Find and correct the errors in the following.
(a) $ (2 x+5)^{2}=4 x^{2}+25 $
(b) $ \left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=x^{2}-\frac{1}{4} $
(c) $ (5 a-b)^{2}=10 a^{2}-5 a b+b^{2} $
(d) $ (p-3)(p-7)=p^{2}+21 $


To do:

We have to correct the errors in the below expressions.

(a) \( (2 x+5)^{2}=4 x^{2}+25 \)
(b) \( \left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=x^{2}-\frac{1}{4} \)
(c) \( (5 a-b)^{2}=10 a^{2}-5 a b+b^{2} \)
(d) \( (p-3)(p-7)=p^{2}+21 \)

Solution:

We know that,

$(a+b)^2=a^2+2ab+b^2$

$(a-b)^2=a^2-2ab+b^2$

Therefore,

(a) LHS $=(2x+5)^2$

$=(2x)^2+2.(2x).(5)+(5)^2$

$=4x^2+20x+25$

RHS $=4x^2+25$

Therefore, we have to add $20x$ on RHS

(b) LHS $=(x-\frac{1}{2})(x-\frac{1}{2})$

$=(x-\frac{1}{x})^2$

$=x^2-2.x.\frac{1}{x}+(\frac{1}{x})^2$

$=x^2-2+\frac{1}{x^2}$

RHS $=x^{2}-\frac{1}{4}$

Therefore, we have to replace $-\frac{1}{4}$ by $-2+\frac{1}{x^2}$ on RHS

(c) LHS $=(5 a-b)^{2}$

$=(5a)^2-2.5a.b+(b)^2$

$=25a^2-10ab+b^2$

RHS $=10 a^{2}-5 a b+b^{2}$

Therefore, we have to replace $10 a^{2}-5 a b$ by $25a^2-10ab$ on RHS.

(d) LHS $=(p-3)(p-7)$

$=p(p)-p(7)-3(p)-3(-7)$

$=p^2-7p-3p+21$

$=p^2-10p+21$

RHS $=p^{2}+21$

Therefore, we have to add $-10p$ on RHS.

Updated on: 10-Oct-2022

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