ग्राफ \(y=x^2+2x-8\) (x)-अक्ष को किन (x) मानों पर काटता है?
At which (x) values does \(y=x^2+2x-8\) cut the (x)-axis?
Explanation opens after your attempt
A. (x=-4,2)
Concept
Set (y=0) on the (x)-axis. From \(x^2+2x-8=0\), (x=-4,2).
Why this answer is correct
The correct answer is A. (x=-4,2). Set (y=0) on the (x)-axis. From \(x^2+2x-8=0\), (x=-4,2).
Exam Tip
(x)-अक्ष पर (y=0) रखें। \(x^2+2x-8=0\) से (x=-4,2) मिलता है।
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