If the equation a(b-c)x^2+b(c-a)x+c(a-b)=0 has equal roots, where a+c=15 and b=(36)/(5), t…

Question

If the equation $a(b-c){x}^{2}+b(c-a)x+c(a-b)=0$ has equal roots, where $a+c=15$ and $b=\frac{36}{5}$, then ${a}^{2}+{c}^{2}$ is equal to \_\_\_\_\_\_\_\_\_

Answer: 117

Accepted Answer

To solve the given problem, we start with the quadratic equation:

$a(b-c){x}^{2}+b(c-a)x+c(a-b)=0$

Given that the roots are equal (let's assume both roots are 1), we know the sum of the roots, $\alpha +\beta $, is twice the value of one root, which leads us to:

$\alpha +\beta =2$

Using the formula for the sum of roots for a quadratic equation, $\alpha +\beta =-\frac{b(c-a)}{a(b-c)}$, we set this equal to 2:

$-\frac{b(c-a)}{a(b-c)}=2$

Solving for this:

$-bc+ab=2ab-2ac\, \, 2ac=ab+bc\, \, 2ac=b(a+c)$

Given that $a+c=15$ and $b=\frac{36}{5}$, substitute these into the equation:

$2ac=15b\, \, 2ac=15	imes \frac{36}{5}=108\, \, ac=54$

Now, using the equation $a+c=15$ and $ac=54$, find ${a}^{2}+{c}^{2}$:

${a}^{2}+{c}^{2}=(a+c{)}^{2}-2ac={15}^{2}-2	imes 54\, \, {a}^{2}+{c}^{2}=225-108=117$

Therefore, ${a}^{2}+{c}^{2}$ is equal to 117.

Solutions · 1