^2 + 5x - 3.
(2x−1)(x+3)=0(2x - 1)(x + 3) = 0
Since the constant term is negative and the middle coefficient is positive, the factors must have opposite signs.
3
Apply the zero product property to find the solutions for the first equation.
x=12x = \frac{1}{2} and x=−3x = -3, yielding the set {−3,12}\{-3, \frac{1}{2}\}
Setting the individual linear factors 2x−12x - 1 and x+3x + 3 to zero gives the solutions.
4
Rearrange the equation 3x2−10x=−83x^2 - 10x = -8 into standard form.
3x2−10x+8=03x^2 - 10x + 8 = 0
Add 88 to both sides to set the right side to zero.
5
Factor the trinomial 3x2−10x+83x^2 - 10x + 8.
(3x−4)(x−2)=0(3x - 4)(x - 2) = 0
The constant term is positive and the middle coefficient is negative, meaning both constant terms in the binomial factors must be negative.
6
Solve for xx using the zero product property.
x=43x = \frac{4}{3} and x=2x = 2, yielding the set {43,2}\{\frac{4}{3}, 2\}
Setting 3x−4=03x - 4 = 0 gives x=43x = \frac{4}{3}, and setting x−2=0x - 2 = 0 gives x=2x = 2.
7
Expand and rearrange the equation x(x−4)=12x(x - 4) = 12 into standard form.
x2−4x−12=0x^2 - 4x - 12 = 0
Distribute the xx on the left side to get x2−4xx^2 - 4x and subtract 1212 from both sides to set the equation to zero.
8
Factor the trinomial x2−4x−12x^2 - 4x - 12.
(x−6)(x+2)=0(x - 6)(x + 2) = 0
Find two numbers that multiply to −12-12 and add to −4-4. Those numbers are −6-6 and +2+2.
9
Solve for xx using the zero product property.
x=6x = 6 and x=−2x = -2, yielding the set {−2,6}\{-2, 6\}
Setting x−6=0x - 6 = 0 gives x=6x = 6, and setting x+2=0x + 2 = 0 gives x=−2x = -2.

Anahtar Kavram

Solving quadratic equations by rewriting them in standard form, factoring the trinomials, and using the zero product property.
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Konular

Solving Quadratic Equations by Factoring

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