How do you solve $2 x^{2} - 3 \leq x$ $2x^2-3<=x$ ?

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How do you solve $2 x^{2} - 3 \leq x$ $2x^2-3<=x$ ?

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Answer 1 · 2016-10-06T19:10:03

X

Explanation:

$2 x^{2} - 3 \leq x$ $2x^2-3<=x$
Think $2 x^{2} - 3 = x$ $2x^2-3=x$
$2 x^{2} - x - 3 = 0$ $2x^2-x-3=0$
Factorise
$(2 x - 3) (x + 1) = 0$ $(2x-3)(x+1)=0$
So actually
$(2 x - 3) (x + 1) \leq 0$ $(2x-3)(x+1)<=0$
When you do this make sure you keep everything on the right sides
Now sketch the graph. Yes please do that!!
It is bucket shaped and crosses the x axis at $\frac{3}{2}$ $3/2$ and -1
So for the values to be less than or equal to 0, x must be greater than or equal to -1 and less than or equal to $\frac{3}{2}$ $3/2$

$- 1 \leq x \leq \frac{3}{2}$ $-1<=x<=3/2$

Answer 2 · 2016-10-06T19:13:45

Sorry $- 1 \leq x \leq \frac{3}{2}$ $-1<=x<=3/2$

Explanation:

See previous answer

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