In a quadratic equation the leading coefficient is 1, aseem reads the coefficient 16 of x as 19 and obtains the roots as -15 and -4. The real roots are
Answers
Answer: (-10) and (-16)
Step-by-step explanation:
The roots obtained are (-15) and (-4). We have to find the real roots.
Do the same steps you do for factorising the equation and obtaining the roots but in the reverse order.
x = -15
∴ x + 15 = 0
x = -4
∴ x + 4 = 0
If (x + 15) and (x + 4) are equal to zero and (-15) and (-4) are the roots of the wrong equation, then the wrong quadratic equation must have been :-
(x + 4)(x + 15) = 0
Now, expand this equation.
(x + 4)(x + 15) = 0
x² + 19x + 60 = 0
We know that she reads the coefficient of x as 19 while the real coefficient is 16. So, just replace the coefficient and factorise the equation.
x² + 16x + 60 = 0
x² + 10x + 6x + 60 = 0
x(x + 10) + 6(x + 10) = 0
(x + 10)(x + 6) = 0
x + 10 = 0
∴ x = -10
x + 6 = 0
∴ x = -6
Therefore, the real roots are (-10) and (-6).