Exercise 4.4 — Class 9 Mathematics Solutions

Solution

Solution

1. Check for common factors.

   Examine the terms: $x^2$, $-8x$, and $16$. The greatest common factor is $1$, so no factor can be taken out.

2. Express as a perfect square.

   Identify the first and last terms as squares of monomials:

   $$\begin{array}{cc}{x}^2& =(x{)}^2\\ 16& =(4{)}^2\end{array}$$

   Check the middle term:

   $$\begin{array}{cc}2ab& =2(x)(4)\\ & =8x\end{array}$$

   Using the algebraic identity $a^2-2ab+b^2=(a-b{)}^2$ where $a=x$ and $b=4$:

   $$\begin{array}{cc}{x}^2-8x+16& =(x{)}^2-2(x)(4)+(4{)}^2\\ & =(x-4{)}^2\end{array}$$

3. Take the square root.

   Taking the square root of the expression:

   $$\begin{array}{cc}\sqrt{x^2-8x+16}& =\pm \sqrt{(x-4{)}^2}\\ & =\pm (x-4)\end{array}$$

Solution

1. Check for common factors.

   Examine the coefficients $9$, $12$, and $4$. The greatest common factor is $1$, so no factor can be taken out.

2. Express as a perfect square.

   Identify the first and last terms as squares of monomials:

   $$\begin{array}{cc}9x^2& =(3x{)}^2\\ 4& =(2{)}^2\end{array}$$

   Check the middle term:

   $$\begin{array}{cc}2ab& =2(3x)(2)\\ & =12x\end{array}$$

   Using the algebraic identity $a^2+2ab+b^2=(a+b{)}^2$ where $a=3x$ and $b=2$:

   $$\begin{array}{cc}9x^2+12x+4& =(3x{)}^2+2(3x)(2)+(2{)}^2\\ & =(3x+2{)}^2\end{array}$$

3. Take the square root.

   Taking the square root of the expression:

   $$\begin{array}{cc}\sqrt{9x^2+12x+4}& =\pm \sqrt{(3x+2{)}^2}\\ & =\pm (3x+2)\end{array}$$

Solution

1. Check for common factors.

   Examine the coefficients $36$, $84$, and $49$. Since $49=7^2$ and $36$ is not divisible by $7$, the greatest common factor is $1$.

2. Express as a perfect square.

   Identify the first and last terms as squares of monomials:

   $$\begin{array}{cc}36a^2& =(6a{)}^2\\ 49& =(7{)}^2\end{array}$$

   Check the middle term:

   $$\begin{array}{cc}2(6a)(7)& =2\times 42a\\ & =84a\end{array}$$

   Using the algebraic identity $A^2+2AB+B^2=(A+B{)}^2$ where $A=6a$ and $B=7$:

   $$\begin{array}{cc}36a^2+84a+49& =(6a{)}^2+2(6a)(7)+(7{)}^2\\ & =(6a+7{)}^2\end{array}$$

3. Take the square root.

   Taking the square root of the expression:

   $$\begin{array}{cc}\sqrt{36a^2+84a+49}& =\pm \sqrt{(6a+7{)}^2}\\ & =\pm (6a+7)\end{array}$$

Solution

1. Factor out the greatest common factor.

   Examine the coefficients $64$, $-32$, and $4$. The greatest common numerical factor is $4$. Factoring out $4$:

   $$\begin{array}{cc}64y^2-32y+4& =4(16y^2-8y+1)\end{array}$$

2. Express the quadratic factor as a perfect square.

   Inside the parentheses, recognize:

   $$\begin{array}{cc}16y^2& =(4y{)}^2\\ 1& =(1{)}^2\end{array}$$

   Verify the middle term:

   $$\begin{array}{cc}2(4y)(1)& =8y\end{array}$$

   Using $a^2-2ab+b^2=(a-b{)}^2$ with $a=4y$ and $b=1$:

   $$\begin{array}{cc}64y^2-32y+4& =4[(4y{)}^2-2(4y)(1)+(1{)}^2]\\ & =4(4y-1{)}^2\\ & ={[2(4y-1)]}^2\\ & =(8y-2{)}^2\end{array}$$

3. Take the square root.

   Taking the square root of the expression:

   $$\begin{array}{cc}\sqrt{64y^2-32y+4}& =\pm \sqrt{4(4y-1{)}^2}\\ & =\pm 2(4y-1)\\ & =\pm (8y-2)\end{array}$$

Solution

1. Factor out the greatest common factor.

   Examine the coefficients $200$, $-120$, and $18$. Notice that neither $200$ nor $18$ is a perfect square. The greatest common numerical factor is $2$. Factoring out $2$:

   $$\begin{array}{cc}200t^2-120t+18& =2(100t^2-60t+9)\end{array}$$

2. Express the quadratic factor as a perfect square.

   Examine the terms inside the parentheses:

   $$\begin{array}{cc}100t^2& =(10t{)}^2\\ 9& =(3{)}^2\end{array}$$

   Check the middle term:

   $$\begin{array}{cc}2ab& =2(10t)(3)\\ & =60t\end{array}$$

   Using the algebraic identity $a^2-2ab+b^2=(a-b{)}^2$ where $a=10t$ and $b=3$:

   $$\begin{array}{cc}100t^2-60t+9& =(10t{)}^2-2(10t)(3)+(3{)}^2\\ & =(10t-3{)}^2\end{array}$$

   Therefore, the polynomial can be written as:

   $$\begin{array}{cc}200t^2-120t+18& =2(10t-3{)}^2\end{array}$$

3. Take the square root.

   Taking the square root of the expression:

   $$\begin{array}{cc}\sqrt{200t^2-120t+18}& =\pm \sqrt{2(10t-3{)}^2}\\ & =\pm \sqrt{2}\cdot \sqrt{(10t-3{)}^2}\\ & =\pm \sqrt{2}(10t-3)\end{array}$$

   [!NOTE] Pedagogical Note on the Form of the Final Answer: In various textbook printings and answer keys, the final result is written in either of two standard algebraic forms:

   1. Radical in front: $\pm \sqrt{2}(10t-3)$
   2. Radical following the binomial: $\pm (10t-3)\sqrt{2}$

   Both representations are mathematically identical by the commutative property of multiplication:

   $$\begin{array}{cc}\pm \sqrt{2}(10t-3)& =\pm (10t-3)\sqrt{2}\end{array}$$

   If expanded, it can also be expressed as $\pm (10\sqrt{2}t-3\sqrt{2})$. Take care not to omit the irrational factor $\sqrt{2}$, as $200$ is not a perfect square ($200=2\times 100$).

Solution

1. Factor out the greatest common factor.

   Examine the coefficients $40$, $120$, and $90$. Notice that neither $40$ nor $90$ is a perfect square. The greatest common numerical factor is $10$. Factoring out $10$:

   $$\begin{array}{cc}40x^2+120x+90& =10(4x^2+12x+9)\end{array}$$

2. Express the quadratic factor as a perfect square.

   Examine the terms inside the parentheses:

   $$\begin{array}{cc}4x^2& =(2x{)}^2\\ 9& =(3{)}^2\end{array}$$

   Check the middle term:

   $$\begin{array}{cc}2ab& =2(2x)(3)\\ & =12x\end{array}$$

   Using the identity $a^2+2ab+b^2=(a+b{)}^2$ with $a=2x$ and $b=3$:

   $$\begin{array}{cc}4x^2+12x+9& =(2x{)}^2+2(2x)(3)+(3{)}^2\\ & =(2x+3{)}^2\end{array}$$

   Therefore, the polynomial can be written as:

   $$\begin{array}{cc}40x^2+120x+90& =10(2x+3{)}^2\end{array}$$

3. Take the square root.

   Taking the square root of the expression:

   $$\begin{array}{cc}\sqrt{40x^2+120x+90}& =\pm \sqrt{10(2x+3{)}^2}\\ & =\pm \sqrt{10}\cdot \sqrt{(2x+3{)}^2}\\ & =\pm \sqrt{10}(2x+3)\end{array}$$

   Because the factored numerical coefficient $10$ is not a perfect square, its square root remains $\pm \sqrt{10}$ outside the linear factor $(2x+3)$.

Solution

Solution

1. The terms are already in descending powers of $x$.

   First term of the root.

   $$\begin{array}{cc}\sqrt{4x^4}& =2x^2\end{array}$$

   Subtract $(2x^2{)}^2=4x^4$. Remainder: $-28x^3+37x^2+42x+9$.

2. Second term of the root.

   Double the root so far: $2(2x^2)=4x^2$. Divide: $-28x^3÷4x^2=-7x$. Multiply the divisor $4x^2-7x$ by $-7x$ and subtract:

   $$\begin{array}{cc}(4x^2-7x)(-7x)& =-28x^3+49x^2\\ \text{Remainder}& =-12x^2+42x+9\end{array}$$

3. Third term of the root.

   Double the root so far: $2(2x^2-7x)=4x^2-14x$. Divide: $-12x^2÷4x^2=-3$. Multiply the divisor $4x^2-14x-3$ by $-3$ and subtract:

   $$\begin{array}{cc}(4x^2-14x-3)(-3)& =-12x^2+42x+9\\ \text{Remainder}& =0\end{array}$$

   The remainder is $0$, so the square root is $\pm (2x^2-7x-3)$.

Solution

1. The terms are already in descending powers of $x$.

   First term of the root.

   $$\begin{array}{cc}\sqrt{121x^4}& =11x^2\end{array}$$

   Subtract $(11x^2{)}^2=121x^4$. Remainder: $-198x^3-183x^2+216x+144$.

2. Second term of the root.

   Double the root so far: $2(11x^2)=22x^2$. Divide: $-198x^3÷22x^2=-9x$. Multiply the divisor $22x^2-9x$ by $-9x$ and subtract:

   $$\begin{array}{cc}(22x^2-9x)(-9x)& =-198x^3+81x^2\\ \text{Remainder}& =-264x^2+216x+144\end{array}$$

3. Third term of the root.

   Double the root so far: $2(11x^2-9x)=22x^2-18x$. Divide: $-264x^2÷22x^2=-12$. Multiply the divisor $22x^2-18x-12$ by $-12$ and subtract:

   $$\begin{array}{cc}(22x^2-18x-12)(-12)& =-264x^2+216x+144\\ \text{Remainder}& =0\end{array}$$

   The remainder is $0$, so the square root is $\pm (11x^2-9x-12)$.

Solution

1. The terms are already in descending powers of $x$.

   First term of the root.

   $$\begin{array}{cc}\sqrt{x^4}& =x^2\end{array}$$

   Subtract $(x^2{)}^2=x^4$. Remainder: $-10x^{3}y+27x^2{y}^2-10x{y}^3+y^4$.

2. Second term of the root.

   Double the root so far: $2(x^2)=2x^2$. Divide: $-10x^{3}y÷2x^2=-5xy$. Multiply the divisor $2x^2-5xy$ by $-5xy$ and subtract:

   $$\begin{array}{cc}(2x^2-5xy)(-5xy)& =-10x^{3}y+25x^2{y}^2\\ \text{Remainder}& =2x^2{y}^2-10x{y}^3+y^4\end{array}$$

3. Third term of the root.

   Double the root so far: $2(x^2-5xy)=2x^2-10xy$. Divide: $2x^2{y}^2÷2x^2=y^2$. Multiply the divisor $2x^2-10xy+y^2$ by $y^2$ and subtract:

   $$\begin{array}{cc}(2x^2-10xy+y^2)(y^2)& =2x^2{y}^2-10x{y}^3+y^4\\ \text{Remainder}& =0\end{array}$$

   The remainder is $0$, so the square root is $\pm (x^2-5xy+y^2)$.

Solution

1. The terms are already in descending powers of $x$.

   First term of the root.

   $$\begin{array}{cc}\sqrt{4x^4}& =2x^2\end{array}$$

   Subtract $(2x^2{)}^2=4x^4$. Remainder: $-12x^3+37x^2-42x+49$.

2. Second term of the root.

   Double the root so far: $2(2x^2)=4x^2$. Divide: $-12x^3÷4x^2=-3x$. Multiply the divisor $4x^2-3x$ by $-3x$ and subtract:

   $$\begin{array}{cc}(4x^2-3x)(-3x)& =-12x^3+9x^2\\ \text{Remainder}& =28x^2-42x+49\end{array}$$

3. Third term of the root.

   Double the root so far: $2(2x^2-3x)=4x^2-6x$. Divide: $28x^2÷4x^2=7$. Multiply the divisor $4x^2-6x+7$ by $7$ and subtract:

   $$\begin{array}{cc}(4x^2-6x+7)(7)& =28x^2-42x+49\\ \text{Remainder}& =0\end{array}$$

   The remainder is $0$, so the square root is $\pm (2x^2-3x+7)$.

Solution

1. Model and Given Equation.

   The investor's return $R(x)$ in rupees is given as a quadratic function of the investment amount $x$ (where $x$ is measured in thousands of rupees):

   $$R(x)=-x^2+6x-8$$

   To factorize conveniently, factor out a negative sign $-1$ so that the leading coefficient becomes positive:

   $$\begin{array}{cc}R(x)& =-x^2+6x-8\\ & =-(x^2-6x+8)\end{array}$$

2. Factorize the quadratic expression.

   To factorize $x^2-6x+8$, find two integers whose product is $(+1)\times (+8)=8$ and whose sum is $-6$. The two numbers are $-2$ and $-4$, since:

   $$(-2)\times (-4)=8\text{and}(-2)+(-4)=-6$$

   Split the middle term:

   $$\begin{array}{cc}R(x)& =-(x^2-6x+8)\\ & =-[x^2-2x-4x+8]\\ & =-[x(x-2)-4(x-2)]\\ & =-(x-2)(x-4)\end{array}$$

3. Find roots and practical interpretation.

   To find the investment levels that yield zero return, set $R(x)=0$:

   $$\begin{array}{cc}-(x-2)(x-4)& =0\\ (x-2)(x-4)& =0\end{array}$$

   By the zero-product property:

   $$\begin{array}{cc}x-2=0& ⟹x=2\\ x-4=0& ⟹x=4\end{array}$$

   Practical Interpretation: Since $x$ represents the investment in thousands of rupees:

   • $x=2$ corresponds to an investment of $2\times \mathrm{1,000}=\text{Rs }\mathrm{2,000}$.
   • $x=4$ corresponds to an investment of $4\times \mathrm{1,000}=\text{Rs }\mathrm{4,000}$.

   Hence, the investment levels that result in zero return are $\text{Rs }\mathrm{2,000}$ and $\text{Rs }\mathrm{4,000}$.

Solution

1. Model and Given Equation.

   The company's profit function in terms of sales volume $x$ units is:

   $$P(x)=x^3-15x^2+75x-125$$

   The break-even point is defined as the production or sales level where total revenue equals total cost, resulting in zero profit:

   $$P(x)=0$$

2. Factorize using the cube of a binomial identity.

   Recall the algebraic identity for the cube of a difference:

   $$(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3$$

   Comparing the given cubic polynomial:

   • The first term is $x^3=(x{)}^3$, so $a=x$.
   • The last term is $125=(5{)}^3$, so $b=5$.

   Verify the intermediate terms:

   $$\begin{array}{cc}-3a^{2}b& =-3(x{)}^2(5)=-15x^2\\ +3a{b}^2& =+3(x)(5{)}^2=+3(x)(25)=+75x\end{array}$$

   Both terms match the given polynomial exactly. Therefore:

   $$\begin{array}{cc}P(x)& =x^3-15x^2+75x-125\\ & =(x{)}^3-3(x{)}^2(5)+3(x)(5{)}^2-(5{)}^3\\ & =(x-5{)}^3\\ & =(x-5)(x-5)(x-5)\end{array}$$

3. Find the break-even point and practical interpretation.

   Set the profit function to zero:

   $$\begin{array}{cc}P(x)& =0\\ (x-5{)}^3& =0\\ x-5& =0\\ x& =5\end{array}$$

   Practical Interpretation:

   • When $x=5$ units are produced and sold, the profit is $\text{Rs }0$ (break-even point).
   • For $x<5$, $P(x)<0$ (the company incurs an operational loss).
   • For $x>5$, $P(x)>0$ (the company yields a positive profit).

   Hence, the break-even point is at $x=5$ units.

Solution

1. Model and Given Equation.

   The potential energy function in the electric field is given as a cubic function of distance $x$:

   $$V(x)=2x^3-6x^2+4x$$

   To find where the potential energy is zero, we must solve the equation:

   $$V(x)=0$$

2. Factorize the cubic expression.

   First, factor out the common monomial factor $2x$ from all terms:

   $$\begin{array}{cc}V(x)& =2x^3-6x^2+4x\\ & =2x(x^2-3x+2)\end{array}$$

   Now factorize the quadratic trinomial $x^2-3x+2$ by splitting the middle term. Find two numbers whose product is $(+1)\times (+2)=2$ and whose sum is $-3$. The numbers are $-1$ and $-2$, since:

   $$(-1)\times (-2)=2\text{and}(-1)+(-2)=-3$$

   Factorizing further:

   $$\begin{array}{cc}V(x)& =2x[x^2-x-2x+2]\\ & =2x[x(x-1)-2(x-1)]\\ & =2x(x-1)(x-2)\end{array}$$

3. Find the roots and practical interpretation.

   Set $V(x)=0$:

   $$2x(x-1)(x-2)=0$$

   By the zero-product property, at least one of the factors must be zero:

   $$\begin{array}{cc}2x=0& ⟹x=0\\ x-1=0& ⟹x=1\\ x-2=0& ⟹x=2\end{array}$$

   Practical Interpretation: In physics, points of zero potential energy represent positions where the electric potential reference level is attained or neutral transition boundaries occur. Therefore, the potential energy is zero at $x=0$, $x=1$, and $x=2$.

Solution

1. Model and Given Equation.

   In structural mechanics, the beam deflection profile is modeled by the quadratic expression:

   $$y(x)=2x^2-8x+6$$

   where $y(x)$ is the vertical displacement at coordinate position $x$ along the beam length. The points of zero deflection are determined by solving:

   $$y(x)=0$$

2. Factorize the quadratic polynomial.

   First, factor out the common numerical coefficient $2$:

   $$\begin{array}{cc}y(x)& =2x^2-8x+6\\ & =2(x^2-4x+3)\end{array}$$

   Next, factorize the quadratic factor $x^2-4x+3$ by splitting the middle term. Find two numbers whose product is $(+1)\times (+3)=3$ and whose sum is $-4$. The numbers are $-1$ and $-3$, since:

   $$(-1)\times (-3)=3\text{and}(-1)+(-3)=-4$$

   Factorizing inside the brackets:

   $$\begin{array}{cc}y(x)& =2[x^2-3x-x+3]\\ & =2[x(x-3)-1(x-3)]\\ & =2(x-1)(x-3)\end{array}$$

3. Determine points of zero deflection and practical interpretation.

   Set the deflection function to zero:

   $$\begin{array}{cc}2(x-1)(x-3)& =0\\ (x-1)(x-3)& =0\end{array}$$

   By the zero-product property:

   $$\begin{array}{cc}x-1=0& ⟹x=1\\ x-3=0& ⟹x=3\end{array}$$

   Practical Interpretation: In structural engineering, locations of zero deflection along a beam typically identify fixed, pinned, or roller support positions, or nodal points where the beam experiences no transverse bending displacement. Therefore, zero deflection occurs at positions $x=1$ and $x=3$.