if roots of the quadratic equation x2+px+mn=0 are real and equal, show that the roots of the quadratic equation x2-2(m+n)x+(m2+n2+2p)=0 are also equal.
Answers
Given, A quadratic equation x² + px + mn = 0 has equal and real roots. We need to show if the roots of the quadratic equation x² - 2(m + n)x + (m² + n² + 2p) = 0 has also equal and real roots.
Comparing both the quadratic equations with the standard form of quadratic equation i.e., ax² + bx + c = 0 , we get values of a , b and c for both quadratic equations as:
a = 1 , b = p , c = mn
a₂ = 1 , b₂ = -2(m + n) , c₂ = (m² + n² + 2p)
Since, the roots of the first quadratic equation ( x² + px + mn = 0 ) are equal and real then the determinant would be equal to 0.
∴ D = 0
⇒ b² - 4ac = 0
⇒ p² - 4×1×mn = 0
⇒ p² = 4mn ...(1)
Similarly, In the second quadratic equation, we need to show that this quadratic equation would have real and equal roots therefore, Its determinant must be equal to 0:
⇒ D = 0
⇒ (b₂)² - 4a₂c₂ = 0
⇒ { -2(m + n) }² = 4×1×(m² + n² + 2p)
⇒ 4m² + 4n² + 8mn = 4m² + 4n² + 8p
⇒ 2p² = 8p
⇒ 2p = 8
⇒ p = 4, Which is also equal to mn { from (1) }
Now, the second quadratic equation becomes:
⇒ x² - 2(m + n)x + (m² + n² + 2mn) = 0
⇒ x² - 2(m + n)x + (m + n)² = 0 [∵ a² + b² + 2ab = (a + b)² ]
Now, The determinant must be equal to zero.
⇒ (b₂)² - 4a₂c₂
⇒ { -(2m + 2n) }² - 4(m + n)²
⇒ 4m² + 4n² + 8mn - 4(m² + n² + 2mn)
⇒ 4m² + 4n² + 8mn - 4m² - 4n² - 8mn
⇒ 0
Since, The determinant is equal to zero, hence it would have real and equal roots.
Hence, Proved.