Question

Find the zeroesof the following quadratic polynomial and verify the realationship betwween the zeroes and the cofficients [1]4x'2-20x+25

Answer 1

Answer:

The roots of the given quadratic equation are

$\boxed{\red{\sf\:x\:=\:\frac{5}{2}}}\:\:\sf\:\&\:\:\:\boxed{\red{\sf\:x\:=\:\frac{5}{2}}}$

Step-by-step-explanation:

The given quadratic equation is

$\sf\:4x^{2}\:-\:20x\:+\:25\:=\:0$

$\therefore\sf\:4x^{2}\:-\:20x\:+\:25\:=\:0\\\\\implies\sf\:4x^{2}\:-\:10x\:-\:10x\:+\:25\:=\:0\\\\\implies\sf\:2x\:(\:2x\:-\:5\:)\:-\:5\:(\:2x\:-\:5\:)\:=\:0\\\\\implies\sf\:(\:2x\:-\:5\:)\:(\:2x\:-\:5\:)\:=\:0\\\\\implies\sf\:(\:2x\:-\:5\:)\:=\:0\:\:\:-\:-\:[\:Considering\:only\:one\:bracket\:]\\\\\implies\sf\:2x\:-\:5\:=\:0\\\\\implies\sf\:2x\:=\:5\\\\\implies\boxed{\red{\sf\:x\:=\:\frac{5}{2}}}\:\:\sf\:\&\:\:\:\boxed{\red{\sf\:x\:=\:\frac{5}{2}}}$

$\sf\:Now,\\\\\sf\:\alpha\:=\:\frac{5}{2}\\\\\sf\:\beta\:=\:\frac{5}{2}\\\\\sf\:Now,\\\\\sf\:Comparing\:4x^{2}\:-\:20x\:+\:25\:with\:ax^{2}\:+\:bx\:+\:c\:=\:0\:,\:we\:get,\\\\\sf\:\bullet\:a\:=\:4\\\\\sf\bullet\:b\:=\:-\:20\\\\\sf\bullet\:c\:=\:25\\\\\sf\:We\:know\:that\\\\\pink{\sf\:\alpha\:+\:\beta\:=\:-\:\frac{b}{a}}\\\\\implies\sf\:\frac{5}{2}\:+\:\frac{5}{2}\\\\\implies\sf\:\dfrac{5\:+\:5}{2}\\\\\implies\sf\:\frac{10}{2}\\\\\implies\sf\:5\:=\:-\:\cancel{\frac{\:-\:20}{4}}\\\\\implies\boxed{\red{\sf\:5\:=\:5}}$

Hence verified!

Now, we know that,

$\pink{\sf\:\alpha\:.\:\beta\:=\:\frac{c}{a}}\\\\\implies\sf\:\frac{5}{2}\:\times\:\frac{5}{2}\\\\\implies\boxed{\red{\sf\:\frac{25}{4}\:=\:\frac{25}{4}}}$

Hence verified!