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Zorluk: OrtaRadical and Rational Equations
An equation is shown below.
2x+7=x−4\sqrt{2x + 7} = x - 4
If xx satisfies the equation above, what is the value of x+3x + 3?
  1. A
    4
  2. B
    2
  3. 12Cevap
  4. D
    6

Cevap

12
To solve the equation, square both sides to obtain 2x+7=x2−8x+162x + 7 = x^2 - 8x + 16. Rearranging terms gives the quadratic equation x2−10x+9=0x^2 - 10x + 9 = 0, which factors as (x−9)(x−1)=0(x - 9)(x - 1) = 0. This yields potential solutions of x=9x = 9 and x=1x = 1. Testing these in the original equation shows that x=9x = 9 is a valid solution because 25=5\sqrt{25} = 5, but x=1x = 1 is extraneous because 9≠−3\sqrt{9} \neq -3. Thus, the only real solution is x=9x = 9. Evaluating x+3x + 3 for this solution gives 9+3=129 + 3 = 12.

Adım Adım Çözüm

1
Square both sides of the equation to eliminate the radical.
2x+7=(x−4)22x + 7 = (x - 4)^2, which expands to 2x+7=x2−8x+162x + 7 = x^2 - 8x + 16.
Squaring both sides is the standard method to solve radical equations.
2
Rearrange the terms to set the quadratic equation to zero.
x2−10x+9=0x^2 - 10x + 9 = 0.
This allows us to solve the quadratic equation by factoring.
3
Factor the quadratic equation.
(x−9)(x−1)=0(x - 9)(x - 1) = 0, giving potential solutions of x=9x = 9 and x=1x = 1.
Factoring helps find the roots of the quadratic equation.
4
Verify each potential solution in the original equation to check for extraneous roots.
x=9x = 9 is valid because 2(9)+7=9−4\sqrt{2(9)+7} = 9-4 is true (5=55=5). x=1x = 1 is extraneous because 2(1)+7=1−4\sqrt{2(1)+7} = 1-4 is false (3≠−33 \neq -3).
Squaring both sides can introduce extraneous solutions that do not satisfy the original equation.
5
Substitute the valid solution x=9x = 9 into the expression x+3x + 3.
9+3=129 + 3 = 12.
The question asks for the value of the expression x+3x + 3 rather than just xx.

Anahtar Kavram

Solving radical equations and checking for extraneous solutions
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