Question

if x=3 is a root of x^2-x-k=0 find the value of p so that the roots of equation x^2-k(2x+k+2)+p=0 becomes real and e​

Answer 1

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Step by step solution:

$Given \ p(x)=x^2-x-k=0 \ and \ x=3\\\\putting \ x=3 \ to \ get \ value \ of \ k\\\\ p(x)=x^2-x-k=0\\\\p(3)=9-3-k=0\\\\6=k$

$we \ have \ to \ find \ value \ of \ p\\\\given \ q(x)=x^2-k(2x+k+2)+p=0 \ for \ real \ and \ equal \ roots\\\\putting \ k=6\\\\q(x)=x^2-k(2x+k+2)+p=0\\\\q(x)=x^2-6(2x+6+2)+p=0\\\\q(x)=x^2-6(2x+8)+p=0\\\\q(x)=x^2-12x-48+p\\\\for \ real \ and \ equal \ roots \ b^2-4ac=0\\\\(-12)^2-4 \times1 \times (-48+p)=0\\\\144=4 \times(-48+p)\\\\(-48+p)=36\\\\p=36+48\\\\p=84$

$Verification\\\\q(x)=x^2-12x-48+p\\\\putting \ p=84 \ here\\\\q(x)=x^2-12x-48+84\\\\q(x)=x^2-12x+36\\\\q(x)=(x)^2+(6)^2-2\times x\times6\\\\q(x)=(x-6)^2\\\\q(x)=(x-6)(x-6)\\\\x-6=0\\\\x=6\\\\So \ there \ are \ two \ zeroes \ which \ are \ equal \ and \ real\\\\Hence \ verified$

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