Question

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

Answer 1

Answer:

$S=\left \{ \frac{-5}{2}, -3\right \}$

Step-by-step 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'':

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

Answer 2

Answer:

The two values of x are -2.5 and -3

Step-by-step 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

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

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

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

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

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

therefore

The two values of x are -2.5 and -3