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$$\sqrt{2 x+6}+4=x+3$$What is the solution set of the equation above?

SAT · Math · previous-year question

  1. A.$\{-1\}$
  2. B.$\{5\}$correct
  3. C.$\{-1,5\}$
  4. D.$\{0,-1,5\}$

Answer

B. $\{5\}$

Explanation

Choice B is correct. Subtracting 4 from both sides of $\sqrt{2 x+6}+4=x+3$ isolates the radical expression on the left side of the equation as follows: $\sqrt{2 x+6}=x-1$. Squaring both sides of $\sqrt{2 x+6}=x-1$ yields $2 x+6=x^{2}-2 x+1$. This equation can be rewritten as a quadratic equation in standard form: $x^{2}-4 x-5=0$. One way to solve this quadratic equation is to factor the expression $x^{2}-4 x-5$ by identifying two numbers with a sum of -4 and a product of -5 . These numbers are -5 and 1 . So the quadratic equation can be factored as $(x-5)(x+1)=0$. It follows that 5 and -1 are

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