Question

The Value of 'a' for which one root of the equation x²-(a + 1)x + a² + a - 8 = 0 exceeds 2 and others is less than 2 , are given by.

Note :- Graph is in attachment
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Answer 1

The graph of a quadratic function is determined by the leading coefficient and the discriminant.

Let us assume,

$\hookrightarrow f(x)=x^{2}-(a+1)x+a^{2}+a-8$

We know that the quadratic function passes the x-axis twice.

$\hookrightarrow \underline{D>0}$

$\hookrightarrow(a+1)^{2}-4(a^{2}+a-8)>0$

$\hookrightarrow(a^{2}+2a+1)-(4a^{2}+4a-32)>0$

$\hookrightarrow-3a^{2}-2a+33>0$

$\hookrightarrow3a^{2}+2a-33<0$

$\hookrightarrow(3a+11)(a-3)<0$

$\large\hookrightarrow\red{\boxed{\red{\bold{-\dfrac{11}{3}<a<3}}}}$

We are given that one root is greater than 2 and the other is less than 2. Then $(2,f(2))$ must go under the x-axis.

$\large\hookrightarrow\red{\boxed{\red{\bold{f(2)<0}}}}$

By this inequality,

$\hookrightarrow \underline{f(2)<0}$

$\hookrightarrow 4-2(a+1)+a^{2}+a-8<0$

$\hookrightarrow 4-2a-2+a^{2}+a-8<0$

$\hookrightarrow a^{2}-a-6<0$

$\hookrightarrow(a+2)(a-3)<0$

$\large\hookrightarrow\red{\boxed{\red{\bold{-2<a<3}}}}$

The union of two inequality is,

$\large\hookrightarrow\red{\boxed{\red{\bold{\red{\underline{-2<a<3}}}}}}$

This is our answer.