Find the real values of m for which f(x) = x2 + mx + 9A) f(x) is a perfect square - Maths - Complex Numbers and Quadratic Equations

Question

Find the real values of m for which f(x) = x2 + mx +
9A) f(x) is a perfect square for all x E RB) f(x) = 0 has real
rootsC) f(x) = 0 has integral roots onlyD) f(x) =0 has unequal real
roots

Priyanka Kedia · Accepted Answer

The given quadratic equation be as,fx=x2+mx+91fx=x2+mx+9       =x2+mx+6x-6x+9       =x+32+mx-6x       =x+32+x6-mfx is a perfect square for all x∈R
 So, coefficient of x will be zero
6-m=0⇒m=62fx=0 has real roots
 So,b2-4ac=0⇒m2-4×9×1=0⇒m2=36⇒m=±63fx=0 has integral roots
 So,b2-4ac=0⇒m2-4×9×1=0⇒m2=36⇒m=±6Since fx=0 has integral roots
 Therefore we can both the values
⇒m=±64fx=0 has unequal and real roots
 So,D>0⇒b2-4ac>0⇒m2-4×9×1>0⇒m2>36⇒m>±6

Find the real values of m for which f(x) = x2 + mx + 9A) f(x) is a perfect square for all x E RB) f(x) = 0 has real rootsC) f(x) = 0 has integral roots onlyD) f(x) =0 has unequal real roots

Find the real values of m for which f(x) = x² + mx + 9A) f(x) is a perfect square for all x E RB) f(x) = 0 has real rootsC) f(x) = 0 has integral roots onlyD) f(x) =0 has unequal real roots