#1**0 **

Have you learned Vieta's formula yet? You could really use it here.

\(\text {root}_1 + \text {root}_2 = 5\\\text {in your situation. Vieta's formula says that this will be equal to} \\-\frac ba \\\text {where }b=-M \text { and } a=x \text { in your case}\)

So this gives us \(-\frac {-M}x = 5 \\\implies \frac Mx = 5 \\\implies \mathbf {M = 5x}\)

So M=5x.

Try to solve the equation now that you know this.

Nacirema Oct 4, 2020

#2**0 **

\(\displaystyle \text{If the equation }\: x^{3}+bx^{2}+cx+d=0,\: \text{has roots}\: \alpha, \beta, \gamma,\\ \text{then} \\ \alpha + \beta + \gamma = -b, \\ \alpha \beta + \beta \gamma + \gamma \alpha = c, \text{ and}\\ \alpha \beta \gamma = -d. \)

\(\displaystyle \text{So, for the equation} \: x^{3}-Mx +15=0,\\ \alpha + \beta + \gamma = 0 \dots \dots(1)\\ \alpha \beta + \beta \gamma +\gamma \alpha = -M \dots \dots(2)\\ \alpha\beta \gamma=-15 \dots \dots(3).\)

\(\displaystyle \text{If now } \: \alpha + \beta = 5, \text{then from (1)}, \gamma = -5.\)

\(\displaystyle \text{Substituting into (2)},\: \alpha \beta -5\beta-5\alpha=\alpha \beta -5(\alpha + \beta)=\alpha \beta -25=-M, \\ \text{from which} \: M =25-\alpha\beta \dots \dots (4).\)

\(\displaystyle \text{From (3)} \: \alpha \beta(-5)=-15\quad\text{so } \: \alpha \beta = 3 \: \: \text{and so from (4)}, \\ M=25-3 = 22.\)

.Guest Oct 4, 2020