How many real roots are in y=2x^2-7x-15?

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Answer 1 · 2018-03-29T06:14:25

The real roots in $y = 2 x^{2} - 7 x - 15$ $y=2x^2-7x-15$ are $x = - \frac{3}{2}$ $x=-3/2$ and $x = 5$ $x=5$ .

$y = 2 x^{2} - 7 x - 15$ $y=2x^2-7x-15$

$y = 0$ $y=0$

$2 x^{2} - 7 x - 15 = 0$ $2x^2-7x-15=0$

Attempting to factorise

$2 x^{2} + 3 x - 10 x - 15 = 0$ $2x^2+3x-10x-15=0$

$x (2 x + 3) - 5 (2 x + 3) = 0$ $x(2x+3)-5(2x+3)=0$

$(x - 5) (2 x + 3) = 0$ $(x-5)(2x+3)=0$

$x - 5 = 0$ $x-5=0$

$x = 5$ $x=5$

$2 x + 3 = 0$ $2x+3=0$

$2 x = - 3$ $2x=-3$

$x = - \frac{3}{2}$ $x=-3/2$

The real roots in $y = 2 x^{2} - 7 x - 15$ $y=2x^2-7x-15$ are $x = - \frac{3}{2}$ $x=-3/2$ and $x = 5$ $x=5$ .