How do you solve $12 x^{2} + 2 = 11 x$ $12x ^ { 2} + 2= 11x$ ?

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How do you solve $12 x^{2} + 2 = 11 x$ $12x ^ { 2} + 2= 11x$ ?

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Answer 1 · 2018-06-01T13:17:07

$x = \frac{2}{3} or x = \frac{1}{4}$ $x=2/3" or "x=1/4$

Explanation:

$rearrange into standard form; a x^{2} + b x + c = 0$ $"rearrange into "color(blue)"standard form";ax^2+bx+c=0$

$subtract 11 x from both sides$ $"subtract "11x" from both sides"$

$12 x^{2} - 11 x + 2 = 0$ $12x^2-11x+2=0$

$using the a-c method to factor the quadratic$ $"using the a-c method to factor the quadratic"$

$the factors of the product 12 \times 2 = 24$ $"the factors of the product "12xx2=24$

$which sum to - 11 are - 8 and - 3$ $"which sum to - 11 are - 8 and - 3"$

$split the middle term using these factors$ $"split the middle term using these factors"$

$12 x^{2} - 8 x - 3 x + 2 = 0 \leftarrow factor by grouping$ $12x^2-8x-3x+2=0larrcolor(blue)"factor by grouping"$

$4 x (3 x - 2) - 1 (3 x - 2) = 0$ $color(red)(4x)(3x-2)color(red)(-1)(3x-2)=0$

$take out the common factor (3 x - 2)$ $"take out the "color(blue)"common factor "(3x-2)$

$(3 x - 2) (4 x - 1) = 0$ $(3x-2)(color(red)(4x-1))=0$

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

$3 x - 2 = 0 \Rightarrow x = \frac{2}{3}$ $3x-2=0rArrx=2/3$

$4 x - 1 = 0 \Rightarrow x = \frac{1}{4}$ $4x-1=0rArrx=1/4$

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How do you solve $12 x^{2} + 2 = 11 x$ $12x ^ { 2} + 2= 11x$ ?

1 Answer

Explanation:

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Humanities

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How do you solve 12x ^ { 2} + 2= 11x12x2+2=11x12x ^ { 2} + 2= 11x?

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Explanation:

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How do you solve $12 x^{2} + 2 = 11 x$ $12x ^ { 2} + 2= 11x$ ?