Question

if the roots of the quadratic equation ax^2+bx+c = 0 are equal,then show that b^2=4ac
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Answer 1

SOLUTION :-

Quadratic equation : ax² + bx + c = 0

We know that,

$\boxed{\bf x= \dfrac{-b\pm\sqrt{b^2-4ac} }{2a}}$

Given that, two roots are equal.

$\longrightarrow \sf \dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{-b-\sqrt{b^2-4ac}}{2a} \\\\\longrightarrow -b+\sqrt{b^2-4ac}=-b-\sqrt{b^2-4ac}\\\\\longrightarrow \sqrt{b^2-4ac}+\sqrt{b^2-4ac}= -b+b\\\\\longrightarrow 2 \ \sqrt{b^2-4ac}=0\\\\\sf Squaring \ both \ the \ sides \ we \ get \ , \\\\\longrightarrow 4 \times ( b^2-4ac) = 0 \\\\\longrightarrow b^2-4ac = 0 \\\\\longrightarrow\bf b^2= 4ac$

Hence proved.

Answer 2

Answer:

To prove:

$\sf{If \ the \ roots \ of \ the \ quadratic \ equation} \\ \\ \sf{ax^{2}+bx+c=0 \ are \ equal, \ then \ show \ that} \\ \\ \sf{b^{2}=4ac}$

Proof:

Method (1):

$\sf{Let \ the \ roots \ be \ \alpha \ and \ \beta} \\ \\ \sf{By \ formula \ method.} \\ \\ \sf{\alpha=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a} \ and \ \beta=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}.} \\ \\ \sf{If \ roots \ are \ equal \ difference \ between \ them \ will \ be \ 0.} \\ \\ \sf{\alpha-\beta=0} \\ \\ \sf{\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}-(\dfrac{-b-\sqrt{b^{2}-4ac}}{2a})=0} \\ \\ \sf{\therefore{\dfrac{-b+b+\sqrt{b^{2}-4ac}+\sqrt{b^{2}-4ac}}{2a}=0}} \\ \\ \sf{\therefore{\dfrac{2\sqrt{b^{2}-4ac}}{2a}=0}} \\ \\ \sf{\therefore{\dfrac{\sqrt{b^{2}-4ac}}{a}=0}} \\ \\ \sf{\therefore{\sqrt{b^{2}-4ac}=0}} \\ \\ {Squaring \ both \ sides} \\ \\ \sf{b^{2}-4ac=0} \\ \\ \sf{\therefore{b^{2}=4ac}} \\ \\ \sf{Hence, \ proved} \\ \\ \purple{\tt{b^{2}=4ac}}$

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Method (2):

$\sf{Let \ the \ roots \ be \ \alpha \ and \ \beta} \\ \\ \sf{Roots \ are \ equal, \ i.e. \ \beta=\alpha} \\ \\ \boxed{\sf{\alpha+\beta=\dfrac{-b}{a}}} \\ \\ \sf{Substitute \ \beta=\alpha} \\ \\ \sf{\therefore{2\alpha=\dfrac{-b}{a}}} \\ \\ \sf{Squaring \ both \ sides} \\ \\ \sf{\therefore{4\alpha^{2}=\dfrac{b^{2}}{a^{2}}...(1)}} \\ \\ \boxed{\sf{\alpha\beta=\dfrac{c}{a}}} \\ \\ \sf{Substitute \ \beta=\alpha} \\ \\ \sf{\alpha^{2}=\dfrac{c}{a}} \\ \\ \sf{Multiply \ both \ sides \ by \ 4} \\ \\ \sf{\therefore{4\alpha^{2}=\dfrac{4c}{a}...(2)}}$

$\sf{From \ (1) \ and \ (2), \ we \ get} \\ \\ \sf{\dfrac{b^{2}}{a^{2}}=\dfrac{4c}{a}} \\ \\ \sf{Multiply \ both \ sides \ by \ a^{2}} \\ \\ \sf{\therefore{b^{2}=4ac}} \\ \\ \sf{Hence, \ proved.} \\ \\ \purple{\tt{b^{2}=4ac}}$