EXERCISE 4.2

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1. Find the roots of the following quadratic equations by factorisation.

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(i) $x^2-3x-10=0$

Solution:

Given, $$ x^2-3x-10=0 $$ Split the middle term. $$ x^2-5x+2x-10=0 $$ Factorise. $$ x(x-5)+2(x-5)=0 $$ $$ (x-5)(x+2)=0 $$ Therefore, $$ x-5=0 \quad\text{or}\quad x+2=0 $$ $$ x=5 \quad\text{or}\quad x=-2 $$

Answer:

$$ \boxed{x=5,\,-2} $$

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

Solution:

Given, $$ 2x^2+x-6=0 $$ Split the middle term. $$ 2x^2+4x-3x-6=0 $$ Factorise. $$ 2x(x+2)-3(x+2)=0 $$ $$ (x+2)(2x-3)=0 $$ Therefore, $$ x+2=0 \quad\text{or}\quad 2x-3=0 $$ $$ x=-2 \quad\text{or}\quad x=\frac32 $$

Answer:

$$ \boxed{x=-2,\;\frac32} $$

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(iii) $\sqrt2\,x^2+7x+5\sqrt2=0$

Solution:

Here, $$ a=\sqrt2,\qquad b=7,\qquad c=5\sqrt2 $$ Since $$ ac=\sqrt2\times5\sqrt2=10 $$ Two numbers whose product is $10$ and sum is $7$ are $$ 5,\;2 $$ Split the middle term. $$ \sqrt2x^2+5x+2x+5\sqrt2=0 $$ Group the terms. $$ (\sqrt2x^2+5x)+(2x+5\sqrt2)=0 $$ $$ x(\sqrt2x+5)+\sqrt2(\sqrt2x+5)=0 $$ $$ (\sqrt2x+5)(x+\sqrt2)=0 $$ Therefore, $$ \sqrt2x+5=0 $$ or $$ x+\sqrt2=0 $$ Hence, $$ x=-\frac5{\sqrt2} =-\frac{5\sqrt2}{2} $$ or $$ x=-\sqrt2 $$

Answer:

$$ \boxed{x=-\frac{5\sqrt2}{2},\;-\sqrt2} $$

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(iv) $2x^2-x+\dfrac18=0$

Solution:

Multiply throughout by $8$. $$ 16x^2-8x+1=0 $$ Factorise. $$ (4x-1)^2=0 $$ Therefore, $$ 4x-1=0 $$ $$ x=\frac14 $$ Since both factors are equal, $$ x=\frac14,\;\frac14 $$

Answer:

$$ \boxed{x=\frac14,\;\frac14} $$

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(v) $100x^2-20x+1=0$

Solution:

Factorise. $$ (10x-1)^2=0 $$ Therefore, $$ 10x-1=0 $$ $$ x=\frac1{10} $$ Hence, $$ x=\frac1{10},\;\frac1{10} $$

Answer:

$$ \boxed{x=\frac1{10},\;\frac1{10}} $$

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2. Solve the problems given in Example 1.

Answer: Solve according to the examples explained in the textbook using the factorisation method.

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3. Find two numbers whose sum is $27$ and product is $182$.

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Solution:

Let the numbers be $$ x\quad\text{and}\quad y $$ Then, $$ x+y=27 $$ $$ xy=182 $$ Substitute $$ y=27-x $$ into $$ xy=182 $$ $$ x(27-x)=182 $$ $$ 27x-x^2=182 $$ $$ x^2-27x+182=0 $$ Factorise. $$ (x-13)(x-14)=0 $$ Therefore, $$ x=13 \quad\text{or}\quad x=14 $$ Hence the two numbers are $$ 13,\;14 $$

Answer:

$$ \boxed{13,\;14} $$

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4. Find two consecutive positive integers, the sum of whose squares is $365$.

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Solution:

Let the first integer be $$ x $$ The next integer is $$ x+1 $$ According to the question, $$ x^2+(x+1)^2=365 $$ Expand. $$ x^2+x^2+2x+1=365 $$ $$ 2x^2+2x-364=0 $$ Divide by $2$. $$ x^2+x-182=0 $$ Factorise. $$ (x-13)(x+14)=0 $$ $$ x=13 \quad\text{or}\quad x=-14 $$ Since the integers are positive, $$ x=13 $$ Therefore, $$ x+1=14 $$

Answer:

$$ \boxed{13,\;14} $$

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5. The altitude of a right triangle is $7$ cm less than its base. If the hypotenuse is $13$ cm, find the other two sides.

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Solution:

Let the base be $$ x\text{ cm} $$ Then the altitude is $$ x-7\text{ cm} $$ Using Pythagoras theorem, $$ x^2+(x-7)^2=13^2 $$ $$ x^2+x^2-14x+49=169 $$ $$ 2x^2-14x-120=0 $$ Divide by $2$. $$ x^2-7x-60=0 $$ Factorise. $$ (x-12)(x+5)=0 $$ $$ x=12 $$ Therefore, $$ x-7=5 $$

Answer:

$$ \boxed{\text{Base}=12\text{ cm},\qquad \text{Altitude}=5\text{ cm}} $$

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6. A cottage industry produces a certain number of pottery articles in a day. The cost of production of each article is ₹3 more than twice the number of articles produced. If the total production cost is ₹90, find the number of articles produced and the cost of each article.

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Solution:

Let the number of articles produced be $$ x $$ Cost of one article $$ 2x+3 $$ Total production cost $$ x(2x+3)=90 $$ $$ 2x^2+3x-90=0 $$ Factorise. $$ 2x^2+15x-12x-90=0 $$ $$ x(2x+15)-6(2x+15)=0 $$ $$ (2x+15)(x-6)=0 $$ Since the number of articles cannot be negative, $$ x=6 $$ Cost of one article $$ 2(6)+3=15 $$

Answer:

$$ \boxed{\text{Number of articles}=6} $$ $$ \boxed{\text{Cost of each article}=₹15} $$

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