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For the quadratic equation \(-4x^2+6x-9=0\), if the coefficient of \(x^2\) is \(a\), coefficient of \(x\) is \(b\), and constant term is \(c\), find \(a-b+c\).द्विघात समीकरण \(-4x^2+6x-9=0\) में यदि \(x^2\) का गुणांक \(a\), \(x\) का गुणांक \(b\) और स्थिर पद \(c\) है, तो \(a-b+c\) का मान ज्ञात कीजिए।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(-19\)\(-19\)
Here \(a=-4\), \(b=6\), \(c=-9\). So \(a-b+c=-4-6+(-9)=-4-6-9=-19\).यहाँ \(a=-4\), \(b=6\), \(c=-9\) है। अतः \(a-b+c=-4-6+(-9)=-4-6-9=-19\).
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