Class 10, Exercise 7(d), Quadratic Equations (Advanced)

Class 10: Quadratic Equations – Exercise 7(d)

Date: August 26, 2017Author: ICSE CBSE ISC Board Mathematics Portal for Students 0 Comments

Solve each of the following equations:

Question 1: $\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$ where $x \neq 3 \ and \ x \neq -\frac{3}{2}$

Answer:

$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$ where $x \neq 3 \ and \ x \neq -\frac{3}{2}$

$\frac{2x(2x+3)+(x-3)}{(x-3)(2x+3)} + \frac{3x+9}{(x-3)(2x+3)}=0$

$4x^2+6x+x-3+3x+9=0$

$4x^2+10x+6=0$

$4x^2+6x+4x+6=0$

$4x(x+1)+6(x+1)=0$

$(4x+6)((x+1)=0$

$\Rightarrow x = -\frac{3}{2}, -1$

$\\$

Question 2: $(2x+3)^2=81$

Answer:

$(2x+3)^2=81$

$4x^2+9+12x=81$

$4x^2+12x-72=0$

$x^2+3x-18=0$

$x^2+6x-3x-18=0$

$x(x+6)-3(x+6)=0$

$(x-3)(x+6)=0$

$\Rightarrow x = 3, -6$

$\\$

Question 3: $a^2x^2-b^2=0$

Answer:

$a^2x^2-b^2=0$

$(ax-b)(ax+b)=0$

$\Rightarrow x = \frac{b}{a}, -\frac{b}{a}$

$\\$

Question 4: $x^2-\frac{11}{4}x+\frac{15}{8}=0$

Answer:

$x^2-\frac{11}{4}x+\frac{15}{8}=0$ Multiplying the equation by 8.

$8x^2-22x+15=0$

$8x^2-10x-12x+15=0$

$4x(2x-3)-5(2x-3)=0$

$(2x-3)(4x-5)=0$

$\Rightarrow x = \frac{3}{2}, \frac{5}{4}$

$\\$

Question 5: $x+\frac{4}{x}=-4; x \neq 0$

Answer:

$x+\frac{4}{x}=-4; x \neq 0$

$x^2+4x+4=0$

$x^2+2x+2x+4=0$

$(x+2)(x+2)=0$

$\Rightarrow x = -2$

$\\$

Question 6: $2x^4-5x^2+3 = 0$

Answer:

$2x^4-5x^2+3 = 0$

Let $x^2=y$

$2y^2-5y+3=0$

$2y^2-3y-2y+3=0$

$2y(y-1)-3(y-1)=0$

$(2y-3)(y-1)=0$

$\Rightarrow y = \frac{3}{2}, 1$

When $y = 1$

$x^2=1 \Rightarrow x =\pm 1$

When $y = \frac{3}{2}$

$x^2=\frac{3}{2} \Rightarrow x = \pm \sqrt{\frac{3}{2}}$

$\\$

Question 7: $x^4-2x^2-3=0$

Answer:

$x^4-2x^2-3=0$

Let $x^2 = y$

$y^2-2y-3=0$

$y^2-3y+y-3=0$

$y(y-3)+(y-3)=0$

$(y+1)(y-3)=0$

$\Rightarrow y = -1, 3$

When $y = -1$ , $x$ is imaginary

When $y = 3$ , then $x= \pm \sqrt{3}$

$\\$

Question 8: $9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})-52=0$

Answer:

$9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})-52=0$

Let $x + \frac{1}{x}=y$

$x^2+\frac{1}{x^2}+2=y^2$

$x^2+\frac{1}{x^2}=y^2-2$

Therefore $9(y^2-2)-9y-52=0$

$9y^2-9y-70=0$

$9y^2-30y+21y-70=0$

$3y(3y-10)+7(3y-10)=0$

$(3y-10)(3y+7)=0 \Rightarrow y = \frac{10}{3}, -\frac{7}{3}$

When $y = \frac{10}{3}$

$x+\frac{1}{x} = \frac{10}{3}$

$3x^2-10x+3=0$

$3x^2-9x-x+3=0$

$3x(x-3)-(x-3)=0$

$(3x-1)(x-3)=0 \Rightarrow x = \frac{1}{3}, 3$

When $y =-\frac{7}{3}$

$x+\frac{1}{x}=-\frac{7}{3}$

$3x^2+7x+3=0$

Therefore $x = \frac{-7 \pm \sqrt{13}}{6}$

$\\$

Question 9: $2(x^2+\frac{1}{x^2})-(x+\frac{1}{x})=11$

Answer:

$2(x^2+\frac{1}{x^2})-(x+\frac{1}{x})=11$

$y = x+\frac{1}{x}$

$y^2 = x^2+\frac{1}{x^2}+2$

$2(y^2-2)-y=11$

$2y^2-y-15=0$

$2y^2-6y+5y-15=0$

$2y(y-3)+5(y-3)=0$

$(y-3)(2y+5)=0 \Rightarrow y = 3, -\frac{5}{2}$

When $y = 3$

$x+\frac{1}{x}=3$

$x^2-3x+1=0 \Rightarrow x = 3 \pm \frac{\sqrt{5}}{2}$

When $y = -\frac{5}{2}$

$x+\frac{1}{x}=-\frac{5}{2}$

$2x^2+5x+2=0$

$2x^2+4x+x+2=0$

$2x(x+2)+(x+2)=0$

$(x+2)(2x+1)=0 \Rightarrow x = -2, -\frac{1}{2}$

$\\$

Question 10: $(x^2+\frac{1}{x^2})-3(x+\frac{1}{x})-2=0$

Answer:

$(x^2+\frac{1}{x^2})-3(x+\frac{1}{x})-2=0$

Let $x -\frac{1}{x}=y$

$x^2+\frac{1}{x^2} = y^2+2$

$(y^2+2)-3y-2=0$

$y^2-3y=0$

$y(y-3)=0 \Rightarrow y = 0, 3$

When $y = 0$

$x-\frac{1}{x}=0$

$x^2-1=0 \Rightarrow x = \pm 1$

When $y =3$

$x-\frac{1}{x}=3$

$x^2-3x-1=0 \Rightarrow \frac{3 \pm \sqrt{5}}{2}$

$\\$

Question 11: $(x^2+5x+4)(x^2+5x+6)=120$

Answer:

$(x^2+5x+4)(x^2+5x+6)=120$

Let $x^2+5x=y$

$(y+4)(y+6)=120$

$y^2+10y-96=0$

$y^2-6y+16y-96=0$

$y(y-6)+16(y-6)=0$

$(y-6)(y+16)=0 \Rightarrow y =6, -16$

When $y = 6$

$x^2+5x-6=0$

$x^2+6x-x-6=0$

$x(x+6)-(x+6)=0$

$(x+6)(x-1)=0 \Rightarrow x = -6, 1$

When $y = -16$

$x^2+5x+16=0 \Rightarrow x$ is imaginary

$\\$

Question 12: $x^2-5x-10=0$ [2005]

Answer:

Comparing $x^2-5x-10=0$ with $ax^2+bx+c=0$ , we get $a = 1, b = -5 \ and \ c =-10$

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

Therefore $x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-10)}}{2(1)}$

Solving we get $x = 6.53, -1.53$

$\\$

Question 13: $3x^2-x-7=0$ [2004]

Answer:

Comparing $3x^2-x-7=0$ with $ax^2+bx+c=0$ , we get $a = 3, b = -1 \ and \ c =-7$

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

Therefore $x = \frac{-(-1) \pm \sqrt{(-1)^2-4(3)(-7)}}{2(3)}$

Solving we get $x = 1.703, -1.3699$

$\\$

Question 14: $(\frac{x}{x+2})^2-7(\frac{x}{x+2})+12=0, x \neq -2$

Answer:

$(\frac{x}{x+2})^2-7(\frac{x}{x+2})+12=0, x \neq -2$

Let $\frac{x}{x+2}=y$

$y^2-7y+12=0$

$y^2-3y-4y+12=0$

$y(y-3)-4(y-3)=0$

$(y-3)(y-4)=0 \Rightarrow y = 3, 4$

When $y = 3$

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

$3x+6=x \Rightarrow x = -3$

When $y =4$

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

$4x+8=x \Rightarrow x = -\frac{8}{3}$

$\\$

Question 15: $x^2-11x-12=0; \ when \ x \in N$

Answer:

$x^2-11x-12=0; \ when \ x \in N$

$x^2-12x+x-12=0$

$x(x-12)+(x-12)=0$

$(x-12)(x+1)=0$

$x=12, -1$ Since $x \in N, x = 12$

$\\$

Question 16: $x^2-4x-12=0; \ when \ x \in I$

Answer:

$x^2-4x-12=0; \ when \ x \in I$

$x^2-6x+2x-12=0$

$x(x-6)+2(x-6)=0$

$(x-6)(x+2)=0 \Rightarrow x = 6, -2$

$\\$

Question 17: $2x^2-9x+10=0; \ when \ x \in Q$

Answer:

$2x^2-9x+10=0; \ when \ x \in Q$

$2x^2-4x-5x+10=0$

$2x(x-2)-5(x-2)=0$

$(x-2)(2x-5)=0 \Rightarrow x = 2, \frac{5}{2}$

$\\$

Question 18: $(a+b)^2x^2-(a+b)x-6=0; a+b \neq 0$

Answer:

$(a+b)^2x^2-(a+b)x-6=0; a+b \neq 0$

Let $(a+b)x = y$

$y^2-y-6=0$

$y^2-3y+2y-6=0$

$y(y-3)+2(y-3)=0$

$(y-3)(y+2)=0 \Rightarrow y = 3, -2$

When $y =3 \Rightarrow (a+b)x = 3 \Rightarrow x = \frac{3}{a+b}$

When $y = -2 \Rightarrow (a+b)x = -2 \Rightarrow x =-\frac{2}{a+b}$

$\\$

Question 19: $\frac{1}{p}+\frac{1}{q}+\frac{1}{x} = \frac{1}{x+p+q}$

Answer:

$\frac{1}{p}+\frac{1}{q}+\frac{1}{x} = \frac{1}{x+p+q}$

$\frac{1}{p}+\frac{1}{q}+\frac{1}{x} - \frac{1}{x+p+q} =0$

$\frac{p+q}{pq}+\frac{pq}{x(x+p+q)}=0$

$x(x+p+q)+pq=0$

$x^2+(p+q)x+pq = 0$

$(x+p)(x+q)=0 \Rightarrow x = -p, -q$

$\\$

Question 20: $x(x+1)+(x+2)(x+3)=42$

Answer:

$x(x+1)+(x+2)(x+3)=42$

$x^2+x+x^2+5x+6 = 42$

$2x^2+6x-36=0$

$x^2+3x-18=0$

$x^2-3x+6x-18=0$

$x(x-3)+6(x-3)=0$

$(x-3)(x+6)=0 \Rightarrow x = 3, -6$

$\\$

Question 21: $\frac{1}{x+1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}$

Answer:

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

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

$\frac{x+2-2x-2}{(x+1)(x+2)} = \frac{3x+12-4x-12}{(x+3)(x+4)}$

$(x+3)(x+4)=(x+1)(x+2)$

$x^2+7x+12=x^2+3x+2$

$4x=-10 \Rightarrow x = -\frac{5}{2}$

$\\$

Find $m \ or \ p$ so that the equation have equal roots:

Question 22: $(m-3)x^2-4x+1=0$

Answer:

Comparing $(m-3)x^2-4x+1=0$ with $ax^2+bx+c=0$ , we get $a = (m-3), b = -4 \ and \ c =1$

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

$(-4)^2-4(m-3)(1)=0$

$16-4m+12=0 \Rightarrow 7$

Solution: $4x^2-4x+1=0$

$(2x-1)(2x-1)=0 \Rightarrow x = \frac{1}{2}$

$\\$

Question 23: $3x^2+12x+(m+7)=0$

Answer:

Comparing $3x^2+12x+(m+7)=0$ with $ax^2+bx+c=0$ , we get $a = 3, b = 12 \ and \ c =(m+7)$

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

$(12)^2-4(3)(m+7)=0$

$144-12m-84=0$

$12m = 60 \Rightarrow m = 5$

Solving the equation

$3x^2+12x+12=0$

$x^2+4x+4=0$

$(x+2)(x+2)=0 \Rightarrow x = -2$

$\\$

Question 24: $x^2-(m+2)x+(m+5)=0$

Answer:

Comparing $x^2-(m+2)x+(m+5)=0$ with $ax^2+bx+c=0$ , we get $a = 1, b = -(m+2) \ and \ c =(m+5)$

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

$(m+2)^2-4(1)(m+5)=0$

$m^2+4+4m-4m-20=0$

$m^2=16 \Rightarrow m =\pm 4$

When $m=4$

$x^2-6x+9=0$

$(x-3)(x-3)=0 \Rightarrow x = 3$

When $m = -4$

$x^2+2x+1=0$

$(x+1)^2 = 0 \Rightarrow x = -1$

$\\$

Question 25: $px^2-4x+3=0$ [2010]

Answer:

Comparing $px^2-4x+3=0$ with $ax^2+bx+c=0$ , we get $a = p, b = -4 \ and \ c =3$

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

$(-4)^2-4(p)(3)=0$

$16-12p=0$

$p=\frac{4}{3}$

$\\$

Question 26: $x^2+2(m-1)x+(m+5)=0$ [2012]

Answer:

Comparing $x^2+2(m-1)x+(m+5)=0$ with $ax^2+bx+c=0$ , we get $a = 1, b = 2(m-1) \ and \ c =(m+5)$

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

$(2(m-1))^2-4(1)(m+5)=0$

$4(m^2+1-2m)-4(m+5)=0$

$4m^2+4-8m-4m-20=0$

$4m^2-12m-16=0$

$m^2-3m-4=0$

$m^2-4m+m-4=0$

$m(m-4)+(m-4)=0$

$(m-4)(m+1)=0 \Rightarrow m = 4, -1$

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