Question

If one zero of the quadratic polynomial : x^2+(3k-5)x+2=0 is negative of the other. Find the value of k :​

Answer 1

Given :

Polynomial -

• $\tt\:x^{2}+(3k-5)x+2=0$

➷ It is also given that one zero of the given polynomial is negative of the other.

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To find :

• Value of k.

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Concept :

For this question, we will first assume the two zeroes of the given polynomial. Then after using the formula which follows :

$\maltese$ $\sf\:Sum~of~the~zeroes~=~\frac{(-b)}{c}$

we would calculate the required value of k.

Hope I sound clear let's do it :D~

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Assumption :

Let of the polynomial be $\bf\:\alpha$

Then according to the question :

• The other root = $\bf\:(-\alpha)$

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Solution :

Polynomial -

$\tt\:x^{2}+(3k-5)x+2=0$

Here :

• a = $\tt\:1$

• b = $\tt\:3k-5$

• c = $\tt\:2$

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Lets now apply the formula :

❒ $\sf\:Sum~of~the~zeroes~=~\frac{(-b)}{c}$

Substituting the values

$\tt:\implies\:\alpha + (-\alpha)=\frac{(-b)}{c}$

$\tt:\implies\:\alpha + (-\alpha)=\frac{-(3k-5)}{2}$

Multiplying and removing the brackets

$\tt:\implies\:\alpha -\alpha=\frac{-3k+5}{2}$

Terms with opposite signs gets cancelled out

$\tt:\implies\:\cancel{\alpha} -\cancel{\alpha}=\frac{-3k+5}{2}$

$\tt:\implies\:0=\frac{-3k+5}{2}$

Cross multiplying

$\tt:\implies\:-3k+5=2 \times 0$

$\tt:\implies\:-3k+5=0$

Transposing +5 from LHS to RHS it becomes -5

$\tt:\implies\:-3k=-5$

Negative signs gets cancelled out from both sides

$\tt:\implies\:\cancel{-}3k=\cancel{-}5$

$\tt:\implies\:3k=5$

Transposing 3 to RHS it goes to the denominator

$\small{\underline{\boxed{\mathrm{:\implies~k~=~\frac{5}{3}}}}}$ ★

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