Question

4.Given that y=2x^2+3x,find the range of values of x for which y>9.pls check my answer and working whether i'm right. @ganeshie8

Answer 1

\[9=2x^2+3x\]\[9<2x^2+3x\]\[0<2x^2+3x-9\]\[2x^2+3x-9>0\]\[(x+3)(2x-3)>0\]

Answer 2

sorry, how 1.5 can be smaller than -3?

Answer 3

it should be written like -3<x<1.5 so the answer is \[x \in (-3;1.5)\]

Answer 4

This inequality states "x is less than -3 but greater than 3/2"

It should be saying "x is less that 3/2 but greater than -3"

How would you write that? @MARC_

Answer 5

can u show me the working from the start? @Jhannybean

Answer 6

\[\begin{align}
2x^2 +3x >9~\text{ Since} ~~y >9
\\&: ~ 2x^2 +3x -9>0
\\&: ~ x^2+3x-18 > 0
\\&: ~ ( x+6)(x-3)>0
\\&: ~ \left(x + \frac{6}{2}\right)\left(x-\frac{3}{2}\right)>0
\\&: ~ (x+3)(2x-3) >0
\\&: ~ x+3 > 0 \implies x > -3
\\&: 2x-3 > 0 \implies x > \frac{3}{2}
\\&: ~ \boxed{ -3 > x >\dfrac{3}{2}}
\end{align}\]

Answer 7

BUM method: \[\color{red}{2}x^2 +3x \color{red}{-9}\]\[x^2 +3x -8\] Multiply the red colored numbers togeher to simplify the quadratic.

That's how I solve these types of quadratics, it's not necessary.

Answer 8

BUM is what?

Answer 9

Whoop,I made a typo, that is not \(-8\) it's supposed to say \(-18\) @MARC_

Answer 10

Im not sure exactly :P

Answer 11

Any more questions? @MARC_

Answer 12

Nope...thnx @Jhannybean @Kamille

Answer 13

Awesome :D

Answer 14

@ganeshie8 do u know what is BUM?

Answer 15

\[(x+3)(2x-3)>0 \]

your factorization is good so far, next all you need to is think of a parabola

Answer 16

what are the x intercepts of the corresponding parabola ?

Answer 17

-3 and 3/2

Answer 18

Yes! and your inequality is > 0 so :