SOLUTION: if the equation x^2-15-k(2x-8)=0has equal roots find the value of k

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Question 629638: if the equation x^2-15-k(2x-8)=0has equal roots find the value of k
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If the roots are equal, the equation looks
like +%28+x+-+a+%29%5E2+=+0+, then there are 2 roots
at +x+=+a+, so I can say
+%28+x+-+a+%29%5E2+=+x%5E2+-+15+-+k%2A%28+2x+-+8+%29+
+x%5E2+-+2a%2Ax+%2B+a%5E2+=+x%5E2+-+2k%2Ax+-+15+%2B+8k+
+x%5E2+-+2a%2Ax+%2B+a%5E2+=+x%5E2+-+2k%2Ax+%2B+%28+8k+-+15+%29+
Matching up coefficients:
+k+=+a+
+a%5E2+=+8k+-+15+
By substitution:
+k%5E2+=+8k+-+15+
+k%5E2+-+8k+%2B+15+=+0+
+%28+k+-+3+%29%2A%28+k+-+5+%29+=+0+
+k+=+3+
+k+=+5+
--------
check answers:
+x%5E2+-+15+-+k%2A%28+2x+-+8+%29+=+0+
+x%5E2+-+15+-+3%2A%28+2x+-+8+%29+=+0+
+x%5E2+-+15+-+6x+%2B+24+=+0+
+x%5E2+-+6x+%2B+9+=+0+
+%28+x+-+3+%29%5E2+=+0+
so, +k+=+3+ works
+x%5E2+-+15+-+k%2A%28+2x+-+8+%29+=+0+
+x%5E2+-+15+-+5%2A%28+2x+-+8+%29+=+0+
+x%5E2+-+15+-+10x+%2B+40+=+0+
+x%5E2+-+10x+%2B+25+=+0+
+%28+x+-+5+%29%5E2+=+0+
+k+=+5+ works too