determine the value of k for which this equation has two real roots: kx^2+6x+5=0
+26400 determine the value of k for which this equation has two real roots: kx^2+6x+5=0
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\\\ D = b^2-4ac\)
discriminant D > 0 ( two real roots )
a = k
b = 6
c = 5
D = 6*6 - 4*k*5
D = 36 -20k
D > 0 ( two real roots ):
\(\begin{array}{rcll} 36 - 20k &>& 0 \qquad &| \qquad -36\\ -20k &>& -36 \qquad &| \qquad \cdot(-1)\\ 20k &<& 36 \qquad &| \qquad :20\\ k &<& \frac{36}{20} \\ k &<& \frac{9}{5} \\ \mathbf{k} &\mathbf{<}& \mathbf{1.8} \\ \end{array}\)
+26400 determine the value of k for which this equation has two real roots: kx^2+6x+5=0
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\\\ D = b^2-4ac\)
discriminant D > 0 ( two real roots )
a = k
b = 6
c = 5
D = 6*6 - 4*k*5
D = 36 -20k
D > 0 ( two real roots ):
\(\begin{array}{rcll} 36 - 20k &>& 0 \qquad &| \qquad -36\\ -20k &>& -36 \qquad &| \qquad \cdot(-1)\\ 20k &<& 36 \qquad &| \qquad :20\\ k &<& \frac{36}{20} \\ k &<& \frac{9}{5} \\ \mathbf{k} &\mathbf{<}& \mathbf{1.8} \\ \end{array}\)
