Question

plz ans ques no 7 nd 8

Answer 1

QUADRATICS EQUATIONS AND EXPRESSION

Answer 2

Hey there !!!!!!!!

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A quadratic equation of the form Ax²+Bx+C=0 has

✪ Imaginary roots  if discriminant  < 0

                          B²-4AC < 0

✪ Real and equal roots if discriminant = 0

                          B²-4AC=0

✪ Real and distinct roos if discriminant > 0

                          B²-4AC >0

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7) x²-2ax+a²-b²-c²= 0 comparing with Ax²+Bx+C = 0

A=1 B= -2a  C= a²-b²-c²

          =B²-4AC

          = 4a²-4(1)(a²-b²-c²)
      
          = 4a²-4a²+4b²+4c²

          =4b²+4c²

Square of a real number is always positive 

4b²+4c² is greater than zero.

ROOTS ARE REAL.

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8) (a-b+c)x²+4(a-b)x+(a-b-c) comparing with Ax²+Bx+C = 0

A=(a-b+c)  B=4(a-b)   C= a-b-c

          =B²-4AC

          =(4(a-b))²-4(a-b+c)(a-b-c))

           =16(a-b)²-4((a-b)²-c²)

           =16(a-b)²-4(a-b)²+4c²

           =12(a-b)²+4c²

Sum of squares is always positive

So (a-b+c)x²+4(a-b)x+(a-b-c) has REAL ROOTS

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Hope this helped you..............