Quadratic Equations — Exercise 4.3

For ax²+bx+c=0, compute D=b²−4ac. If D0 there are two distinct real roots. For D≥0 use x=(−b±√D)/(2a).

(i) 2x²−3x+5=0

\[D=(-3)^2-4(2)(5)=9-40=-31\]

(ii) 3x²−4√3 x+4=0

\[D=(-4\sqrt3)^2-4(3)(4)=48-48=0\]

\[x=\frac{4\sqrt3}{6}=\frac{2\sqrt3}{3}\]

(iii) 2x²−6x+3=0

\[D=(-6)^2-4(2)(3)=36-24=12>0\]

\[x=\frac{6\pm\sqrt{12}}4=\frac{3\pm\sqrt3}{2}\]

(i) 2x²+kx+3=0

Equal roots require D=0.

\[k^2-4(2)(3)=0\;\Rightarrow\;k^2=24\]

(ii) kx(x−2)+6=0

The equation is kx²−2kx+6=0. For a quadratic, k≠0. Set its discriminant to zero:

\[(-2k)^2-4(k)(6)=0\]

\[4k(k-6)=0\]

The algebraic possibilities are k=0 and k=6. But k=0 changes the equation to 6=0, which has no roots and is not quadratic.

Let breadth be b>0 metres, so length is 2b. Then:

\[2b^2=800\;\Rightarrow\;b^2=400\]

The roots are b=20 and b=−20; a physical breadth must be positive.

Let their present ages be x and 20−x. Four years ago they were x−4 and 16−x, so:

\[(x-4)(16-x)=48\]

\[-x^2+20x-64=48\;\Rightarrow\;x^2-20x+112=0\]

\[D=(-20)^2-4(1)(112)=400-448=-48\]

Ages are real quantities, but this equation has no real roots. Another check: four years ago their ages summed to 12, whose greatest possible product is 6×6=36, below 48.

Let length be l and breadth w. The perimeter gives l+w=40, so w=40−l. Use the area:

\[l(40-l)=400\;\Rightarrow\;l^2-40l+400=0\]

\[(l-20)^2=0\]

The repeated root gives l=20 and w=20. A square is a special rectangle.