Photo AI
19. Solve the equation: rac{1}{x - rac{1}{2}} = rac{1}{ rac{1}{x} - rac{1}{2}} (1) Expand and rearrange to get the quadratic form - Edexcel - GCSE Maths - Question 19 - 2022 - Paper 1
Question 19
19. Solve the equation: rac{1}{x - rac{1}{2}} = rac{1}{ rac{1}{x} - rac{1}{2}} (1) Expand and rearrange to get the quadratic form. (2) For values of $x... show full transcript
Worked Solution & Example Answer:19. Solve the equation: rac{1}{x - rac{1}{2}} = rac{1}{ rac{1}{x} - rac{1}{2}} (1) Expand and rearrange to get the quadratic form - Edexcel - GCSE Maths - Question 19 - 2022 - Paper 1
Step 1
Solve the equation: rac{1}{x - rac{1}{2}} = rac{1}{ rac{1}{x} - rac{1}{2}}
Answer
To solve the equation, start by using a common denominator on both sides. Rewriting the left side:
egin{align*}
rac{1}{x - rac{1}{2}} &= rac{1}{ rac{2x - 1}{2x}}
&= rac{2x}{2x - 1}.
ext{Therefore, we have: }
\ rac{1}{(x - rac{1}{2}) (2x - 1)} = 1.
ext{Cross-multiplying leads to: }
(x - rac{1}{2}) = (2x - 1).
\text{On simplifying,}
\x - rac{1}{2} - 2x + 1 = 0
\rightarrow -x + rac{1}{2} = 0.
\text{Thus, } x = rac{1}{2}.
ext{Check if any extraneous solutions arise.}
\x = rac{1}{2} ext{ makes the denominator zero, hence is invalid. }
\So, return to quadratic form.
\ ext{Let's rewrite }
\ rac{1}{x - rac{1}{2}} = rac{1}{ rac{1}{x} - rac{1}{2}}.
\ ext{This yields a polynomial expression in } x.
\ ext{We can simplify and equate into } ax^2 + bx + c = 0.
\
Step 2
Expand and rearrange to get the quadratic form.
Answer
For the expanded form, we need to find a common denominator for both sides:
egin{align*}
rac{1}{x - rac{1}{2}} &= rac{2x}{(2x - 1)}.
ext{Cross-multiplying gives: }
\1 imes (2x - 1) = (x - rac{1}{2}) imes 2x
&= 2x^2 - x.
\ ext{Rearranging leads to: } 0 = 2x^2 - 3x + 1.
ext{This means the quadratic equation is } 2x^2 - 3x + 1 = 0.
\
Step 3
For values of $x$, solve $ rac{1}{x} = rac{1}{2}$ and state fractions.
Answer
For this part,
egin{align*}
rac{1}{x} = rac{1}{2}
\Rightarrow x = 2.
ext{Alternatively, solving the quadratic } 2x^2 - 3x + 1 = 0 ext{ via the quadratic formula yields: }
\x = rac{-(-3) ext{±} ext{sqrt}((-3)^2 - 4 imes 2 imes 1)}{2 imes 2} = rac{3 ext{±} ext{sqrt}(9 - 8)}{4}.
\n\text{Thus: } x = rac{3 ext{±} 1}{4}
\Rightarrow x = 1 ext{ or } x = rac{1}{2}.
\ ext{Consolidating results, we find: } x = 2 ext{ and } x = 1 ext{ different fractions.}
\