# Solve the following quadratic equation by factorization, whole no root is:

\[7x+\dfrac{3}{x}=35\dfrac{3}{5}\]

Answer

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Hint: Use the basic rule of factorization of quadratic equation i.e. split the middle terms when multiplied which gives the same as multiplication of first and last term.

We have the quadratic equation as \[7x+\dfrac{3}{x}=35\dfrac{3}{5}\]

\[\dfrac{7{{x}^{2}}+3}{x}=\dfrac{35\times 5+3}{5}\]

\[\left( 7{{x}^{2}}+3 \right)5=x\left( 175+3 \right)\]

\[35{{x}^{2}}+15=178x\]

\[35{{x}^{2}}-178x+15=0\]

Now, as we know any quadratic equation is factored by splitting the coefficient of \[x\] in addition of two numbers such that if we multiply them, that should be equal to the multiplication of coefficient of \[{{x}^{2}}\] and constant term.

Hence, let us assume we have quadratic equation as \[A{{x}^{2}}+Bx+C=0\]

Now, let \[B\] be expressed as \[\left( \alpha +\beta \right)\] summation then, \[\alpha \beta \] should be equal to \[AC\];

So, that we have

\[{{x}^{2}}+\left( \alpha +\beta \right)x+\alpha \beta =0\]

\[{{x}^{2}}+\alpha x+\beta x+\alpha \beta =0\]

\[x\left( x+\alpha \right)+\beta \left( x+\alpha \right)=0\]

\[\left( x+\alpha \right)\left( x+\beta \right)=0\]

\[x=-\alpha ,-\beta \]

In the same way, let us try to factorize the given quadratic equation –

\[35{{x}^{2}}-178x+15=0\]

Comparing with equation \[A{{x}^{2}}+Bx+C\]

\[AC=35\times 15=525\]

Now \[B\] is equal to \[\left( -178 \right)\].

We need to express it into two terms \[-\alpha \] and \[-\beta \]so that \[\alpha \beta =525\]

Hence, \[-\left( \alpha +\beta \right)=-178\]

\[\alpha +\beta =178\]

\[178=177+1\Rightarrow 177\times 1=177\]

\[178=176+2\Rightarrow 176\times 2=352\]

\[178=175+3\Rightarrow 175\times 3=525\]

Hence we can break \[-178\]as \[-175\] and \[-3\] so that addition of them is \[-178\] and multiplication will be \[525\].

Therefore, we can write the given quadratic equation as

\[35{{x}^{2}}-178x+15=0\]

\[35{{x}^{2}}-175x-3x+15=0\]

\[35x\left( x-5 \right)-3\left( x-5 \right)=0\]

\[\left( 35x-3 \right)\left( x-5 \right)=0\]

\[x=5,\dfrac{3}{35}\]

Hence, the roots of the equation \[\left( 5,\dfrac{3}{35} \right)\].

Note: As we know the form of quadratic equation is \[{{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0\] or \[{{x}^{2}}-\left( \text{Sum of roots} \right)x+\text{ Multiplication roots }=0\]. Factorization method uses the same concept. For the verification of splitting we can check the brackets formed step 3 of factorization as \[35x\left( x-5 \right)-3\left( x-5 \right);\] If the two brackets formed after taking common parts from first two terms and last two terms; these should be equal. If the two brackets formed are not equal, then splitting has gone wrong and we need to check the splitting step once again. Students make mistakes during taking common between the first two and last two terms. We need to take the common HCF of the first two and last two terms respectively, then we will get two brackets formed which is equal.

We have the quadratic equation as \[7x+\dfrac{3}{x}=35\dfrac{3}{5}\]

\[\dfrac{7{{x}^{2}}+3}{x}=\dfrac{35\times 5+3}{5}\]

\[\left( 7{{x}^{2}}+3 \right)5=x\left( 175+3 \right)\]

\[35{{x}^{2}}+15=178x\]

\[35{{x}^{2}}-178x+15=0\]

Now, as we know any quadratic equation is factored by splitting the coefficient of \[x\] in addition of two numbers such that if we multiply them, that should be equal to the multiplication of coefficient of \[{{x}^{2}}\] and constant term.

Hence, let us assume we have quadratic equation as \[A{{x}^{2}}+Bx+C=0\]

Now, let \[B\] be expressed as \[\left( \alpha +\beta \right)\] summation then, \[\alpha \beta \] should be equal to \[AC\];

So, that we have

\[{{x}^{2}}+\left( \alpha +\beta \right)x+\alpha \beta =0\]

\[{{x}^{2}}+\alpha x+\beta x+\alpha \beta =0\]

\[x\left( x+\alpha \right)+\beta \left( x+\alpha \right)=0\]

\[\left( x+\alpha \right)\left( x+\beta \right)=0\]

\[x=-\alpha ,-\beta \]

In the same way, let us try to factorize the given quadratic equation –

\[35{{x}^{2}}-178x+15=0\]

Comparing with equation \[A{{x}^{2}}+Bx+C\]

\[AC=35\times 15=525\]

Now \[B\] is equal to \[\left( -178 \right)\].

We need to express it into two terms \[-\alpha \] and \[-\beta \]so that \[\alpha \beta =525\]

Hence, \[-\left( \alpha +\beta \right)=-178\]

\[\alpha +\beta =178\]

\[178=177+1\Rightarrow 177\times 1=177\]

\[178=176+2\Rightarrow 176\times 2=352\]

\[178=175+3\Rightarrow 175\times 3=525\]

Hence we can break \[-178\]as \[-175\] and \[-3\] so that addition of them is \[-178\] and multiplication will be \[525\].

Therefore, we can write the given quadratic equation as

\[35{{x}^{2}}-178x+15=0\]

\[35{{x}^{2}}-175x-3x+15=0\]

\[35x\left( x-5 \right)-3\left( x-5 \right)=0\]

\[\left( 35x-3 \right)\left( x-5 \right)=0\]

\[x=5,\dfrac{3}{35}\]

Hence, the roots of the equation \[\left( 5,\dfrac{3}{35} \right)\].

Note: As we know the form of quadratic equation is \[{{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0\] or \[{{x}^{2}}-\left( \text{Sum of roots} \right)x+\text{ Multiplication roots }=0\]. Factorization method uses the same concept. For the verification of splitting we can check the brackets formed step 3 of factorization as \[35x\left( x-5 \right)-3\left( x-5 \right);\] If the two brackets formed after taking common parts from first two terms and last two terms; these should be equal. If the two brackets formed are not equal, then splitting has gone wrong and we need to check the splitting step once again. Students make mistakes during taking common between the first two and last two terms. We need to take the common HCF of the first two and last two terms respectively, then we will get two brackets formed which is equal.

Last updated date: 27th Sep 2023

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