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# algebra

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Two of the real roots of x^3 - Mx + 15 = 0 sum to 5.  What is M?

Oct 3, 2020

#1
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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.

Oct 4, 2020
#2
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$$\displaystyle \text{If the equation }\: x^{3}+bx^{2}+cx+d=0,\: \text{has roots}\: \alpha, \beta, \gamma,\\ \text{then} \\ \alpha + \beta + \gamma = -b, \\ \alpha \beta + \beta \gamma + \gamma \alpha = c, \text{ and}\\ \alpha \beta \gamma = -d.$$

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

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

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

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

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Oct 4, 2020