How do you find the value of the discriminant and state the type of solutions given $5 b^{2} + b - 2 = 0$ $5b^2+b-2=0$ ?

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Answer 1 · 2017-09-02T10:18:25

$D = 41$ $D=41$ , two real solutions and $b \approx - 0.74, b \approx 0.54$ $b ~~ -0.74 , b ~~ 0.54$ .

$5 b^{2} + b - 2 = 0$ $5b^2+b-2 =0$ . Comparing with standard quadratic equation

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ we get here $a = 5, b = 1, c = - 2$ $a=5 ,b =1 ,c =-2$

Discriminant $D = b^{2} - 4 \cdot a \cdot c = 1 - 4 \cdot 5 \cdot (- 2) = 1 + 40 = 41$ $D = b^2-4*a*c = 1 -4*5*(-2)=1+40=41$

If $D$ $D$ is positive, we will get two real solutions, if it is zero we get

just one solution, and if it is negative we get complex solutions.

Here $D$ $D$ is positive , so there are two real solution.

$b = \frac{- b \pm \sqrt{D}}{2 a} or b = \frac{- 1 \pm \sqrt{41}}{10}$ $b= (-b +- sqrt(D))/(2a) or b= (-1+- sqrt41)/10$ or

$:. b ~~ -0.74(2 dp) , b ~~ 0.54(2dp)$ [Ans]

How do you find the value of the discriminant and state the type of solutions given 5b^2+b-2=05b2+b−2=05b^2+b-2=0?