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 , then a 2 + c 2 is equal to \ \ \ \ \

Mathematics · Quadratic Equations

JEE Main 2025 — 23 January, Morning Shift — Question 25

If the equation $a(b-c) x^{2}+b(c-a) x+c(a-b)=0$ has equal roots, where $\mathrm{a}+\mathrm{c}=15$ and

$\mathrm{b}=\frac{36}{5}$ , then $\mathrm{a}^{2}+\mathrm{c}^{2}$ is equal to $\_\_\_\_\_$

Numerical answer — enter this value.

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

$\mathrm{x}=1$ is root

$\therefore$ other root is 1

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

$\Rightarrow-\mathrm{bc}+\mathrm{ab}=2 \mathrm{ab}-2 \mathrm{ac}$

$\Rightarrow 2 \mathrm{ac}=\mathrm{ab}+\mathrm{bc}$

$\Rightarrow 2 \mathrm{ac}=\mathrm{b}(\mathrm{a}+\mathrm{c})$

$\Rightarrow 2 \mathrm{ac}=15 \mathrm{~b}$

$\Rightarrow 2 \mathrm{ac}=15\left(\frac{36}{5}\right)=108$

$\Rightarrow \mathrm{ac}=54$ , $a+c=15$

$\mathrm{a}^{2}+\mathrm{c}^{2}+2 \mathrm{ac}=225$

$\mathrm{a}^{2}+\mathrm{c}^{2}=225-108=117$

Answer key and solution verified before publishing.

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