ग्राफ \(y=x^2-8x+15\) (x)-अक्ष को किन (x) मानों पर काटता है?
At which (x) values does the graph \(y=x^2-8x+15\) cut the (x)-axis?
Explanation opens after your attempt
A. (x=3,5)
Concept
Set (y=0) on the (x)-axis. From \(x^2-8x+15=0\), we get (x=3,5).
Why this answer is correct
The correct answer is A. (x=3,5). Set (y=0) on the (x)-axis. From \(x^2-8x+15=0\), we get (x=3,5).
Exam Tip
(x)-अक्ष पर (y=0) रखें। \(x^2-8x+15=0\) से (x=3,5) मिलता है।
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