Class 10: Quadratic Equations

Solve each of the following use quadratic formula

$\displaystyle \textbf{Question 1: } \text{Solve using the quadratic formula: }x^2-6x=27.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-6x-27=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-6,\ c=-27$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(-27)}}{2(1)}$
$\displaystyle =\frac{6\pm\sqrt{36+108}}{2}$
$\displaystyle =\frac{6\pm\sqrt{144}}{2}$
$\displaystyle =\frac{6\pm12}{2}$
$\displaystyle \Rightarrow x=9\text{ or }-3$
$\\$

$\displaystyle \textbf{Question 2: } \text{Solve using the quadratic formula: }x^2-10x+21=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-10,\ c=21$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(21)}}{2(1)}$
$\displaystyle =\frac{10\pm\sqrt{100-84}}{2}$
$\displaystyle =\frac{10\pm\sqrt{16}}{2}$
$\displaystyle =\frac{10\pm4}{2}$
$\displaystyle \Rightarrow x=3\text{ or }7$
$\\$

$\displaystyle \textbf{Question 3: } \text{Solve using the quadratic formula: }x^2+6x-10=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=6,\ c=-10$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-6\pm\sqrt{6^2-4(1)(-10)}}{2(1)}$
$\displaystyle =\frac{-6\pm\sqrt{36+40}}{2}$
$\displaystyle =\frac{-6\pm\sqrt{76}}{2}$
$\displaystyle =\frac{-6\pm2\sqrt{19}}{2}$
$\displaystyle \Rightarrow x=-3\pm\sqrt{19}$
$\\$

$\displaystyle \textbf{Question 4: } \text{Solve using the quadratic formula: }x^2+2x-6=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=2,\ c=-6$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-2\pm\sqrt{2^2-4(1)(-6)}}{2(1)}$
$\displaystyle =\frac{-2\pm\sqrt{4+24}}{2}$
$\displaystyle =\frac{-2\pm\sqrt{28}}{2}$
$\displaystyle =\frac{-2\pm2\sqrt{7}}{2}$
$\displaystyle \Rightarrow x=-1\pm\sqrt{7}$
$\\$

$\displaystyle \textbf{Question 5: } \text{Solve using the quadratic formula: }3x^2+2x-1=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=3,\ b=2,\ c=-1$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-2\pm\sqrt{2^2-4(3)(-1)}}{2(3)}$
$\displaystyle =\frac{-2\pm\sqrt{4+12}}{6}$
$\displaystyle =\frac{-2\pm\sqrt{16}}{6}$
$\displaystyle =\frac{-2\pm4}{6}$
$\displaystyle \Rightarrow x=-1\text{ or }\frac{1}{3}$
$\\$

$\displaystyle \textbf{Question 6: } \text{Solve using the quadratic formula: }2x^2+7x+5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=2,\ b=7,\ c=5$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-7\pm\sqrt{7^2-4(2)(5)}}{2(2)}$
$\displaystyle =\frac{-7\pm\sqrt{49-40}}{4}$
$\displaystyle =\frac{-7\pm\sqrt{9}}{4}$
$\displaystyle =\frac{-7\pm3}{4}$
$\displaystyle \Rightarrow x=-1\text{ or }-\frac{5}{2}$
$\\$

$\displaystyle \textbf{Question 7: } \text{Solve using the quadratic formula: }\frac{2}{3}x=-\frac{1}{6}x^2-\frac{1}{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{2}{3}x=-\frac{1}{6}x^2-\frac{1}{3}$
$\displaystyle \text{Multiplying both sides by }6,$
$\displaystyle 4x=-x^2-2$
$\displaystyle x^2+4x+2=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=4,\ c=2$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-4\pm\sqrt{4^2-4(1)(2)}}{2(1)}$
$\displaystyle =\frac{-4\pm\sqrt{16-8}}{2}$
$\displaystyle =\frac{-4\pm\sqrt{8}}{2}$
$\displaystyle =\frac{-4\pm2\sqrt{2}}{2}$
$\displaystyle \Rightarrow x=-2\pm\sqrt{2}$
$\\$

$\displaystyle \textbf{Question 8: } \text{Solve using the quadratic formula: }\frac{1}{15}x^2+\frac{5}{3}=\frac{2}{3}x.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{15}x^2+\frac{5}{3}=\frac{2}{3}x$
$\displaystyle \text{Multiplying both sides by }15,$
$\displaystyle x^2+25=10x$
$\displaystyle x^2-10x+25=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-10,\ c=25$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(25)}}{2(1)}$
$\displaystyle =\frac{10\pm\sqrt{100-100}}{2}$
$\displaystyle =\frac{10}{2}$
$\displaystyle \Rightarrow x=5$
$\\$

$\displaystyle \textbf{Question 9: } \text{Solve using the quadratic formula: }x^2-6x=2\sqrt{2}x.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-6x-2\sqrt{2}x=0$
$\displaystyle x^2-(6+2\sqrt{2})x=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-(6+2\sqrt{2}),\ c=0$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{6+2\sqrt{2}\pm\sqrt{(6+2\sqrt{2})^2}}{2}$
$\displaystyle =\frac{6+2\sqrt{2}\pm(6+2\sqrt{2})}{2}$
$\displaystyle \Rightarrow x=3+\sqrt{2}\pm(3+\sqrt{2})$
$\displaystyle \Rightarrow x=3\sqrt{2}\text{ or }-\sqrt{2}$
$\\$

$\displaystyle \textbf{Question 10: } \text{Solve using the quadratic formula: }\frac{4}{x}-3=\frac{5}{2x+3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{4}{x}-3=\frac{5}{2x+3}$
$\displaystyle \frac{4-3x}{x}=\frac{5}{2x+3}$
$\displaystyle (4-3x)(2x+3)=5x$
$\displaystyle 8x+12-6x^2-9x=5x$
$\displaystyle -6x^2-x+12=5x$
$\displaystyle 6x^2+6x-12=0$
$\displaystyle x^2+x-2=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=1,\ c=-2$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-1\pm\sqrt{1^2-4(1)(-2)}}{2(1)}$
$\displaystyle =\frac{-1\pm\sqrt{9}}{2}$
$\displaystyle =\frac{-1\pm3}{2}$
$\displaystyle \Rightarrow x=1\text{ or }-2$
$\\$

$\displaystyle \textbf{Question 11: } \text{Solve using the quadratic formula: }\frac{2x+3}{x+3}=\frac{x+4}{x+2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{2x+3}{x+3}=\frac{x+4}{x+2}$
$\displaystyle (2x+3)(x+2)=(x+4)(x+3)$
$\displaystyle 2x^2+7x+6=x^2+7x+12$
$\displaystyle x^2-6=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=0,\ c=-6$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-0\pm\sqrt{0^2-4(1)(-6)}}{2(1)}$
$\displaystyle =\frac{\pm\sqrt{24}}{2}$
$\displaystyle =\pm\sqrt{6}$
$\displaystyle \Rightarrow x=\pm\sqrt{6}$
$\\$

$\displaystyle \textbf{Question 12: } \text{Solve using the quadratic formula: }\sqrt{6}x^2-4x-2\sqrt{6}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{6}x^2-4x-2\sqrt{6}=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=\sqrt{6},\ b=-4,\ c=-2\sqrt{6}$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-4)\pm\sqrt{(-4)^2-4(\sqrt{6})(-2\sqrt{6})}}{2\sqrt{6}}$
$\displaystyle =\frac{4\pm\sqrt{16+48}}{2\sqrt{6}}$
$\displaystyle =\frac{4\pm\sqrt{64}}{2\sqrt{6}}$
$\displaystyle =\frac{4\pm8}{2\sqrt{6}}$
$\displaystyle \Rightarrow x=\sqrt{6}\text{ or }-\frac{\sqrt{6}}{3}$
$\\$

$\displaystyle \textbf{Question 13: } \text{Solve using the quadratic formula: }\frac{2x}{x-4}+\frac{2x-5}{x-3}=8\frac{1}{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{2x}{x-4}+\frac{2x-5}{x-3}=8\frac{1}{3}$
$\displaystyle \frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3}$
$\displaystyle 3[2x(x-3)+(2x-5)(x-4)]=25(x-4)(x-3)$
$\displaystyle 3[2x^2-6x+2x^2-13x+20]=25(x^2-7x+12)$
$\displaystyle 12x^2-57x+60=25x^2-175x+300$
$\displaystyle 13x^2-118x+240=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=13,\ b=-118,\ c=240$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-118)\pm\sqrt{(-118)^2-4(13)(240)}}{2(13)}$
$\displaystyle =\frac{118\pm\sqrt{13924-12480}}{26}$
$\displaystyle =\frac{118\pm\sqrt{1444}}{26}$
$\displaystyle =\frac{118\pm38}{26}$
$\displaystyle \Rightarrow x=6\text{ or }\frac{40}{13}$
$\\$

$\displaystyle \textbf{Question 14: } \text{Solve using the quadratic formula: }\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}$
$\displaystyle \frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}$
$\displaystyle 3[(x-1)(x-4)+(x-2)(x-3)]=10(x-2)(x-4)$
$\displaystyle 3[x^2-5x+4+x^2-5x+6]=10(x^2-6x+8)$
$\displaystyle 3(2x^2-10x+10)=10x^2-60x+80$
$\displaystyle 6x^2-30x+30=10x^2-60x+80$
$\displaystyle 4x^2-30x+50=0$
$\displaystyle 2x^2-15x+25=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=2,\ b=-15,\ c=25$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-15)\pm\sqrt{(-15)^2-4(2)(25)}}{2(2)}$
$\displaystyle =\frac{15\pm\sqrt{225-200}}{4}$
$\displaystyle =\frac{15\pm5}{4}$
$\displaystyle \Rightarrow x=5\text{ or }\frac{5}{2}$
$\\$

Without solving, comment on the roots of each of the equations

Notes: The nature of the roots depends on the value of the $\displaystyle b^2-4ac$ . Also in the equation $\displaystyle ax^2+bx+c=0$ , $\displaystyle a, b, and c$ are real and $\displaystyle a \leq 0$ . The following cases can happen:

i) $\displaystyle b^2-4ac = 0$ : In this case the roots would be real and equal.

ii) $\displaystyle b^2-4ac > 0$ : In this case the roots would be real and unequal.

iii) $\displaystyle b^2-4ac < 0$ : In this case the roots would be imaginary numbers

$\displaystyle \textbf{Question 15: } \text{Without solving, comment on the roots of }7x^2-9x+2=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=7,\ b=-9,\ c=2$
$\displaystyle b^2-4ac=(-9)^2-4(7)(2)=81-56=25>0$
$\displaystyle \therefore \text{Since }b^2-4ac>0,\text{ the roots are real and unequal.}$
$\\$

$\displaystyle \textbf{Question 16: } \text{Without solving, comment on the roots of }6x^2-13x+4=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=6,\ b=-13,\ c=4$
$\displaystyle b^2-4ac=(-13)^2-4(6)(4)=169-96=73>0$
$\displaystyle \therefore \text{Since }b^2-4ac>0,\text{ the roots are real and unequal.}$
$\\$

$\displaystyle \textbf{Question 17: } \text{Without solving, comment on the roots of }25x^2-10x+1=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=25,\ b=-10,\ c=1$
$\displaystyle b^2-4ac=(-10)^2-4(25)(1)=100-100=0$
$\displaystyle \therefore \text{Since }b^2-4ac=0,\text{ the roots are real and equal.}$
$\\$

$\displaystyle \textbf{Question 18: } \text{Without solving, comment on the roots of }x^2+2\sqrt{3}x-9=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=2\sqrt{3},\ c=-9$
$\displaystyle b^2-4ac=(2\sqrt{3})^2-4(1)(-9)=12+36=48>0$
$\displaystyle \therefore \text{Since }b^2-4ac>0,\text{ the roots are real and unequal.}$
$\\$

$\displaystyle \textbf{Question 19: } \text{Without solving, comment on the roots of }x^2-ax-b^2=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }Ax^2+Bx+C=0,\ A=1,\ B=-a,\ C=-b^2$
$\displaystyle B^2-4AC=(-a)^2-4(1)(-b^2)=a^2+4b^2>0$
$\displaystyle \therefore \text{Since }B^2-4AC>0,\text{ the roots are real and unequal.}$
$\\$

$\displaystyle \textbf{Question 20: } \text{Without solving, comment on the roots of }2x^2+8x+9=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }Ax^2+Bx+C=0,\ A=2,\ B=8,\ C=9$
$\displaystyle B^2-4AC=8^2-4(2)(9)=64-72=-8<0$
$\displaystyle \therefore \text{Since }B^2-4AC<0,\text{ the roots are imaginary.}$
$\\$

Find the value of $\displaystyle p$ if the quadratic equations have equal roots:

$\displaystyle \textbf{Question 21: } \text{Find the value of }p\text{ if }4x^2-(p-2)x+1=0\text{ has equal roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=4,\ b=-(p-2),\ c=1$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle [-(p-2)]^2-4(4)(1)=0$
$\displaystyle (p-2)^2-16=0$
$\displaystyle p^2-4p+4-16=0$
$\displaystyle p^2-4p-12=0$
$\displaystyle p^2-6p+2p-12=0$
$\displaystyle p(p-6)+2(p-6)=0$
$\displaystyle (p+2)(p-6)=0$
$\displaystyle \Rightarrow p=6\text{ or }-2$
$\\$

$\displaystyle \textbf{Question 22: } \text{Find the value of }p\text{ if }x^2+(p-3)x+p=0\text{ has equal roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=p-3,\ c=p$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle (p-3)^2-4(1)(p)=0$
$\displaystyle p^2-6p+9-4p=0$
$\displaystyle p^2-10p+9=0$
$\displaystyle p^2-9p-p+9=0$
$\displaystyle p(p-9)-1(p-9)=0$
$\displaystyle (p-1)(p-9)=0$
$\displaystyle \Rightarrow p=1\text{ or }9$
$\\$

$\displaystyle \textbf{Question 23: } \text{Find the value of }p\text{ if }3x^2-12x+(p-5)=0\text{ has equal roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=3,\ b=-12,\ c=p-5$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle (-12)^2-4(3)(p-5)=0$
$\displaystyle 144-12(p-5)=0$
$\displaystyle 144-12p+60=0$
$\displaystyle 204-12p=0$
$\displaystyle 12p=204$
$\displaystyle \Rightarrow p=17$
$\\$

$\displaystyle \textbf{Question 24: } \text{Find the value of }p\text{ if }(p-2)x^2-(5+p)x+16=0\text{ has equal roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=p-2,\ b=-(5+p),\ c=16$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle [-(5+p)]^2-4(p-2)(16)=0$
$\displaystyle (5+p)^2-64(p-2)=0$
$\displaystyle p^2+10p+25-64p+128=0$
$\displaystyle p^2-54p+153=0$
$\displaystyle p^2-51p-3p+153=0$
$\displaystyle p(p-51)-3(p-51)=0$
$\displaystyle (p-3)(p-51)=0$
$\displaystyle \Rightarrow p=3\text{ or }51$
$\\$