����ý

For the quadratic function \(y=2x^{2}-7x-15\) determine the nature of the roots.

\(y = 2{x^2} - 7x - 15\)

\({b^2} - 4ac\) where a = 2, b = -7 and c = -15

\(= {( - 7)^2} - (4 \times 2 \times - 15)\)

= \(49-(-120)\)

= \(49+120\)

= \(169\) which is \(\textgreater0\) therefore there are two real roots.

Find the value of \(k\) if the quadratic function \(y=x^{2}+6x+k\) has one real root .

If the quadratic has one real root, then \(b^{2}-4ac=0\)

\(b^{2}-4ac=0\)

\(36-(4\times 1\times k)=0\)

\(36-4k=0\)

\(36=4k\)

\(k=9\)