QUADRATIC EQUATIONS
A quadratic polynomial $ax^2+bx+c,\:where\:a \ne 0$ equated to zero becomes a quadratic equation.Therefore, $ax^2+bx+c=0$.
Any equation of the form $p(x)=0$, where $p(x)$ is a polynomial of degree 2, is a quadratic equation.
Standard form of a quadratic equation:
$ax^2+bx+c=0, \: where\: a \ne 0$.Example 1
Represent the following situations mathematically.(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Solution:
Let the number of marbles John and Jivanti had be x and 45-x respectively.
After losing 5 marbles each,
John had $x-5$ marbles and
Jivanti had $45-x-5=40-x$ marbles.
Given that: $(x-5) \times (40-x)=124$
$40x-x^2-200+5x=124$
$-x^2+45x-200-124=0$
$-x^2+45x-324=0$
$x^2-45x+324=0$
$(x-36)(x-9)=0$
$x-36=0, \implies x=36$
$x-9=0, \implies x=9$
Therefore, they had 36 and 9 marbles initially.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.
Solution:
Let the number of toys produced in a day be x.
Then, the cost of production = $55-x$
$(x)(55-x)=750$
$55x-x^2=750$
$x^2-55x+750=0$
$(x-25)(x-30)=0$
$x-25=0, \implies x=25$
$x-30=0, \implies x=30$
Therefore, the number of toys produced on that day is 25 or 30.
Example 2
Check whether the following are quadratic equations:(i) $(x-2)^2+1=2x-3$
Solution:
$(x-2)^2+1=2x-3$
$(a-b)^2=a^2-2ab+b^2$
$x^2-4x+4+1=2x-3$
$x^2-4x+4+1-2x+3=0$
$x^2-6x+8=0$
It is of the form $ax^2+bx+c=0$
Therefore, the given equation is a quadratic equation.
(ii) $x(x+1)+8=(x+2)(x-2)$
Solution:
$x(x+1)+8=(x+2)(x-2)$
$(a+b)(a-b)=a^2-b^2$
$x^2+x+8=x^2-4$
$x^2+x+8-x^2+4=0$
$x+12=0$
It is not of the form $ax^2+bx+c=0$
Therefore, the given equation is not a quadratic equation.
(iii) $x(2x+3)=x^2+1$
Solution:
$x(2x+3)=x^2+1$
$2x^2+3x=x^2+1$
$2x^2+3x-x^2-1=0$
$x^2+3x-1=0$
It is of the form $ax^2+bx+c=0$
Therefore, the given equation is a quadratic equation.
(iv) $(x+2)^3=x^3-4$
Solution:
$(x+2)^3=x^3-4$
$(a+b)^3=a^3+3a^2b+3ab^2+b^3$
$x^3+6x^2+12x+8=x^3-4$
$x^3+6x^2+12x+8-x^3+4=0$
$6x^2+12x+12=0$
Dividing by 6, we get, $x^2+2x+2=0$
It is of the form $ax^2+bx+c=0$
Therefore, the given equation is a quadratic equation.
EXERCISE 4.1
1. Check whether the following are quadratic equations :(i) $(x+1)^2=2(x–3)$
Solution:
$(x+1)^2=2(x–3)$
$(a+b)^2=a^2+2ab+b^2$
$x^2+2x+1=2x-6$
$x^2+2x+1-2x+6=0$
$x^2+7=0$
It is of the form $ax^2+bx+c=0$
Therefore, the given equation is a quadratic equation.
(ii) $x^2–2x=(–2)(3–x)$
Solution:
$x^2–2x=(–2)(3–x)$
$x^2–2x=-6+2x$
$x^2–2x+6-2x=0$
$x^2–4x+6=0$
It is of the form $ax^2+bx+c=0$
Therefore, the given equation is a quadratic equation.
(iii) $(x–2)(x+1)=(x–1)(x+3)$
Solution:
$(x–2)(x+1)=(x–1)(x+3) $
$x^2+x-2x-2=x^2+3x-x-3$
$x^2+x-2x-2-x^2-3x+x+3=0$
$-3x+1=0$
It is not of the form $ax^2+bx+c=0$
Therefore, the given equation is not a quadratic equation.
(iv) $(x–3)(2x+1)=x(x+5)$
Solution:
$(x–3)(2x+1)=x(x+5)$
$2x^2+x-6x-3=x^2+5x$
$2x^2+x-6x-3-x^2-5x=0$
$x^2-10x-3=0$
It is of the form $ax^2+bx+c=0$
Therefore, the given equation is a quadratic equation.
(v) $(2x–1)(x–3)=(x+5)(x–1)$
Solution:
$(2x–1)(x–3)=(x+5)(x–1)$
$2x^2-6x-x+3=x^2-x+5x-5$
$2x^2-6x-x+3-x^2+x-5x+5=0$
$x^2-11x+8=0$
It is of the form $ax^2+bx+c=0$
Therefore, the given equation is a quadratic equation.
(vi) $x^2+3x+1=(x–2)^2 $
Solution:
$x^2+3x+1=(x–2)^2 $
$(a-b)^2=a^2-2ab+b^2$
$x^2+3x+1=x^2-4x+4 $
$x^2+3x+1-x^2+4x-4=0 $
$7x-3=0 $
It is not of the form $ax^2+bx+c=0$
Therefore, the given equation is not a quadratic equation.
(vii) $(x+2)^3=2x(x^2–1)$
Solution:
$(x+2)^3=2x(x^2–1)$
$(a+b)^3=a^3+3a^2b+3ab^2+b^3$
$x^3+6x^2+12x+8=2x^3-2x$
$x^3+6x^2+12x+8-2x^3+2x=0$
$-x^3+6x^2+14x+8=0$
It is not of the form $ax^2+bx+c=0$
Therefore, the given equation is not a quadratic equation.
(viii) $x^3–4x^2–x+1=(x–2)^3 $
Solution:
$x^3–4x^2–x+1=(x–2)^3 $
$(a-b)^3=a^3-3a^2b+3ab^2-b^3$
$x^3–4x^2–x+1=x^3-6x^2+12x-8 $
$x^3–4x^2–x+1-x^3+6x^2-12x+8=0 $
$2x^2–13x+9=0 $
It is of the form $ax^2+bx+c=0$
Therefore, the given equation is a quadratic equation.
2. Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is $528 \:m^2$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
$l=2b+1$
$l \times b=528 \: m^2$
$(2b+1)(b)=528$
$2b^2+b=528$
$2b^2+b-528=0$
$(b-16)(2b+33)=0$
$b-16=0, \implies b=16$
$2b+33=0, \implies b=\frac{-33}{2}$. this is not possible.
Therefore, $b=16\:m$
$\implies l=2(16)+1$
$\implies l=32+1=33\:m$
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let the two consecutive integers be x and x+1.
$(x)(x+1)=306$
$x^2+x=306$
$x^2+x-306=0$
$(x+18)(x-17)=0$
$x+18=0,\implies x=-18$, not possible.
$x-17=0, \implies x=17$
The integers are 17 and 18.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Solution:
Let Rohan's age be x years.
Rohan's mothers age is $x+26$
3 years from now,
Rohan's age is $x+3$.
Rohan's mother's age is $x+29$
$(x+3)(x+29)=360$
$x^2+29x+3x+87=360$
$x^2+29x+3x+87-360=0$
$x^2+32x-273=0$
$(x+39)(x-7)=0$
$x+39=0,\implies x=-39$, not possible.
$x-7=0, \implies x=7$
Therefore, Rohan's present age is 7.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
Let the two consecutive integers be x and x+1.
$(x)(x+1)=306$
$x^2+x=306$
$x^2+x-306=0$
$(x+18)(x-17)=0$
$x+18=0,\implies x=-18$, not possible.
$x-17=0, \implies x=17$
The integers are 17 and 18.