Class 10: Quadratic Equations – Exercise 7(b) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Solve each of the following use quadratic formula

Question 1: $\displaystyle x^2-6x=27$

Answer:

Comparing $\displaystyle x^2-6x=27$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -6 \text{ and } c =-27$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(-6) \pm \sqrt{(-6)^2-4(1)(-27)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 9, -3$

$\displaystyle \\$

Question 2: $\displaystyle x^2-10x+21=0$

Answer:

Comparing $\displaystyle x^2-10x+21=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -10 \text{ and } c =21$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(-10) \pm \sqrt{(-10)^2-4(1)(21)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 3, 7$

$\displaystyle \\$

Question 3: $\displaystyle x^2+6x-10=0$

Answer:

Comparing $\displaystyle x^2+6x-10=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = 6 \text{ and } c =-10$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(6) \pm \sqrt{(6)^2-4(1)(-10)}}{2(1)}$

$\displaystyle \text{Solving we get } x = -3 \pm \sqrt{19}$

$\displaystyle \\$

Question 4: $\displaystyle x^2+2x-6=0$

Answer:

Comparing $\displaystyle x^2+2x-6=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = 2 \text{ and } c =-6$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(2) \pm \sqrt{(2)^2-4(1)(-6)}}{2(1)}$

$\displaystyle \text{Solving we get } x = -1 \pm \sqrt{7}$

$\displaystyle \\$

Question 5: $\displaystyle 3x^2+2x-1=0$

Answer:

Comparing $\displaystyle 3x^2+2x-1=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 3, b = 2 \text{ and } c =-1$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(2) \pm \sqrt{(2)^2-4(3)(-1)}}{2(3)}$

$\displaystyle \text{Solving we get } x = -1, \frac{1}{3}$

$\displaystyle \\$

Question 6: $\displaystyle 2x^2+7x+5=0$

Answer:

Comparing $\displaystyle 2x^2+7x+5=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 2, b = 7 \text{ and } c =5$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(7) \pm \sqrt{(7)^2-4(2)(5)}}{2(2)}$

$\displaystyle \text{Solving we get } x = -1, - \frac{5}{2}$

$\displaystyle \\$

$\displaystyle \text{Question 7: } \frac{2}{3} x=- \frac{1}{6} x^2- \frac{1}{3}$

Answer:

$\displaystyle \frac{2}{3} x=- \frac{1}{6} x^2- \frac{1}{3}$

Multiplying by 6

$\displaystyle 4x = -x^2-2$ or $\displaystyle x^2+4x+2=0$

Comparing $\displaystyle x^2+4x+2=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = 4 \text{ and } c =2$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(4) \pm \sqrt{(4)^2-4(1)(2)}}{2(1)}$

$\displaystyle \text{Solving we get } x = -2 \pm \sqrt{2}$

$\displaystyle \\$

$\displaystyle \text{Question 8: } \frac{1}{15} x^2+ \frac{5}{3} = \frac{2}{3} x$

Answer:

$\displaystyle \frac{1}{15} x^2+ \frac{5}{3} = \frac{2}{3} x$

Multiplying by 15 and simplifying we get

$\displaystyle x^2+25=10x$ or $\displaystyle x^2-10x+25=0$

Comparing $\displaystyle x^2-10x+25=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -10 \text{ and } c =25$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(-10) \pm \sqrt{(-10)^2-4(1)(25)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 5, 5$

$\displaystyle \\$

Question 9: $\displaystyle x^2-6x=2\sqrt{2}x$

Answer:

$\displaystyle x^2-6x=2\sqrt{2}x$

or $\displaystyle x^2-2\sqrt{2}x-6x=0$

Comparing $\displaystyle x^2-2\sqrt{2}x-6x=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -2\sqrt{2} \text{ and } c =-6$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(-2\sqrt{2}) \pm \sqrt{(-2\sqrt{2})^2-4(1)(-6)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 3\sqrt{2}, -\sqrt{2}$

$\displaystyle \\$

$\displaystyle \text{Question 10: } \frac{4}{x} -3= \frac{5}{2x+3}$

Answer:

$\displaystyle \frac{2x+3}{x+3} = \frac{x+4}{x+2}$

$\displaystyle (2x+3)(4-3x)=5x$

$\displaystyle 8x+12-6x^2-9x = 5x$

$\displaystyle 6x^2+6x-12=0$ or $\displaystyle x^2+x-2=0$

Comparing $\displaystyle x^2-2\sqrt{2}x-6x=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = 1 \text{ and } c =-2$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(1) \pm \sqrt{(1)^2-4(1)(-2)}}{2(1)}$

$\displaystyle \text{Solving we get } x = -2, 1$

$\displaystyle \\$

$\displaystyle \text{Question 11: } \frac{2x+3}{x+3} = \frac{x+4}{x+2}$

Answer:

$\displaystyle \frac{2x+3}{x+3} = \frac{x+4}{x+2}$

$\displaystyle (2x+3)(x+2)=(x+4)(x+3)$

$\displaystyle 2x^2+7x+6 = x^2+7x+12$

$\displaystyle x^2-6=0$

Comparing $\displaystyle x^2-6=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = 0 \text{ and } c =-6$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(0) \pm \sqrt{(0)^2-4(1)(-6)}}{2(1)}$

$\displaystyle \text{Solving we get } x = \pm \sqrt{6}$

$\displaystyle \\$

Question 12: $\displaystyle \sqrt{6} x^2-4x-2\sqrt{6}=0$

Answer:

$\displaystyle \sqrt{6} x^2-4x-2\sqrt{6}=0$

Comparing $\displaystyle \sqrt{6} x^2-4x-2\sqrt{6}=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = \sqrt{6}, b = -4 \text{ and } c =-2\sqrt{6}$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(-4) \pm \sqrt{(-4)^2-4(\sqrt{6})(-2\sqrt{6})}}{2(\sqrt{6})}$

$\displaystyle \text{Solving we get } x = \sqrt{6}, - \frac{\sqrt{6}}{3}$

$\displaystyle \\$

$\displaystyle \text{Question 13: } \frac{2x}{x-4} + \frac{2x-5}{x-3} =8 \frac{1}{3}$

Answer:

$\displaystyle \frac{2x}{x-4} + \frac{2x-5}{x-3} =8 \frac{1}{3}$

$\displaystyle 3[2x(x-3)+(2x-5)(x-4)]=25(x-4)(x-3)$

$\displaystyle 3(2x^2-6x+2x^2-13x+20) = 25(x^2-7x+12)$

$\displaystyle 12x^2-57x+60=25x^2-175x+300$

$\displaystyle 13x^2-118x+240=0$

Comparing $\displaystyle 13x^2-118x+240=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 13, b = -118 \text{ and } c =240$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(-118) \pm \sqrt{(-118)^2-4(13)(240)}}{2(13)}$

$\displaystyle \text{Solving we get } x = 6, \frac{40}{13}$

$\displaystyle \\$

$\displaystyle \text{Question 14: } \frac{x-1}{x-2} + \frac{x-3}{x-4} =3 \frac{1}{3}$

Answer:

$\displaystyle \frac{x-1}{x-2} + \frac{x-3}{x-4} =3 \frac{1}{3}$

$\displaystyle 3[(x-1)(x-4)+(x-2)(x-3)]=10(x-2)(x-4)$

$\displaystyle 3(2x^2-10x+10)=10(x^2-6x+8)$

$\displaystyle 6x^2-30x+30=10x^2-60x+80$

$\displaystyle 4x^2-30x+50=0 or 2x^2-15x+25=0$

Comparing $\displaystyle 2x^2-15x+25=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 2, b = -15 \text{ and } c =25$

$\displaystyle \text{Since } x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } x = \frac{-(-15) \pm \sqrt{(-15)^2-4(2)(25)}}{2(2)}$

$\displaystyle \text{Solving we get } x = 5, \frac{5}{2}$

$\displaystyle \\$

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

Question 15: $\displaystyle 7x^2-9x+2=0$

Answer:

Comparing $\displaystyle 7x^2-9x+2=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 7, b = -9 \text{ and } c =2$

$\displaystyle \text{Therefore } b^2-4ac = (-9)^2-4(7)(2) = 81 - 56 = 25 > 0$

Hence $\displaystyle \text{Since } b^2-4ac > 0$ : In this case the roots would be real and unequal.

$\displaystyle \\$

Question 16: $\displaystyle 6x^2-13x+4=0$

Answer:

Comparing $\displaystyle 6x^2-13x+4=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 6, b = -13 \text{ and } c =4$

$\displaystyle \text{Therefore } b^2-4ac = (-13)^2-4(6)(4) = 169 - 96 = 73 > 0$

Hence $\displaystyle \text{Since } b^2-4ac > 0$ : In this case the roots would be real and unequal.

$\displaystyle \\$

Question 17: $\displaystyle 25x^2-10x+1=0$

Answer:

Comparing $\displaystyle 25x^2-10x+1=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 25, b = -10 \text{ and } c =1$

$\displaystyle \text{Therefore } b^2-4ac = (-10)^2-4(25)(1) = 100 - 100 = 0$

Hence $\displaystyle \text{Since } b^2-4ac = 0$ : In this case the roots would be real and equal

$\displaystyle \\$

Question 18: $\displaystyle x^2+2\sqrt{3}-9=0$

Answer:

Comparing $\displaystyle x^2+2\sqrt{3}-9=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = 2\sqrt{3} \text{ and } c =-9$

$\displaystyle \text{Therefore } b^2-4ac = (-9)^2-4(1)(2\sqrt{3}) = 81 - 8\sqrt{3} > 0$

Hence $\displaystyle \text{Since } b^2-4ac > 0$ : In this case the roots would be real and unequal

$\displaystyle \\$

Question 19: $\displaystyle x^2-ax-b^2=0$

Answer:

Comparing $\displaystyle x^2-ax-b^2=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -a \text{ and } c =-b^2$

$\displaystyle \text{Therefore } b^2-4ac = (-a)^2-4(1)(-b^2) = a^2+b^2 > 0$

Hence $\displaystyle \text{Since } b^2-4ac > 0$ : In this case the roots would be real and unequal

$\displaystyle \\$

Question 20: $\displaystyle 2x^2+8x+9=0$

Answer:

Comparing $\displaystyle x^2-ax-b^2=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -a \text{ and } c =-b^2$

$\displaystyle \text{Therefore } b^2-4ac = (-a)^2-4(1)(-b^2) = a^2+b^2 > 0$

Hence $\displaystyle \text{Since } b^2-4ac > 0$ : In this case the roots would be real and unequal

$\displaystyle \\$

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

Question 21: $\displaystyle 4x^2-(p-2)x+1=0$

Answer:

Comparing $\displaystyle 4x^2-(p-2)x+1=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 4, b = -(p-2) \text{ and } c =1$

For roots to be equal, we should have $\displaystyle b^2-4ac = 0$

Substituting

$\displaystyle (-(p-2)^2)-4(4)(1) = 0$

$\displaystyle p^2+4-4p-16=0$

$\displaystyle p^2-4p-12=0$

$\displaystyle p^2-6p+2p-12=0$

$\displaystyle p(p-6)+2(p-=0$

$\displaystyle (p+2)(p-6)=0 \Rightarrow p = 6 or p = -2$

$\displaystyle \\$

Question 22: $\displaystyle x^2+(p-3)x+p=0$

Answer:

Comparing $\displaystyle x^2+(p-3)x+p=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = (p-3) \text{ and } c =p$

For roots to be equal, we should have $\displaystyle b^2-4ac = 0$

Substituting

$\displaystyle ((p-3)^2)-4(1)(p) = 0$

$\displaystyle p^2+9-6p-4p=0$

$\displaystyle p^2-10p+9=0$

$\displaystyle p^2-9p-p+9=0$

$\displaystyle (p-1)(p-9)=0 \Rightarrow p = 1 or p = 9$

$\displaystyle \\$

Question 23: $\displaystyle 3x^2-12x+(p-5)=0$

Answer:

$\displaystyle 3x^2-12x+(p-5)=0$

Comparing $\displaystyle 3x^2-12x+(p-5)=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 3, b = -12 \text{ and } c =(n-5)$

For roots to be equal, we should have $\displaystyle b^2-4ac = 0$

$\displaystyle (-12)^2-4(3)(n-5)=0$

$\displaystyle 144-12n+60=0$

$\displaystyle 12n = 84$

or $\displaystyle n = 6$

$\displaystyle \\$

Question 24: $\displaystyle (p-2)x^2-(5+p)x+16=0$

Answer:

$\displaystyle (p-2)x^2-(5+p)x+16=0$

Comparing $\displaystyle (p-2)x^2-(5+p)x+16=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = (p-2), b = -(5+p) \text{ and } c =16$

For roots to be equal, we should have $\displaystyle b^2-4ac = 0$

$\displaystyle (-(5+p)^2)-4(p-2)(16)=0$

$\displaystyle 25+p^2+10p-64p+128=0$

$\displaystyle p^2-54p+153=0$

$\displaystyle p^2-51p-3p+153=0$

$\displaystyle p(p-51)-3(p-51)=0$

$\displaystyle (p-3)(p-51)=0 \Rightarrow p = 3 or 51$

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Class 10: Quadratic Equations – Exercise 7(b)

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