Question

Select the two values of x that are roots of this equation.
2x2 + 11x+15= 0

Answer 1

The two values of x are -2.5 and -3

Explanation:

we have

`2x^(2)+11x+15=0`

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`a=2\\b=11\\c=15`

substitute in the formula

`x=\frac{-11(+/-)\sqrt{11^(2)-4(2)(15)}} {2(2)}`

`x=\frac{-11(+/-)√(1)} {4}`

`x=\frac{-11(+/-)1} {4}`

`x_1=\frac{-11(+)1} {4}=-2.5`

`x_2=\frac{-11(-)1} {4}=-3`

therefore

The two values of x are -2.5 and -3

Answer 2

`S=\left \{ (-5)/(2), -3\right \}`

Explanation:

Solving it by factoring. Firstly by multiplying the parameter a (2) by c(15) =30 then find two numbers whose product is 30 and their sum is 11. Here we have: 6 and 5. 6x5=30 6+5=11. Rewrite the b parameter as 6x+5x replacing 11x as it follows, then factor by grouping:

`2x^(2)+11x+15\\2x^(2)+6x+5x+15\Rightarrow (2x^(2)+6x)+(5x+15)\Rightarrow 2x(x+3)+5(x+3)\Rightarrow (2x+5)(x+3)`

Solving each factor separately as a linear equation to find x' and x'':

`2x+5=0\Rightarrow 2x=-5\Rightarrow x=(-5)/(2)\\(x+3)=0\Rightarrow x=-3\\S=\left \{ (-5)/(2), -3\right \}`