CBSE Class 10 Mathematics Quadratic Equations MCQs Set 26

Answer: (b) \(\dfrac15\) and \(\dfrac45\)

Explanation:
1. Put \(t=\sqrt{x/(1-x)}\). Then \(t+\dfrac1t=\dfrac52\), so \(2t^2-5t+2=0\) and \(t=2\) or \(\dfrac12\).
2. \(\dfrac{x}{1-x}=4\) gives \(x=\dfrac45\), and \(\dfrac{x}{1-x}=\dfrac14\) gives \(x=\dfrac15\).
3. Both values satisfy the original equation.

Teacher's Note:
1. A substitution \(t+1/t=k\) turns this into a quadratic.
2. Check that the values keep the expressions under the roots positive.

Answer: (c) length 46 m, breadth 31 m

Explanation:
1. Let the breadth be \(b\) m, so the length is \(b+15\).
2. New area \((b+23)(b-3)=b(b+15)+86\) gives \(b^2+20b-69=b^2+15b+86\).
3. So \(5b=155\), \(b=31\) and the length is 46 m.

Teacher's Note:
1. This is actually linear because the \(b^2\) terms cancel.
2. Check: \(46\times31=1426\) and \(54\times28=1512\), a difference of 86.

Answer: (c) 2550 and 2450

Explanation:
1. Let the first pump deliver \(q\) litres a minute. Then \(10q+10(q-10)=5000\).
2. \(20q=5100\), so \(q=255\) litres a minute.
3. In 10 minutes the pumps deliver 2550 litres and 2450 litres.

Teacher's Note:
1. Half of 10 000 litres is 5000 litres.
2. The second pump gives 245 litres a minute.

Answer: (d) 6, 7, 8

Explanation:
1. Let the numbers be \(n-1,n,n+1\). Product \(=n(n^2-1)\) and sum \(=3n\).
2. \(n(n^2-1)=48n\), so \(n^2=49\) (taking \(n\neq0\)).
3. For positive integers \(n=7\), giving 6, 7, 8.

Teacher's Note:
1. Taking the middle number as \(n\) keeps the working short.
2. The integers \(-8,-7,-6\) and \(-1,0,1\) also satisfy the condition, which is why the question asks for positive integers.

Answer: (c) 12 sides

Explanation:
1. \(\dfrac{n(n-3)}{2}=54\) gives \(n^2-3n-108=0\).
2. \((n-12)(n+9)=0\), so \(n=12\) (the negative root is rejected).

Teacher's Note:
1. The number of sides must be a positive integer.
2. Check: \(12\times9/2=54\).

Answer: (c) 12 and 5 units

Explanation:
1. Let the sides be \(l\) and \(b\). Then \(l+b=17\).
2. The diagonal is the diameter, so \(l^2+b^2=169\).
3. \(lb=\dfrac{289-169}{2}=60\), so \(l,b\) are roots of \(t^2-17t+60=0\), that is 12 and 5.

Teacher's Note:
1. The diagonal of a rectangle inscribed in a circle is a diameter.
2. Use \((l+b)^2=l^2+b^2+2lb\).

Answer: (d) \(x=3\) and \(x=\dfrac{10}{13}\)

Explanation:
1. Multiply through by \(2(x-2)(x-1)\): \(8x(x-1)-6x(x-2)=15(x-2)(x-1)\).
2. \(2x^2+4x=15x^2-45x+30\), so \(13x^2-49x+30=0\).
3. \((x-3)(13x-10)=0\), so \(x=3\) or \(\dfrac{10}{13}\).

Teacher's Note:
1. Values making a denominator zero (\(x=1,2\)) must be rejected.
2. Both roots found are allowed.

Answer: (a) \(\dfrac35\)

Explanation:
1. Multiply by \(x(x+1)(3x-1)\): \(3(x+1)(3x-1)=2x(3x-1)+3x(x+1)\).
2. LHS \(=9x^2+6x-3\) and RHS \(=6x^2-2x+3x^2+3x=9x^2+x\).
3. So \(5x=3\) and \(x=\dfrac35\).

Teacher's Note:
1. The \(x^2\) terms cancel, so the equation is really linear.
2. Check: LHS \(=5\) and RHS \(=\dfrac{2}{8/5}+\dfrac{3}{4/5}=1.25+3.75=5\).

Answer: (b) \(x=3\)

Explanation:
1. Isolate: \(\sqrt{2x-5}=3-\sqrt{x+1}\). Squaring, \(2x-5=9-6\sqrt{x+1}+x+1\), so \(6\sqrt{x+1}=15-x\).
2. Squaring again: \(36(x+1)=225-30x+x^2\), \(x^2-66x+189=0\), \((x-3)(x-63)=0\).
3. \(x=63\) fails the original equation (\(8+11\neq3\)); \(x=3\) gives \(2+1=3\).

Teacher's Note:
1. Squaring can add false roots, so always check.
2. Here \(x=63\) is the extraneous root.

Answer: (b) 1

Explanation:
1. Let \(u=5^{x-1}\). Then \(u+\dfrac1u=2\), so \((u-1)^2=0\) and \(u=1\).
2. \(5^{x-1}=1\) gives \(x-1=0\), so \(x=1\).

Teacher's Note:
1. \(u+1/u\ge2\) for positive \(u\), with equality only at \(u=1\).
2. This is the AM-GM inequality.

Answer: (b) 35 m and 25 m

Explanation:
1. Let the breadth be \(b\). Then \(b(b+10)=875\), so \(b^2+10b-875=0\).
2. \((b-25)(b+35)=0\), so \(b=25\) (the negative root is rejected).
3. The length is 35 m.

Teacher's Note:
1. Always reject a negative length.
2. Check: \(35\times25=875\).

Answer: (d) 30 m and 40 m

Explanation:
1. Let the sides be \(a\) and \(a+10\). Then \(a^2+(a+10)^2=2500\).
2. \(2a^2+20a-2400=0\), so \(a^2+10a-1200=0\) and \((a-30)(a+40)=0\).
3. \(a=30\), and the sides are 30 m and 40 m.

Teacher's Note:
1. This is a multiple of the 3-4-5 triangle.
2. Check: \(900+1600=2500\).

Answer: (c) \(x=3\) and \(x=-\dfrac43\)

Explanation:
1. Combine: \(\dfrac{2x}{x^2-4}=\dfrac65\), so \(10x=6x^2-24\).
2. \(3x^2-5x-12=0\), \((x-3)(3x+4)=0\).
3. So \(x=3\) or \(-\dfrac43\).

Teacher's Note:
1. Exclude \(x=\pm2\) as they make the denominators zero.
2. Both roots are valid.

Answer: (c) \(x=2\) and \(x=4\)

Explanation:
1. Put \(y=-3\): \(x^2-6x+8=0\).
2. \((x-2)(x-4)=0\), so \(x=2\) or \(4\).

Teacher's Note:
1. On the graph these are the points where the line \(y=-3\) cuts the parabola.
2. They are symmetric about \(x=3\).

Answer: (c) \(x=3\) and \(x=-\dfrac23\)

Explanation:
1. \(3x^2-7x-6=0\) means \(3x^2=7x+6\), the intersections of \(y=3x^2\) with \(y=7x+6\).
2. Factorising, \((x-3)(3x+2)=0\).
3. So \(x=3\) or \(-\dfrac23\).

Teacher's Note:
1. Draw one standard parabola and a straight line to solve many quadratics.
2. Check: \(27-21-6=0\).

Answer: (d) true, because \(x^3=x\cdot x^2=x(x+1)=x^2+x=2x+1\)

Explanation:
1. Multiply \(x^2=x+1\) by \(x\): \(x^3=x^2+x\).
2. Replace \(x^2\) by \(x+1\): \(x^3=2x+1\).

Teacher's Note:
1. Reducing powers using the given relation is a standard trick.
2. The same method gives \(x^4=3x+2\).

Answer: (a) not real (complex)

Explanation:
1. Discriminant \(D=(-8)^2-4(5)(9)=64-180=-116<0\).
2. So the roots are not real.

Teacher's Note:
1. \(D<0\) means no real roots.
2. \(D=0\) gives equal roots and \(D>0\) gives two distinct real roots.

Answer: (c) 25

Explanation:
1. \(4^x-4^{x-1}=4^{x-1}(4-1)=3\cdot4^{x-1}=24\), so \(4^{x-1}=8\).
2. \(2^{2x-2}=2^3\), so \(2x=5\).
3. \((2x)^2=25\).

Teacher's Note:
1. Factor out the smaller power of 4.
2. Here \(2x=5\) directly.

Answer: (a) a multiple of 11

Explanation:
1. Let the digits be \(a,\ a+c,\ c\). The number is \(100a+10(a+c)+c=110a+11c\).
2. This equals \(11(10a+c)\), a multiple of 11.

Teacher's Note:
1. The alternating digit sum \(a-(a+c)+c=0\) is the divisibility test for 11.
2. Such numbers are 110, 121, 132, ...

Answer: (a) 9, 18, 27 (such numbers exist)

Explanation:
1. Let the numbers be \(k,2k,3k\). Then \(k^2+4k^2+9k^2=14k^2=1134\).
2. \(k^2=81\), so \(k=9\) (or \(-9\)).
3. The numbers are 9, 18, 27.

Teacher's Note:
1. Check: \(81+324+729=1134\).
2. The negative choice \(-9,-18,-27\) also works.

Answer: (a) \(\dfrac32\)

Explanation:
1. \(\dfrac{p^2+p+1}{p^2+1}=1+\dfrac{p}{p^2+1}\).
2. Since \(p^2+1\ge2|p|\), \(\dfrac{p}{p^2+1}\le\dfrac12\), with equality at \(p=1\).
3. The greatest value is \(1+\dfrac12=\dfrac32\).

Teacher's Note:
1. The expression is therefore at most \(\dfrac32\) for all real \(p\).
2. Use \(p^2+1\ge2p\).

Answer: (a) \(0\) and \(\dfrac{18}{5}\)

Explanation:
1. From \(5x+9=0\), \(x=-\dfrac95\).
2. Substitute: \(5\cdot\dfrac{81}{25}-\dfrac95(9+4m)+2m^2=0\), i.e. \(\dfrac{81}{5}-\dfrac{81}{5}-\dfrac{36m}{5}+2m^2=0\).
3. \(2m^2-\dfrac{36}{5}m=0\), so \(m=0\) or \(m=\dfrac{18}{5}\).

Teacher's Note:
1. A common root satisfies both equations.
2. Always substitute the root of the simpler equation.

Answer: (d) 33 m

Explanation:
1. Let the breadth be \(b\). \((b+7)(b-1)=b(b+3)+83\).
2. \(b^2+6b-7=b^2+3b+83\), so \(3b=90\) and \(b=30\).
3. The original length is \(b+3=33\) m.

Teacher's Note:
1. The \(b^2\) terms cancel.
2. Check: \(33\times30=990\) and \(37\times29=1073\), a rise of 83.

Answer: (d) \(-7\) and \(12\)

Explanation:
1. For \(ax^2+bx+c=0\), sum \(=-b/a=-7\) and product \(=c/a=12\).
2. The roots are \(-3\) and \(-4\), which agree.

Teacher's Note:
1. Sum of roots is \(-b/a\), product is \(c/a\).
2. The signs trip up many students.

Answer: (c) \(m=1\) or \(m=6\)

Explanation:
1. \((2m-5)(m-4)=2m^2-13m+20=m+8\).
2. \(2m^2-14m+12=0\), so \(m^2-7m+6=0\).
3. \((m-1)(m-6)=0\).

Teacher's Note:
1. Bring all terms to one side before factorising.
2. Check \(m=6\): \(7\times2=14=6+8\).

Answer: (a) Rs. 6

Explanation:
1. Let the price be \(p\). \(\dfrac{240}{p}-\dfrac{240}{p+2}=10\).
2. \(240\times2=10p(p+2)\), so \(p^2+2p-48=0\).
3. \((p-6)(p+8)=0\), so \(p=6\).

Teacher's Note:
1. Check: 40 books at Rs. 6 and 30 books at Rs. 8.
2. The number of books fewer is 10.

Answer: (b) \(-3\)

Explanation:
1. Sum of roots \(=-\dfrac ba\) and product \(=\dfrac3a\).
2. Equal sum and product give \(-b=3\), so \(b=-3\).

Teacher's Note:
1. The value of \(a\) is not fixed by this condition (it must only be non-zero).
2. Roots exist as real numbers only if \(9-12a\ge0\).

Answer: (c) \(\dfrac18\) and \(-\dfrac98\)

Explanation:
1. Let \(t=\sqrt{x/(1+x)}\). Then \(t+\dfrac1t=\dfrac{10}{3}\), so \(3t^2-10t+3=0\) and \(t=3\) or \(\dfrac13\).
2. \(\dfrac{x}{1+x}=9\) gives \(x=-\dfrac98\); \(\dfrac{x}{1+x}=\dfrac19\) gives \(x=\dfrac18\).
3. Both give positive values under the roots.

Teacher's Note:
1. For \(x=-\dfrac98\), \(1+x=-\dfrac18\), so both the ratio and its reciprocal are positive.
2. Always check the domain.