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For the quadratic equation \(x^2-10x+r=0\) to have two equal real roots, what is the value of \(r\)?द्विघात समीकरण \(x^2-10x+r=0\) के दो बराबर (equal) वास्तविक मूल (real roots) होने के लिए \(r\) का मान क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(r=25\)\(r=25\)
Equal real roots require \(\Delta=0\). Here \(a=1,\;b=-10,\;c=r\). So \(\Delta=b^2-4ac=(-10)^2-4(1)(r)=100-4r\). Set \(100-4r=0\Rightarrow 4r=100\Rightarrow r=25\).बराबर वास्तविक मूलों (equal real roots) के लिए \(\Delta=0\) होना चाहिए। यहाँ \(a=1,\;b=-10,\;c=r\) है। इसलिए \(\Delta=b^2-4ac=(-10)^2-4(1)(r)=100-4r\)। अब \(100-4r=0\Rightarrow 4r=100\Rightarrow r=25\)।
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