(k - 5) * x ^ 2 + 2(k - 5) * x + 2 = 0 have equal roots?
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[tex]\rm (k-5)x^2+2(k-5)x+2=0[/tex]
Compare it with [tex]\rm ax^2+bx+c=0[/tex]
[tex]\rm a=k-5 \quad b=2(k-5) \quad c=2[/tex]
When a quadratic equation is having two equal roots,
[tex]\implies \rm b^2-4ac=0[/tex]
[tex]\implies \rm [2(k-5)]^2-4(k-5)(2)=0[/tex]
[tex]\implies \rm 4(k-5)^2-8(k-5)=0[/tex]
[tex]\implies \rm [k-5][4(k-5)-8]=0[/tex]
[tex]\implies \rm 4(k-5)-8=0 \quad \& \quad k-5=0[/tex]
[tex]\implies \rm 4k-20-8=0 \quad \& \quad \bf{k=5}[/tex]
[tex]\implies \rm 4k=28[/tex]
[tex]\implies \bf k=7[/tex]