How do you solve $2 x^{2} \div 3 x = 14$ $2x^2 -: 3x = 14$ ?

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Answer 1 · 2016-08-20T13:40:30

See explanation...

I think the question should have had subtraction rather than division, but I will answer both possibilities:

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How to solve $\frac{2 x^{2}}{3 x} = 14$ $bb((2x^2)/(3x)=14)$

Multiply both sides by $3 x$ $3x$ to get:

$2 x^{2} = 42 x$ $2x^2 = 42x$

Divide both sides by $2 x$ $2x$ to find:

$x = 21$ $x = 21$

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How to solve $bb(2x^2-3x=14)$

Subtract $14$ from both sides to get:

$2x^2-3x-14 = 0$

Use an AC method to factorise:

Look for a pair of factors of $AC=2*14=28$ which differ by $B=3$ .

The pair $7, 4$ works.

Use this to split the middle term then factor by grouping:

$0 = 2x^2-3x-14$

$=2x^2-7x+4x-14$

$=(2x^2-7x)+(4x-14)$

$=x(2x-7)+2(2x-7)$

$=(x+2)(2x-7)$

Hence $x=-2$ or $x=7/2$

How do you solve 2x^2 -: 3x = 142x2÷3x=142x^2 -: 3x = 14 ?