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Zorluk: KolayQuadratic Equations and Polynomial Factoring

What is the sum of all real solutions to the equation x(x−3)=10x(x - 3) = 10?

  1. 3Cevap
  2. B
    5
  3. C
    7
  4. D
    -3
  5. E
    10

Cevap

The sum of all real solutions to the equation is 3.
Expanding the equation x(x−3)=10x(x - 3) = 10 yields x2−3x−10=0x^2 - 3x - 10 = 0. Factoring this expression gives (x−5)(x+2)=0(x - 5)(x + 2) = 0, yielding solutions x=5x = 5 and x=−2x = -2. The sum of these two real solutions is 5+(−2)=35 + (-2) = 3. Alternatively, applying Vieta's formula directly to x2−3x−10=0x^2 - 3x - 10 = 0 gives the sum of roots as −ba=−−31=3-\frac{b}{a} = -\frac{-3}{1} = 3.

Adım Adım Çözüm

1
Expand the left side of the equation.
x2−3x=10x^2 - 3x = 10
Multiply xx across the binomial (x−3)(x - 3) to convert to standard polynomial form.
2
Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−3x−10=0x^2 - 3x - 10 = 0
Subtract 10 from both sides so the equation equals zero.
3
Factor the quadratic expression.
(x−5)(x+2)=0(x - 5)(x + 2) = 0
Find two numbers that multiply to −10-10 and add up to −3-3, which are −5-5 and 22.
4
Solve for the roots of the equation.
x=5x = 5 or x=−2x = -2
Apply the zero-product property: set each factor equal to zero.
5
Calculate the sum of the solutions.
Sum =5+(−2)=3= 5 + (-2) = 3
Add the two real solutions together. Alternatively, by Vieta's formulas, the sum of roots is −ba=−−31=3-\frac{b}{a} = -\frac{-3}{1} = 3.

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Quadratic Equations and Polynomial Factoring
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