1. Check whether the following are quadratic equations

For each equation I rewrite it in the form “polynomial = 0” and check the degree.

(i) $(x+1{)}^2=2(x-3)$
LHS − RHS = $(x+1{)}^2-2(x-3)=x^2+7$.
So $x^2+7=0$. Degree 2 → Yes, quadratic.

(ii) $x^2-2x=(-2)(3-x)$
LHS − RHS = $x^2-2x-(-2)(3-x)=x^2-4x+6$.
So $x^2-4x+6=0$. Degree 2 → Yes, quadratic.

(iii) $(x-2)(x+1)=(x-1)(x+3)$
LHS − RHS = $(x-2)(x+1)-(x-1)(x+3)=1-3x$.
So $1-3x=0$ or $-3x+1=0$. Degree 1 → Not a quadratic (it’s linear).

(iv) $(x-3)(2x+1)=x(x+5)$
LHS − RHS = $(x-3)(2x+1)-x(x+5)=x^2-10x-3$.
So $x^2-10x-3=0$. Degree 2 → Yes, quadratic.

(v) $(2x-1)(x-3)=(x+5)(x-1)$
LHS − RHS = $(2x-1)(x-3)-(x+5)(x-1)=x^2-11x+8$.
So $x^2-11x+8=0$. Degree 2 → Yes, quadratic.

(vi) $x^2+3x+1=(x-2{)}^2$
LHS − RHS = $x^2+3x+1-(x-2{)}^2=7x-3$.
So $7x-3=0$. Degree 1 → Not a quadratic (linear).

(vii) $(x+2{)}^3=2x(x^2-1)$
LHS − RHS = $(x+2{)}^3-2x(x^2-1)=-x^3+6x^2+14x+8$.
Degree 3 → Not a quadratic (cubic).

(viii) $x^3-4x^2-x+1=(x-2{)}^3$
LHS − RHS = $x^3-4x^2-x+1-(x-2{)}^3=2x^2-13x+9$.
So $2x^2-13x+9=0$. Degree 2 → Yes, quadratic.

Summary (yes = quadratic): (i) yes, (ii) yes, (iii) no, (iv) yes, (v) yes, (vi) no, (vii) no, (viii) yes.

jemh104

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2. Represent (and solve) the situations as quadratic equations

I’ll write the equation and solve each (you asked “solve all the questions” — so I solved them).

(i) The area of a rectangular plot is 528 m2. 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:- The area of a rectangular plot is $528{\text{m}}^2$. Length = one more than twice breadth.
Let breadth = $b$ (m). Then length = $2b+1$
Equation: $(2b+1)\cdot b=528$
So $2b^2+b-528=0$. Divide or solve:

Solve: $2b^2+b-528=0\Rightarrow b=16$
$b=-33\mathrm{/}2$. Breadth must be positive → breadth = 16 m.
Length = $2(16)+1=33$.

Answer: Equation $2b^2+b-528=0$. Dimensions: 16 m by 33 m.

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(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Solution :- Product of two consecutive positive integers is 306.
Let the smaller integer be $n$. Then $n(n+1)=306$.
Equation: $n^2+n-306=0$

Solve: $n^2+n-306=0\Rightarrow n=17$ or $n=-18$. Positive solution → integers are 17 and 18.

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(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 :- Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. Find Rohan’s present age.
Let Rohan’s present age = $x$. Mother = $x+26$. In 3 years: ages $x+3$ and $x+29$.
Equation: $(x+3)(x+29)=360$. Expand: $x^2+32x+87=360\Rightarrow x^2+32x-273=0$

Solve: $x^2+32x-273=0\Rightarrow x=7$ or $x=-39$. Reject negative → Rohan is 7 years old, mother is $7+26=33$ years.

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(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 :- A train travels 480 km at uniform speed. If the speed had been 8 km/h less it would have taken 3 hours more. Find the speed.
Let speed = $s$ km/h. Regular time = $480\mathrm{/}s$. Slower speed time = $480\mathrm{/}(s-8)$. Given: $480\mathrm{/}(s-8)=480\mathrm{/}s+3$

Multiply through and simplify → quadratic: $3s^2-120s-3840=0$ (after clearing denominators and simplifying) which reduces to $s^2-40s-1280=0$. Solving gives $s=40$ or $s=-32$. Reject negative → speed = 40 km/h.