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Solve the equation 
2x^(2)+15 x+35=x^(2) to the nearest tenth.
Answer: 
x=

Solve the equation 2x2+15x+35=x2 2 x^{2}+15 x+35=x^{2} to the nearest tenth.\newlineAnswer: x= x=

Full solution

Q. Solve the equation 2x2+15x+35=x2 2 x^{2}+15 x+35=x^{2} to the nearest tenth.\newlineAnswer: x= x=
  1. Move Terms, Set Equal: First, we need to move all terms to one side of the equation to set it equal to zero. We will subtract x2x^2 from both sides of the equation.\newline2x2+15x+35x2=x2x22x^2 + 15x + 35 - x^2 = x^2 - x^2\newlineThis simplifies to:\newlinex2+15x+35=0x^2 + 15x + 35 = 0
  2. Factor Quadratic Equation: Next, we need to factor the quadratic equation if possible. We look for two numbers that multiply to 3535 and add up to 1515. The numbers 55 and 77 fit this requirement.\newline(x+5)(x+7)=0(x + 5)(x + 7) = 0
  3. Apply Zero Product Property: Now, we apply the zero product property which states that if a product of two factors is zero, then at least one of the factors must be zero.\newlineSo, we set each factor equal to zero:\newlinex+5=0x + 5 = 0 or x+7=0x + 7 = 0
  4. Solve for x: We solve each equation for x:\newlineFor x+5=0x + 5 = 0, we subtract 55 from both sides:\newlinex=5x = -5\newlineFor x+7=0x + 7 = 0, we subtract 77 from both sides:\newlinex=7x = -7\newlineThese are the two solutions to the equation.

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