Question

15. Find a quadratic equation whose roots are 3 less than
the roots of the equation x²-14x+29=0​

Answer 1

${\large {\underline {\blue {\bf {Question}}}}}$

• Find a quadratic equation whose roots are 3 less than the roots of the equation x²-14x+29=0.

${\large {\underline {\blue {\bf {Answer}}}}}$

$\sf \: Let \: the \: roots \: of \: {x}^{2} - 14x + 29 \: be \: \alpha \: and \: \beta.$

• We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

OR

$\boxed{\purple{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$

Therefore,

$\bf \longmapsto \: \bf \: \alpha + \beta = - \: \dfrac{ - 14}{1}$

$\boxed{ \pink{\bf :\implies\: \alpha + \beta = 14}} - - (1)$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

OR

$\boxed{\purple{\tt Product\ of\ the\ zeroes=\frac{c}{a}}}$

Therefore,

$\bf \longmapsto \: \bf \: \alpha \beta = \dfrac{29}{1}$

$\boxed{ \pink{\bf :\implies\: \alpha \beta = 49 }}$

Now,

We have to find a quadratic equation whose roots are

$\bf \: \alpha \: - 3 \: and \: \beta \: - \: 3$

So,

$\bf \longmapsto \: \bf \:Sum \: of \: zeroes = \alpha - 3 + \beta - 3$

$\bf \longmapsto \: \bf \:Sum \: of \: zeroes = \alpha + \beta - 6$

$\bf \longmapsto \: \bf \:Sum \: of \: zeroes = 14 - 6$

$\boxed{ \red{\bf \longmapsto \: \bf \:Sum \: of \: zeroes \: = 8}}$

Now,

$\bf \longmapsto \: \bf \:Product \: of \: zeroes = ( \alpha - 3)( \beta - 3)$

$\bf \longmapsto \: \bf \:Product \: of \: zeroes = \alpha \beta - 3 \alpha - 3 \beta + 9$

$\bf \longmapsto \: \bf \:Product \: of \: zeroes = 29 - 3( \alpha + \beta ) + 9$

$\bf \longmapsto \: \bf \:Product \: of \: zeroes = 38 - 3 \times 14$

$\bf \longmapsto \: \bf \:Product \: of \: zeroes = 38 - 42$

$\boxed{\purple{ \bf \longmapsto \: \bf \: Product\ of\ the\ zeroes= - 4}}$

Hence,

The required Quadratic polynomial is given by

$\bf \longmapsto \: \bf \: {x}^{2} - (Sum \: of \: zeroes )x + Product \: of \: zeroes = 0$

$\bf \longmapsto \: \bf \: {x}^{2} - 8x - 4= 0$