The roots of a quadratic equation (x-7)^{2}-1 | vedclass

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(B) Given equation: $(x-7)^{2}-16=0$Method $1$: Using the identity $a^2 - b^2 = (a-b)(a+b)$$(x-7)^2 - 4^2 = 0$$(x-7-4)(x-7+4) = 0$$(x-11)(x-3) = 0$The

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$3$ and $11$

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(B) Given equation: $(x-7)^{2}-16=0$Method $1$: Using the identity $a^2 - b^2 = (a-b)(a+b)$$(x-7)^2 - 4^2 = 0$$(x-7-4)(x-7+4) = 0$$(x-11)(x-3) = 0$Therefore,$x-11=0$ or $x-3=0$,which gives $x=11$ or $x=3$.Method $2$: Expanding the square$x^2 - 14x + 49 - 16 = 0$$x^2 - 14x + 33 = 0$$x^2 - 11x - 3x + 33 = 0$$x(x-11) - 3(x-11) = 0$$(x-11)(x-3) = 0$Thus,the roots are $3$ and $11$.

The roots of a quadratic equation $(x-7)^{2}-16=0$ are ...... .

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