Let us work out 1.5 | WB Class 10

Ganit Prakash - Class-X - Let us work out 1.5

Let us work out 1.5 Solutions Step-by-Step Approach

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📗 Let us work out 1.5

Q1 to Q11

Question 1

Let us write the nature of two roots of the following quadratic equations:
(i) \(2x^2 + 7x + 3 = 0\)      (ii) \(3x^2 - 2\sqrt{6}x + 2 = 0\)
(iii) \(2x^2 - 7x + 9 = 0\)     (iv) \(\frac{2}{5}x^2 - \frac{2}{3}x + 1 = 0\)

The nature of roots depends on the discriminant \(D = b^2 - 4ac\).

(i) \(2x^2 + 7x + 3 = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=2\), \(b=7\), \(c=3\).

\(D = (7)^2 - 4(2)(3) = 49 - 24 = 25\).

Since \(D > 0\), the roots are real and unequal.

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(ii) \(3x^2 - 2\sqrt{6}x + 2 = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=3\), \(b=-2\sqrt{6}\), \(c=2\).

\(D = (-2\sqrt{6})^2 - 4(3)(2) = 24 - 24 = 0\).

Since \(D = 0\), the roots are real and equal.

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(iii) \(2x^2 - 7x + 9 = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=2\), \(b=-7\), \(c=9\).

\(D = (-7)^2 - 4(2)(9) = 49 - 72 = -23\).

Since \(D < 0\), there are no real roots.

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(iv) \(\frac{2}{5}x^2 - \frac{2}{3}x + 1 = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=\frac{2}{5}\), \(b=-\frac{2}{3}\), \(c=1\).

\(D = \left(-\frac{2}{3}\right)^2 - 4\left(\frac{2}{5}\right)(1) = \frac{4}{9} - \frac{8}{5} = \frac{20 - 72}{45} = -\frac{52}{45}\).

Since \(D < 0\), there are no real roots.

Question 2

By calculating, let us write the value(s) of \(k\) for which each of the following quadratic equations has real and equal roots —
(i) \(49x^2 + kx + 1 = 0\)     (ii) \(3x^2 - 5x + 2k = 0\)
(iii) \(9x^2 - 24x + k = 0\)     (iv) \(2x^2 + 3x + k = 0\)
(v) \(x^2 - 2(5 + 2k)x + 3(7 + 10k) = 0\)
(vi) \((3k + 1)x^2 + 2(k + 1)x + k = 0\)

For real and equal roots, the discriminant \(D = b^2 - 4ac = 0\).

(i) \(49x^2 + kx + 1 = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=49\), \(b=k\), \(c=1\).

\(k^2 - 4(49)(1) = 0 \implies k^2 - 196 = 0 \implies k^2 = 196 \implies\) \(k = \pm 14\).

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(ii) \(3x^2 - 5x + 2k = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=3\), \(b=-5\), \(c=2k\).

\((-5)^2 - 4(3)(2k) = 0 \implies 25 - 24k = 0 \implies 24k = 25 \implies\) \(k = \frac{25}{24}\).

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(iii) \(9x^2 - 24x + k = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=9\), \(b=-24\), \(c=k\).

\((-24)^2 - 4(9)(k) = 0 \implies 576 - 36k = 0 \implies 36k = 576 \implies\) \(k = 16\).

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(iv) \(2x^2 + 3x + k = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=2\), \(b=3\), \(c=k\).

\((3)^2 - 4(2)(k) = 0 \implies 9 - 8k = 0 \implies 8k = 9 \implies\) \(k = \frac{9}{8}\).

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(v) \(x^2 - 2(5 + 2k)x + 3(7 + 10k) = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=1\), \(b=-2(5+2k)\), \(c=3(7+10k)\).

\([-2(5 + 2k)]^2 - 4(1)[3(7 + 10k)] = 0\)

\(4(25 + 20k + 4k^2) - 12(7 + 10k) = 0\)

\(100 + 80k + 16k^2 - 84 - 120k = 0\)

\(16k^2 - 40k + 16 = 0\)

Dividing by 8: \(2k^2 - 5k + 2 = 0\)

\(2k^2 - 4k - k + 2 = 0 \implies 2k(k - 2) - 1(k - 2) = 0 \implies (2k - 1)(k - 2) = 0\).

Answer: \(k = \frac{1}{2}, 2\)

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(vi) \((3k + 1)x^2 + 2(k + 1)x + k = 0\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=3k+1\), \(b=2(k+1)\), \(c=k\).

\([2(k + 1)]^2 - 4(3k + 1)(k) = 0\)

\(4(k^2 + 2k + 1) - 4(3k^2 + k) = 0\)

Dividing by 4: \(k^2 + 2k + 1 - 3k^2 - k = 0\)

\(-2k^2 + k + 1 = 0 \implies 2k^2 - k - 1 = 0\)

\(2k^2 - 2k + k - 1 = 0 \implies 2k(k - 1) + 1(k - 1) = 0 \implies (2k + 1)(k - 1) = 0\).

Answer: \(k = 1, -\frac{1}{2}\)

Question 3

Let us form the quadratic equations from two roots given below —
(i) \(4, 2\)     (ii) \(-4, -3\)     (iii) \(-4, 3\)     (iv) \(5, -3\)

A quadratic equation with roots \(\alpha\) and \(\beta\) is given by: \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\).

(i) Roots: 4, 2

\(x^2 - (4 + 2)x + (4)(2) = 0 \implies \) \(x^2 - 6x + 8 = 0\)

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(ii) Roots: -4, -3

\(x^2 - (-4 - 3)x + (-4)(-3) = 0 \implies x^2 - (-7)x + 12 = 0 \implies \) \(x^2 + 7x + 12 = 0\)

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(iii) Roots: -4, 3

\(x^2 - (-4 + 3)x + (-4)(3) = 0 \implies x^2 - (-1)x - 12 = 0 \implies \) \(x^2 + x - 12 = 0\)

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(iv) Roots: 5, -3

\(x^2 - (5 - 3)x + (5)(-3) = 0 \implies \) \(x^2 - 2x - 15 = 0\)

Question 4

Find \(m\) for which the two roots of the quadratic equation \(4x^2 + 4(3m - 1)x + (m + 7) = 0\) are reciprocal to each other.

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=4\), \(b=4(3m-1)\), \(c=m+7\).

Let the roots be \(\alpha\) and \(\frac{1}{\alpha}\).

Product of the roots = \(\alpha \times \frac{1}{\alpha} = 1\).

From the equation, Product of roots = \(\frac{c}{a} = \frac{m + 7}{4}\).

Therefore, \(\frac{m + 7}{4} = 1\)

\(m + 7 = 4 \implies m = 4 - 7 \implies m = -3\).

Answer: \(m = -3\)

Question 5

If two roots of the quadratic equation \((b - c)x^2 + (c - a)x + (a - b) = 0\) are equal, then let us prove that, \(2b = a + c\).

Comparing the given quadratic equation with \(Ax^2+Bx+C=0\), we get \(A=b-c\), \(B=c-a\), \(C=a-b\).

Since the roots are equal, the discriminant must be zero: \(D = 0\).

\((c - a)^2 - 4(b - c)(a - b) = 0\)

\(c^2 - 2ac + a^2 - 4(ab - b^2 - ac + bc) = 0\)

\(a^2 + c^2 - 2ac - 4ab + 4b^2 + 4ac - 4bc = 0\)

\(a^2 + 4b^2 + c^2 + 2ac - 4ab - 4bc = 0\)

This expression can be rewritten using the identity \((x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx\):

\((a)^2 + (-2b)^2 + (c)^2 + 2(a)(-2b) + 2(-2b)(c) + 2(a)(c) = 0\)

\((a - 2b + c)^2 = 0\)

Taking square root on both sides:

\(a - 2b + c = 0\)

\(\implies 2b = a + c\). [Proved]

Question 6

If two roots of the quadratic equation \((a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0\) are equal, then let us prove that \(\frac{a}{b} = \frac{c}{d}\).

Comparing the given quadratic equation with \(Ax^2+Bx+C=0\), we get \(A=a^2+b^2\), \(B=-2(ac+bd)\), \(C=c^2+d^2\).

Since the roots are equal, the discriminant \(D = 0\).

\([-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0\)

\(4(ac + bd)^2 - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0\)

Dividing by 4:

\((a^2c^2 + 2abcd + b^2d^2) - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0\)

\(2abcd - a^2d^2 - b^2c^2 = 0\)

Multiplying by -1:

\(a^2d^2 - 2abcd + b^2c^2 = 0\)

\((ad - bc)^2 = 0\)

\(ad - bc = 0 \implies ad = bc\)

Dividing both sides by \(bd\):

\(\frac{a}{b} = \frac{c}{d}\). [Proved]

Question 7

Let us prove that the quadratic equation \(2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0\) has no real root, if \(a \neq b\).

Comparing the given quadratic equation with \(Ax^2+Bx+C=0\), we get \(A=2(a^2+b^2)\), \(B=2(a+b)\), \(C=1\).

Let's find the discriminant \(D = B^2 - 4AC\).

\(D = [2(a + b)]^2 - 4[2(a^2 + b^2)](1)\)

\(D = 4(a + b)^2 - 8(a^2 + b^2)\)

\(D = 4(a^2 + 2ab + b^2) - 8a^2 - 8b^2\)

\(D = 4a^2 + 8ab + 4b^2 - 8a^2 - 8b^2\)

\(D = -4a^2 + 8ab - 4b^2\)

\(D = -4(a^2 - 2ab + b^2)\)

\(D = -4(a - b)^2\)

For any real values of \(a\) and \(b\) where \(a \neq b\), the square \((a - b)^2\) is always positive (i.e., \(> 0\)).

Therefore, \(-4(a - b)^2\) will always be strictly less than zero (\(D < 0\)).

Since the discriminant is negative, the equation has no real root. [Proved]

Question 8

If two roots of the quadratic equation \(5x^2 + 2x - 3 = 0\) are \(\alpha\) and \(\beta\), then let us determine the value of,
(i) \(\alpha^2 + \beta^2\)     (ii) \(\alpha^3 + \beta^3\)     (iii) \(\frac{1}{\alpha} + \frac{1}{\beta}\)     (iv) \(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=5\), \(b=2\), \(c=-3\).

Sum of roots: \(\alpha + \beta = -\frac{b}{a} = -\frac{2}{5}\)

Product of roots: \(\alpha\beta = \frac{c}{a} = -\frac{3}{5}\)

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(i) \(\alpha^2 + \beta^2\)

\(= (\alpha + \beta)^2 - 2\alpha\beta\)

\(= \left(-\frac{2}{5}\right)^2 - 2\left(-\frac{3}{5}\right) = \frac{4}{25} + \frac{6}{5} = \frac{4 + 30}{25} = \) \(\frac{34}{25}\)

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(ii) \(\alpha^3 + \beta^3\)

\(= (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)

\(= \left(-\frac{2}{5}\right)^3 - 3\left(-\frac{3}{5}\right)\left(-\frac{2}{5}\right)\)

\(= -\frac{8}{125} - \frac{18}{25} = \frac{-8 - 90}{125} = \) \(-\frac{98}{125}\)

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(iii) \(\frac{1}{\alpha} + \frac{1}{\beta}\)

\(= \frac{\alpha + \beta}{\alpha\beta}\)

\(= \frac{-\frac{2}{5}}{-\frac{3}{5}} = \frac{2}{5} \times \frac{5}{3} = \) \(\frac{2}{3}\)

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(iv) \(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\)

\(= \frac{\alpha^3 + \beta^3}{\alpha\beta}\)

\(= \frac{-\frac{98}{125}}{-\frac{3}{5}} = \frac{98}{125} \times \frac{5}{3} = \frac{98}{25 \times 3} = \) \(\frac{98}{75}\)

Question 9

If one root of the equation \(ax^2 + bx + c = 0\) is twice of the other, then let us show that, \(2b^2 = 9ac\).

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get coefficients as \(a, b, c\).

Let one root be \(\alpha\). Then the other root is \(2\alpha\).

Sum of roots: \(\alpha + 2\alpha = -\frac{b}{a} \implies 3\alpha = -\frac{b}{a} \implies \alpha = -\frac{b}{3a}\) ---- (1)

Product of roots: \(\alpha \times 2\alpha = \frac{c}{a} \implies 2\alpha^2 = \frac{c}{a}\) ---- (2)

Substitute the value of \(\alpha\) from (1) into (2):

\(2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a}\)

\(2\left(\frac{b^2}{9a^2}\right) = \frac{c}{a}\)

\(\frac{2b^2}{9a} = c\)     (cancelling one \(a\) from both denominators)

\(2b^2 = 9ac\). [Proved]

Question 10

Let us form the equation whose roots are reciprocals to the roots of the equation \(x^2 + px + 1 = 0\).

Let \(\alpha\) and \(\beta\) be the roots of \(x^2 + px + 1 = 0\).

We need an equation whose roots are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).

Let \(y = \frac{1}{x} \implies x = \frac{1}{y}\).

Substitute \(x = \frac{1}{y}\) into the given equation:

\(\left(\frac{1}{y}\right)^2 + p\left(\frac{1}{y}\right) + 1 = 0\)

\(\frac{1}{y^2} + \frac{p}{y} + 1 = 0\)

Multiplying by \(y^2\):

\(1 + py + y^2 = 0 \implies y^2 + py + 1 = 0\)

Replacing \(y\) with \(x\) for the standard variable form:

Answer: \(x^2 + px + 1 = 0\).

Question 11

Let us determine the equation whose roots are square of the roots of the equation \(x^2 + x + 1 = 0\).

Let \(\alpha\) and \(\beta\) be the roots of \(x^2 + x + 1 = 0\).

We want the equation with roots \(\alpha^2\) and \(\beta^2\).

Let \(y = x^2\). From the given equation, \(x^2 + 1 = -x\).

Squaring both sides:

\((x^2 + 1)^2 = (-x)^2\)

\((x^2)^2 + 2x^2 + 1 = x^2\)

Substitute \(y = x^2\):

\(y^2 + 2y + 1 = y \implies y^2 + y + 1 = 0\)

Replacing \(y\) with \(x\) for standard notation:

Answer: \(x^2 + x + 1 = 0\).

📗 12. V.S.A.

MCQ, True/False & Fill Blanks

12. (A) M.C.Q. - (i)

The sum of the two roots of the equation \(x^2 - 6x + 2 = 0\) is
(a) \(2\)     (b) \(-2\)     (c) \(6\)     (d) \(-6\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=1\), \(b=-6\), \(c=2\).

Sum of roots \(= -\frac{b}{a} = -\frac{-6}{1} = 6\).

Answer: (c) 6

12. (A) M.C.Q. - (ii)

If the product of the two roots of the equation \(x^2 - 3x + k = 10\) is \(-2\), then the value of \(k\) is
(a) \(-2\)     (b) \(-8\)     (c) \(8\)     (d) \(12\)

Rewrite equation: \(x^2 - 3x + (k - 10) = 0\).

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=1\), \(b=-3\), \(c=(k-10)\).

Product of roots \(= \frac{c}{a} = k - 10 = -2 \implies k = -2 + 10 = 8\).

Answer: (c) 8

12. (A) M.C.Q. - (iii)

If the two roots of the equation \(ax^2 + bx + c = 0\) (\(a \neq 0\)) are real and unequal, then \(b^2 - 4ac\) will be
(a) \(>0\)     (b) \(=0\)     (c) \(<0\)     (d) none of these

For real and unequal roots, the discriminant must be strictly greater than zero.

Answer: (a) \(>0\)

12. (A) M.C.Q. - (iv)

If the two roots of the equation \(ax^2 + bx + c = 0\) (\(a \neq 0\)) be equal, then
(a) \(c = -\frac{b}{2a}\)     (b) \(c = \frac{b}{2a}\)     (c) \(c = \frac{-b^2}{4a}\)     (d) \(c = \frac{b^2}{4a}\)

For equal roots, \(b^2 - 4ac = 0 \implies 4ac = b^2 \implies c = \frac{b^2}{4a}\).

Answer: (d) \(c = \frac{b^2}{4a}\)

12. (A) M.C.Q. - (v)

If the two roots of the equation \(3x^2 + 8x + 2 = 0\) be \(\alpha\) and \(\beta\), then the value of \((\frac{1}{\alpha} + \frac{1}{\beta})\) is
(a) \(-\frac{3}{8}\)     (b) \(\frac{2}{3}\)     (c) \(-4\)     (d) \(4\)

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=3\), \(b=8\), \(c=2\).

\(\alpha + \beta = -\frac{8}{3}\) and \(\alpha\beta = \frac{2}{3}\).

\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-\frac{8}{3}}{\frac{2}{3}} = -4\).

Answer: (c) \(-4\)

12. (B) True or false - (i)

The two roots of the equation \(x^2 + x + 1 = 0\) are real.

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=1\), \(b=1\), \(c=1\).

\(D = 1^2 - 4(1)(1) = 1 - 4 = -3 < 0\). Since D is negative, roots are not real.

Answer: False

12. (B) True or false - (ii)

The two roots of the equation \(x^2 - x + 2 = 0\) are not real.

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=1\), \(b=-1\), \(c=2\).

\(D = (-1)^2 - 4(1)(2) = 1 - 8 = -7 < 0\). Roots are indeed not real.

Answer: True

12. (C) Fill in the blanks - (i)

The ratio of the sum and the product of two roots of the equation \(7x^2 - 12x + 18 = 0\) is ...............

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=7\), \(b=-12\), \(c=18\).

Sum \(= \frac{12}{7}\). Product \(= \frac{18}{7}\).

Ratio \(= \frac{12}{7} : \frac{18}{7} = 12 : 18 = 2 : 3\).

Answer: \(2 : 3\)

12. (C) Fill in the blanks - (ii)

If two roots of the equation \(ax^2 + bx + c = 0\) (\(a \neq 0\)) are reciprocal to each other, then \(c =\) ...............

Let roots be \(\alpha\) and \(\frac{1}{\alpha}\). Product \(= 1 \implies \frac{c}{a} = 1 \implies c = a\).

Answer: \(a\)

12. (C) Fill in the blanks - (iii)

If two roots of the equation \(ax^2 + bx + c = 0\) (\(a \neq 0\)) are reciprocal to each other and opposite in sign, then \(a + c =\) ...............

Let roots be \(\alpha\) and \(-\frac{1}{\alpha}\). Product \(= -1 \implies \frac{c}{a} = -1 \implies c = -a \implies a + c = 0\).

Answer: \(0\)

📗 13. S.A.

Short Answer Type

Question 13 (i)

Let us write the quadratic equation if sum of its roots is 14 and the product of them is 24.

The equation is \(x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0\).

Answer: \(x^2 - 14x + 24 = 0\)

Question 13 (ii)

If the sum and the product of the two roots of the equation \(kx^2 + 2x + 3k = 0\) (\(k \neq 0\)) are equal, let us write the value of \(k\).

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=k\), \(b=2\), \(c=3k\).

Sum of roots \(= -\frac{2}{k}\).

Product of roots \(= \frac{3k}{k} = 3\).

Given they are equal: \(-\frac{2}{k} = 3 \implies 3k = -2 \implies k = -\frac{2}{3}\).

Answer: \(k = -\frac{2}{3}\)

Question 13 (iii)

If the two roots of the equation \(x^2 - 22x + 105 = 0\) are \(\alpha\) and \(\beta\), let us write the value of \((\alpha - \beta)\).

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=1\), \(b=-22\), \(c=105\).

\(\alpha + \beta = 22\) and \(\alpha\beta = 105\).

\((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)

\((\alpha - \beta)^2 = (22)^2 - 4(105) = 484 - 420 = 64\).

\(\alpha - \beta = \pm \sqrt{64} = \pm 8\).

Answer: \(\pm 8\)

Question 13 (iv)

If the sum of the two roots of the equation \(x^2 - x = k(2x - 1)\) is zero, let us write the value of \(k\).

Rewrite the equation: \(x^2 - x - 2kx + k = 0 \implies x^2 - (2k + 1)x + k = 0\).

Comparing the given quadratic equation with \(ax^2+bx+c=0\), we get \(a=1\), \(b=-(2k+1)\), \(c=k\).

Sum of roots \(= 2k + 1\).

Given sum is zero: \(2k + 1 = 0 \implies k = -\frac{1}{2}\).

Answer: \(k = -\frac{1}{2}\)

Question 13 (v)

If one of the roots of the two equations \(x^2 + bx + 12 = 0\) and \(x^2 + bx + q = 0\) is \(2\), let us write the value of \(q\).

Since 2 is a root of \(x^2 + bx + 12 = 0\), substitute \(x = 2\):

\((2)^2 + b(2) + 12 = 0 \implies 4 + 2b + 12 = 0 \implies 2b = -16 \implies b = -8\).

Since 2 is also a root of \(x^2 + bx + q = 0\), substitute \(x = 2\) and \(b = -8\):

\((2)^2 + (-8)(2) + q = 0 \implies 4 - 16 + q = 0 \implies -12 + q = 0 \implies q = 12\).

Answer: \(q = 12\)

Maths With Prasanta Sir

WBBSE Board — Class 10 Quadratic Equations - Ganit Prakash - Let us work out 1.5 Solutions.

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