Question

I. $15x^2-29x-14=0$
II. $6y^2-5y-25=0$

• A. x > y
• B. $x\ge y$
• C. x < y
• D. $x\le y$
• E. Relationship between x and y cannot be determined

Answer 1

The correct option is D $x\le y$

I. $15x^2$ -29x - 14 = 0
$x_1=\frac{29+\sqrt{841+60\times 14}}{30}$
$=\frac{29+41}{30}=\frac{70}{30}{x}_2=\frac{29-\sqrt{1681}}{30}{x}_2=\frac{29-41}{30}=\frac{-12}{30}x=\frac{7}{3}.\frac{-2}{5}II.6y^2-5y-25=0y_1=\frac{5+\sqrt{25-4\times 6\times -25}}{12}=\frac{5+\sqrt{625}}{12}=\frac{30}{12}{y}_2=\frac{5-\sqrt{25-6\times 6\times -25}}{12}\Rightarrow y_2=\frac{5-\sqrt{625}}{12}=\frac{-20}{12}y=\frac{5}{2},\frac{-5}{3}$
So, relationship between x and y can't be determined.