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Accedmychevron_right11thchevron_rightmathchevron_rightTheory Of Quadratic Functionschevron_rightExercise 3.2

Solution

(i) 13x+4x6=1,  x≠0\dfrac{1}{3x}+\dfrac{4x}{6}=1,\; x\ne 0

13x+2x3=1\frac{1}{3x}+\frac{2x}{3}=1

Multiply by 3x3x:

1+2x2=3x1+2x^2=3x 2x2−3x+1=02x^2-3x+1=0 (2x−1)(x−1)=0(2x-1)(x-1)=0 x=12,  x=1\boxed{x=\frac{1}{2},\; x=1}

(ii) xx+1+x+1x=52,  x≠−1,0\dfrac{x}{x+1}+\dfrac{x+1}{x}=\dfrac{5}{2},\; x\ne -1,0

Multiply by 2x(x+1)2x(x+1):

2x(x+1)(xx+1+x+1x)=2x(x+1)⋅522x(x+1)\left(\frac{x}{x+1}+\frac{x+1}{x}\right)=2x(x+1)\cdot\frac{5}{2} 2x2+2(x+1)2=5x(x+1)2x^2+2(x+1)^2=5x(x+1) 2x2+2(x2+2x+1)=5x2+5x2x^2+2(x^2+2x+1)=5x^2+5x 4x2+4x+2=5x2+5x4x^2+4x+2=5x^2+5x x2+x−2=0x^2+x-2=0 (x+2)(x−1)=0(x+2)(x-1)=0 x=−2,  x=1\boxed{x=-2,\; x=1}

(iii) 1x+1+2x+2=7x+5,  x≠−1,−2,−5\dfrac{1}{x+1}+\dfrac{2}{x+2}=\dfrac{7}{x+5},\; x\ne -1,-2,-5

Multiply by (x+1)(x+2)(x+5)(x+1)(x+2)(x+5):

(x+2)(x+5)+2(x+1)(x+5)=7(x+1)(x+2)(x+2)(x+5)+2(x+1)(x+5)=7(x+1)(x+2)

Expand:

(x2+7x+10)+(2x2+12x+10)=7x2+21x+14(x^2+7x+10)+(2x^2+12x+10)=7x^2+21x+14 3x2+19x+20=7x2+21x+143x^2+19x+20=7x^2+21x+14 4x2+2x−6=04x^2+2x-6=0 2x2+x−3=02x^2+x-3=0 (2x+3)(x−1)=0(2x+3)(x-1)=0 x=−32,  x=1\boxed{x=-\frac{3}{2},\; x=1}

(iv) aax−1+bbx−1=a+b,  x≠1a,1b\dfrac{a}{ax-1}+\dfrac{b}{bx-1}=a+b,\; x\ne \dfrac{1}{a},\dfrac{1}{b}

Multiply by (ax−1)(bx−1)(ax-1)(bx-1):

a(bx−1)+b(ax−1)=(a+b)(ax−1)(bx−1)2abx−(a+b)=(a+b)(abx2−(a+b)x+1)\begin{aligned} a(bx-1)+b(ax-1)&=(a+b)(ax-1)(bx-1)\\ 2abx-(a+b)&=(a+b)(abx^2-(a+b)x+1) \end{aligned}

This simplifies to a quadratic in xx (with parameters a,ba,b). The solutions are:

x=2a+b,  x=a+bab\boxed{x=\frac{2}{a+b},\; x=\frac{a+b}{ab}}

(v) 3x2+15x−2x2+5x+1=23x^2+15x-2\sqrt{x^2+5x+1}=2

Let t=x2+5x+1t=\sqrt{x^2+5x+1}, so t2=x2+5x+1t^2=x^2+5x+1.

Note 3x2+15x=3(x2+5x)=3(t2−1)3x^2+15x=3(x^2+5x)=3(t^2-1).

3(t2−1)−2t=23(t^2-1)-2t=2 3t2−2t−5=03t^2-2t-5=0 (3t−5)(t+1)=0(3t-5)(t+1)=0

Since t≥0t\ge 0, take t=53t=\frac{5}{3}.

x2+5x+1=53\sqrt{x^2+5x+1}=\frac{5}{3}

Square:

x2+5x+1=259x^2+5x+1=\frac{25}{9} 9x2+45x−16=09x^2+45x-16=0 x=−45±260118=−45±5118x=\frac{-45\pm\sqrt{2601}}{18}=\frac{-45\pm 51}{18} x=13,  x=−163\boxed{x=\frac{1}{3},\; x=-\frac{16}{3}}

(vi) 2x+8+x+5=7\sqrt{2x+8}+\sqrt{x+5}=7

Isolate and square:

2x+8=7−x+5\sqrt{2x+8}=7-\sqrt{x+5} 2x+8=49−14x+5+x+52x+8=49-14\sqrt{x+5}+x+5 46−x=14x+546-x=14\sqrt{x+5}

Square again:

(46−x)2=196(x+5)(46-x)^2=196(x+5) x2−288x+1136=0x^2-288x+1136=0 (x−4)(x−284)=0(x-4)(x-284)=0

Check in original: x=284x=284 is extraneous.

x=4\boxed{x=4}

(vii) 3x+4=2+2x−4\sqrt{3x+4}=2+\sqrt{2x-4}

Square:

3x+4=4+42x−4+2x−43x+4=4+4\sqrt{2x-4}+2x-4 x+4=42x−4x+4=4\sqrt{2x-4}

Square again:

(x+4)2=16(2x−4)(x+4)^2=16(2x-4) x2−24x+80=0x^2-24x+80=0 (x−20)(x−4)=0(x-20)(x-4)=0

Both satisfy the original equation.

x=4,  20\boxed{x=4,\; 20}

(viii) x+5−x−3=2\sqrt{x+5}-\sqrt{x-3}=2

x+5=2+x−3\sqrt{x+5}=2+\sqrt{x-3}

Square:

x+5=4+4x−3+x−3x+5=4+4\sqrt{x-3}+x-3 4=4x−34=4\sqrt{x-3} x−3=1⇒x=4\sqrt{x-3}=1\Rightarrow x=4 x=4\boxed{x=4}
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