Question 75

The value of $$x$$ for which the quadratic expression $$x^2-10x+21$$ attains its minimum value is the root of the quadratic equation $$x^2+kx-45=0$$. Find the value of $$k$$.

Solution

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We know a quadratic $$ax^2+bx+c$$ attains its minimum value at $$x=-\dfrac{b}{2a}$$

For the given quadratic $$x^2-10x+21$$

$$-\dfrac{b}{2a}=-\left(\dfrac{-10}{2}\right)=5$$

So, $$x=5$$ is the root of the quadratic equation $$x^2+kx-45=0$$

Putting $$x=5$$ we get

$$5^2+5k-45=0$$

$$5k=20$$

$$k=4$$

Hence, the answer is 4.

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