Conditions for roots of quadratic equationIf I have a quadratic function f ( x ) = a x 2 + b x + c, what are the conditions that should the numbers a,b and c satisfy so that the equation f ( x ) = 0 has real roots x 1 and x 2 such that x 1 &lt; 6 &lt; x 2 &lt; 10?My answer is the following system: f ( 6 ) . f ( 10 ) &lt; 0 (so that one of the roots is between 6 and 10) − b 2 a &lt; 10.However, there is no such answer in the answer sheet.

Question

Conditions for roots of quadratic equationIf I have a quadratic function   f  (  x  )  =  a      x    2    +  b  x  +  c, what are the conditions that should the numbers a,b and c satisfy so that the equation   f  (  x  )  =  0 has real roots       x    1   and       x    2   such that       x    1    &amp;lt;  6  &amp;lt;      x    2    &amp;lt;  10?My answer is the following system:  f  (  6  )  .  f  (  10  )  &amp;lt;  0 (so that one of the roots is between 6 and 10)  −      b          2      a        &amp;lt;  10.However, there is no such answer in the answer sheet.

Kristin Myers · Accepted Answer

Step 1The general criterion is that   α separates the roots (their existence is presupposed) if an only if   a  f  (  x  )  &amp;lt;  0. Everything is based on this remark. If   a  f  (  α  )  &amp;gt;  0,   α is either greater or smaller than both roots.Here the conditions would be   a  f  (  6  )  &amp;lt;  0 and   a  f  (  10  )  &amp;gt;  0 plus checking whether 10 is greater or smaller than both roots. But since 6 is greater than one root, and   10  &amp;gt;  6, 10 can only be greater, so that there&#039;s nothing to check.Step 2Finally, the conditions would be:   a  f  (  6  )  &amp;lt;  0.The last condition alone only means that one of   6  =  6,10 separates the roots, the other no, but you don&#039;t know which one. This ambiguity can also be raised with considerin the arithmetic mean of the roots, as you propose.

Raiden Barr · Accepted Answer

Step 1The requirements are as you did one part, and this gives the other part to complete it:1.   f  (  6  )  ⋅  f  (  10  )  &amp;lt;  0, andStep 22.       |    6    +                  b                  2          a                      |    &amp;lt;                              |                          x          2                −                  x          1                          |                    2          ⟺    36  +                    6        b            a        +                    b        2                    4        a              &amp;lt;                    (                  x          1                +                  x          2                          )          2                −        4                  x          1                          x          2                    4        =                    b        2                    4                  a          2                      −            c      a

Answered question

Answer & Explanation

New Questions in Algebra I