Found 225 Articles for Class 8

Given:$2x + 3y = 14$ and $2x - 3y = 2$To do:We have to find the value of $xy$.Solution:The given expressions are $2x + 3y = 14$ and $2x - 3y = 2$. Here, we have to find the value of $xy$. So, by squaring and subtracting using the identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$, we can find the required value.$(a+b)^2=a^2+2ab+b^2$.............(I)$(a-b)^2=a^2-2ab+b^2$.............(II)Let us consider, $2x + 3y = 14$Squaring on both sides, we get, $(2x + 3y)^2 = (14)^2$$(2x)^2+2(2x)(3y)+(3y)^2=196$            [Using (I)]$4x^2+12xy+9y^2=196$..........(III)Now, $2x - 3y = 2$Squaring on both sides, we get, $(2x - 3y)^2 = (2)^2$$(2x)^2-2(2x)(3y)+(3y)^2=4$            [Using (II)]$4x^2-12xy+9y^2=4$..........(IV)Subtracting ... Read More

Given:$x + \frac{1}{x} = 12$To do:We have to find the value of $x - \frac{1}{x}$.Solution:The given expression is $x + \frac{1}{x} = $. Here, we have to find the value of $x^4 + \frac{1}{x^4}$. So, by squaring the given expression and using the identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$, we can find the required value.$(a+b)^2=a^2+2ab+b^2$.............(I)$(a-b)^2=a^2-2ab+b^2$.............(II)Let us consider, $x + \frac{1}{x} = 12$Squaring on both sides, we get, $(x + \frac{1}{x})^2 = (12)^2$$x^2+2\times x \times \frac{1}{x}+\frac{1}{x^2}=144$            [Using (I)]$x^2+2+\frac{1}{x^2}=144$$x^2+\frac{1}{x^2}=144-2$                       (Transposing $2$ to RHS)$x^2+\frac{1}{x^2}=142$Now, $x^2+\frac{1}{x^2}=142$Subtracting 2 from both sides, we get, $x^2+\frac{1}{x^2}-2=142-2$$x^2-2\times ... Read More

Given:$x + \frac{1}{x} = 9$To do:We have to find the value of $x^4 + \frac{1}{x^4}$.Solution:The given expression is $x + \frac{1}{x} = 9$. Here, we have to find the value of $x^4 + \frac{1}{x^4}$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the required value.$(a+b)^2=a^2+2ab+b^2$...................(i)Let us consider, $x + \frac{1}{x} = 9$Squaring on both sides, we get, $(x + \frac{1}{x})^2 = 9^2$$x^2+2\times x \times \frac{1}{x}+\frac{1}{x^2}=81$            [Using (I)]$x^2+2+\frac{1}{x^2}=81$$x^2+\frac{1}{x^2}=81-2$                       (Transposing $2$ to RHS)$x^2+\frac{1}{x^2}=79$Now, $x^2+\frac{1}{x^2}=79$Squaring on both sides, we get, $(x^2+\frac{1}{x^2})^2 = ... Read More

To do:We have to find the values of the given expressions.Solution:Here, we have to find the values of the given expressions. So, simplifying the given expressions using the identities $(a+b)^2=a^2+2ab+b^2$.............(I) and $(a-b)^2=a^2-2ab+b^2$.............(II) and substituting the values of $x$ and $y$, we can find the required values.(i) The given expression is $16x^2 + 24x + 9$.$16x^2 + 24x + 9=(4x)^2+2\times 4x \times3+(3)^2$               [$24x=2\times 4x \times3$]$16x^2 + 24x + 9=(4x+3)^2$                (Using (I), $a=4x$ and $b=3$)Substituting $x = \frac{7}{4}$ in $(4x+3)^2$, we get, $(4x+3)^2=[4\times\frac{7}{4}+3]^2$ $(4x+3)^2=(7+3)^2$ $(4x+3)^2=(10)^2$ $(4x+3)^2=100$The value ... Read More

Given:$3x + 5y = 11$ and $xy = 2$To do:We have to find the value of $9x^2+25y^2$.Solution:The given expressions are $3x + 5y = 11$ and $xy = 2$. Here, we have to find the value of $9x^2+25y^2$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the required value.$xy = 2$............(i)$(a+b)^2=a^2+2ab+b^2$.............(ii)Now, $3x + 5y = 11$Squaring on both sides, we get, $(3x + 5y)^2 = (11)^2$                 [Using (ii)]$(3x)^2+2(3x)(5y)+(5y)^2=121$$9x^2+30xy+25y^2=121$$9x^2+30(2)+25y^2=121$                     [Using (i)]$9x^2+60+25y^2=121$$9x^2+25y^2=121-60$              (Transposing $60$ ... Read More

Given:$x – y = 7$ and $xy = 9$To do:We have to find the value of $x^2+y^2$.Solution:The given expressions are $x – y = 7$ and $xy = 9$. Here, we have to find the value of $x^2 + y^2$. So, by squaring the given expression and using the identity $(a-b)^2=a^2-2ab+b^2$, we can find the required value.$xy = 9$............(i)$(a-b)^2=a^2-2ab+b^2$.............(ii)Now, $x – y = 7$Squaring on both sides, we get, $(x – y)^2 = 7^2$                 [Using (ii)]$x^2-2xy+y^2=49$$x^2-2(9)+y^2=49$                     [Using (i)]$x^2-18+y^2=49$$x^2+y^2=49+18$              ... Read More

Given:$x + y = 4$ and $xy = 2$To do:We have to find the value of $x^2 + y^2$.Solution:The given expressions are $x + y = 4$ and $xy = 2$. Here, we have to find the value of $x^2 + y^2$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the required value.$xy = 2$...........(i)$(a+b)^2=a^2+2ab+b^2$...........(ii)Now, $x + y = 4$Squaring on both sides, we get, $(x + y)^2 = 4^2$$x^2+2 \times x \times y+y^2=16$               [Using (ii)]$x^2+2xy+y^2=16$$x^2+2(2)+y^2=16$                          [Using ... Read More

Given:$x^2 + \frac{1}{x^2} = 18$To do:We have to find the values of $x + \frac{1}{x}$ and $x - \frac{1}{x}$.Solution:The given expression is $x^2 + \frac{1}{x^2} = 18$. Here, we have to find the values of $x + \frac{1}{x}$ and $x - \frac{1}{x}$. So, by using the identities $(a+b)^2=a^2+2ab+b^2$...................(i) and $(a-b)^2=a^2-2ab+b^2$.............(ii), we can find the required values.Now, $x^2 + \frac{1}{x^2} = 18$Adding $2$ on both sides, we get, $x^2 + \frac{1}{x^2} + 2 = 18+2$$x^2 + \frac{1}{x^2} + 2 \times x \times \frac{1}{x} = 20$               (Since $2\times x \times \frac{1}{x}=2$)$(x+\frac{1}{x})^2=20$            ... Read More

Given:$x - \frac{1}{x} = 3$To do:We have to find the values of $x^2 + \frac{1}{x^2}$ and $x^4 + \frac{1}{x^4}$.Solution:The given expression is $x - \frac{1}{x} = 3$. Here, we have to find the values of $x^2 + \frac{1}{x^2}$ and $x^4 + \frac{1}{x^4}$. So, by squaring the given expression and using the identities $(a+b)^2=a^2+2ab+b^2$...................(i) and $(a-b)^2=a^2-2ab+b^2$.............(ii), we can find the required values. Let us consider, $x - \frac{1}{x} = 3$Squaring on both sides, we get, $(x - \frac{1}{x})^2 = 3^2$                 [Using (ii)]$x^2-2\times x \times \frac{1}{x}+\frac{1}{x^2}=9$$x^2-2+\frac{1}{x^2}=9$$x^2+\frac{1}{x^2}=9+2$                     (Transposing $-2$ ... Read More

Given:$x + \frac{1}{x} =20$To do:We have to find the value of $x^2 + \frac{1}{x^2}$.Solution:The given expression is $x + \frac{1}{x} =20$. Here, we have to find the value of $x^2 + \frac{1}{x^2}$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the value of $x^2 + \frac{1}{x^2}$.$(a+b)^2=a^2+2ab+b^2$...................(i)Now, $x + \frac{1}{x} =20$Squaring on both sides, we get, $(x+\frac{1}{x})^2=(20)^2$$x^2+2\times x \times \frac{1}{x}+(\frac{1}{x})^2=400$           [Using (i)]$x^2+2+\frac{1}{x^2}=400$$x^2+\frac{1}{x^2}=400-2$                 (Transposing $2$ to RHS)$x^2+\frac{1}{x^2}=398$Hence, the value of $x^2+\frac{1}{x^2}$ is $398$.Read More